## Saturday, March 3, 2018

### Free Particle Propagator from the Path Integral Formulation

Free Particle Kernel from the Path Integral Formulation

In my post on the deriving the non-relativistic path integral, I motivated our discussion by the question of finding the transition probability amplitude of a particle existing in a position eigenstate, $|x_a\rangle$, to a new position eigenstate, $|x_b\rangle$. The power of the path integral, however, comes not in finding the transition probability amplitude, but in finding the Green's function (or "kernel") that describes the propagation of some arbitrary quantum mechanical system under some arbitrary Hamiltonian to a different state (by state, I am actually referring to the notion of a "state" in quantum mechanics). I must reiterate the last sentence because it is important - once we find the propagator, all we need to do is specify the initial state (the initial conditions). The boundary conditions are already taken care of - they are tucked away in the Hermitian nature of the operators that we deal with in quantum mechanics, which manifest themselves in the differential equations of quantum mechanics and finally manifest themselves in the integral form of those differential equations that involve convolutions of the propagators that we are solving for in the path integral formulation with our initial states. Let us then steal the result from my blog post on deriving the non-relativistic path integral and utilize it to do the most simple calculation of a propagator in the path integral formulation that we can do - the propagator for a free particle. Before beginning this post, I must mention that a large portion of the following derivations follow from Shankar's "Principles of Quantum Mechanics" with results from Feynman and Hibbs' "Quantum Mechanics and Path Integrals" and Schwartz's "Quantum Field Theory and the Standard Model". Let us now give the Lagrangian for a free particle: $$L_{\textrm{free}}=\frac{1}{2}m\dot{x}^2.$$ Then from my blog post on the path integral, the propagation amplitude for our particle to go from one point in space to another is $$\langle x_b | x_a \rangle = N^n \int \textrm{D}x \textrm{ }e^{iS/\hbar},$$ where $N$ is some normalization that we have already figured out in our derivation of the path integral. Let us first derive the propagator. In a later blog post, I will use this to show that the standard deviation in position of the ground state wave function for the harmonic oscillator increases if we let the state evolve in free space. The manner in which we will derive the propagator will involve us first going to a discrete domain, then moving to the continuum at the end of our derivation. Hence, let us first discretize the free particle action. This is a straight-forward action if we think of the integral in terms of the Reimann definition: $$S=\int \textrm{d}t \textrm{ }L_{\textrm{free}}=\lim_{\epsilon\rightarrow 0}\sum_{j=0}^{n-1}\frac{m}{2}\bigg(\frac{x_{j+1}-x_j}{\epsilon}\bigg)^2\epsilon,$$ where the time derivative has been discretized using a simple forward difference scheme as $\textrm{d}t\rightarrow \epsilon$. We can now plug this into our expression for the the kernel. We have an issue, however - what is the normalization constant, N? If you take a gander at my blog post on the derivation of the non-relativistic path integral, I give it explicitly as a consequence of utilizing a Gaussian integral. Its full form is $$N=\sqrt{\frac{m}{i2\pi\hbar \epsilon}}.$$ Now we insert our discretized action into the kernel: $$\langle x_b|x_a\rangle =\lim_{\epsilon\rightarrow 0}\bigg(\frac{m}{i2\pi\hbar \epsilon}\bigg)^{n/2} \int \textrm{D}x\exp{\bigg[i\sum_{j=0}^{n-1}\frac{m}{2}\frac{(x_{j+1}-x_j)^2}{\hbar\epsilon}}\bigg].$$ We can now pass the measure of our functional integral into a functional analogue of a discrete domain. The following drawing illustrates what is going at each evaluation point.

What were doing with our functional integration is integrating over every possible position that the particle can occupy at $t=j\epsilon$. Hence, at each time integration, we are integrating over an infinite number of positions, each of which are weighted by $\exp(iS/\hbar)$, which is minimum for the path (and hence the set of possible positions) that the particle will actually take. We should therefore be convinced that integration over $\exp{(iS/\hbar)}$ will cancel out for all positions that are not near the set of positions that extremize (minimize) $S$. Passing to a discrete functional domain, our integral for the kernel goes to $$\langle x_a|x_b\rangle = \lim_{\epsilon\rightarrow 0}\sqrt{\frac{m}{i2\pi\hbar \epsilon}}\int_{-\infty}^{\infty} \times \cdots$$ $$\cdots\times \int_{-\infty}^{\infty}\textrm{d}x_1\cdots \textrm{d}x_{n-1}\exp{\bigg[i\sum_{j=0}^{n-1}\frac{m}{2}\frac{(x_{j+1}-x_j)^2}{\hbar\epsilon}}\bigg],$$ where I have suppressed the limit. Now we perform a variable change that will make our full calculation of this integral easier to see. The following variable change was done in Shankar and hence much of the following derivation goes along the same lines as Shankar does in his "Principles of Quantum Mechanics" book. In fact, our future exploration of the harmonic oscillator wave function's evolution in free space is motivated by Shankar's chapter on 1-d problems in quantum mechanics. Hence, let us switch variables to $$y_j = \bigg(\frac{m}{2\hbar \epsilon}\bigg)^{1/2}x_j,$$ where $N^n$ now goes as $(2\hbar\epsilon/m)^{(n-1)/2}N^n=A$. With this variable substitution, our path integral is now $$\langle x_b|x_a\rangle =A\int_{-\infty}^{\infty}\times \cdots \times \int_{-\infty}^{\infty}\textrm{d}x_1\cdots \textrm{d}x_{n-1} \exp{\bigg[-\sum_{j=0}^{n-1}\frac{(y_{j+1}-y_j)^2}{i}\bigg]}.$$ Let us consider an integration over $\textrm{d}x_j$ after $j-1$ integrals have already been performed. This will evaluate to $$\xi\int_{-\infty}^{\infty}\times \cdots\times\int_{-\infty}^{\infty}\textrm{d}x_j\cdots \textrm{d}x_{n-1}\exp{\bigg[-\frac{(y_{j+1}-y_j)^2+(y_j-{y_{0}})^2}{i}\bigg]}\cdots$$ $$\cdots\exp\bigg[-\frac{(y_{n}-y_{n-1})^2}{i}\bigg]$$ $$=A\int_{-\infty}^{\infty}\times\cdots\times\int_{-\infty}^{\infty}\textrm{d}x_{j+1}\cdots\textrm{d}x_{n-1}\sqrt{\frac{(i\pi)^j}{j+1}}\exp\bigg[-\frac{(y_{j+1}-y_{0})^2}{(j+1)i}\bigg]\cdots$$ $$\cdots\exp\bigg[-\frac{(y_{j+2}-y_{j+1})^2}{i}\bigg]\cdots\exp\bigg[-\frac{(y_{n}-y_{n-1})^2}{i}\bigg]$$ using another Gaussian integral. We have included $\xi$ to remind the reader that we still have to worry about $A$, but also that our various $j-1$ integrations gives us an overall factor of $\sqrt{(i\pi)^{j-1}/j}$ before we perform our integral. Shankar gives an explicit calculation of this in his text up to $j=2$ (the above calculation is not completely obvious and I highly recommend taking a look at Shankar's approach or the approach taken in Feynman and Hibbs' "Quantum Mechanics and Path Integrals"). We should therefore expect that, if we go over our integrations for each value $x_j$ between $j=1$ and $j=n$ that our final integral will evaluate to $$\langle x_b|x_a\rangle =A \frac{(i\pi)^{(n-1)/2}}{\sqrt{n}}\exp\bigg[-\frac{(y_{n}-y_0)^2}{ni}\bigg].$$ Now if we make our variable change back from $y_j$ to $x_j$, we obtain $$\langle x_b|x_a\rangle = \bigg(\frac{m}{i2\pi \hbar n\epsilon}\bigg)^{1/2}\exp\bigg[\frac{im(x_{n}-x_0)}{2\hbar n\epsilon}\bigg].$$ This is, of course, is not our final answer, as we have suppressed the limit. As $\epsilon\rightarrow 0$, $n\rightarrow \infty$, so $n\epsilon\rightarrow t_b-t_a$. Moreover, $x_{n}-x_0\rightarrow x_b-x_a$, appropriately. Therefore, taking the limit: $$\langle x_b|x_a\rangle =\bigg(\frac{m}{i2\pi\hbar (t_b-t_a)}\bigg)^{1/2}\exp\bigg[\frac{im(x_b-x_a)^2}{2\hbar (t_b-t_a)}\bigg].$$ Great, so there are other, much more simple, ways that we could have obtained this bad boy without the path integral, but we will not discuss them in this blog post. We can now find the solution to the free particle Shrodinger equation, $$\hat{H}|\psi\rangle = -\frac{\hbar^2}{2m}\frac{d^2}{dx^2}|\psi\rangle=\frac{p^2}{2m}|\psi\rangle,$$ as an initial value problem using the kernel that we have obtained, such that $$\psi(x_b,t_b)=\int \textrm{d}x\textrm{ } \langle x_b,t_b|x,t\rangle \psi(x,t),$$ where $\psi(x,t)$ is the initial wave function that we must specify.

Resources:

Feynman, Richard Phillips, et al. Quantum mechanics and path integrals. Dover Publications, 2014.

Schwartz, Matthew Dean. Quantum field theory and the standard model. Cambridge University Press, 2017.

Shankar, Ramamurti. Principles of quantum mechanics. Springer, 2014.

## Friday, February 9, 2018

### Derivation of Electronic Transition Rate

We have a particle that is in a vacuum, meaning that the Fock state representing the photon is in its ground state. In making our calculations, we assume that the photonic ground state is the asymptotic state and that the interaction occurs for only a finite amount of time (this is the basis of perturbative quantum field theory). We represent the combined state of the photon and the electron's initial state as a direct product, such that $$|\Psi \rangle = |p_n^{(i)} \rangle \bigotimes |0\rangle = |p_n^{(i)} \rangle |0\rangle$$ Similarly, we represent the final combined state in its asymptotic limit as $$|\Psi \rangle = |p_n^{(f)} \rangle \bigotimes |0\rangle = |p_n^{(f)} \rangle |0\rangle$$ Great, Zetilli represents the quantized free electromagnetic field as a discrete sum of momenta over a large volume (the difference between Zetilli and many other quantum texts being that they let the spectrum be discrete [confining the photons to a finite volume], whereas books on QFT and higher level quantum mechanics typically treat the spectrum as continuous). The full quantized field is $$\hat{\mathbf{A}} = \sum_{\mathbf{k}}\sum_{\lambda = 1}^2 \sqrt{\frac{2\pi \hbar c^2}{\omega_k V}}\bigg[\hat{a}_{\mathbf{k},\lambda}e^{i( \mathbf{k}\cdot \mathbf{r}-\omega_k t)}+\hat{a}_{\mathbf{k},\lambda}^\dagger e^{-i( \mathbf{k}\cdot \mathbf{r}-\omega_k t)}\bigg]$$ which results from search for a set of canonical variables to quantized the classical free electromagnetic field with. Luckily, even the classical field can be expressed as a sum over simple harmonic oscillator Hamiltonians, so the proper quantization falls straight out of what we already know about quantum mechanics simple harmonic oscillators. It is quite beautiful. The first sum is over each of the momenta, while, because we are dealing with light that is plane polarized, the second sum is over each of the orthogonal oscillations. Recall that we can obtain the Hamiltonian for this interacting system via a Legendre transformation of the well-known classical electromagnetic Lagrangian (which is, indeed, related to the Lagrangian representing the interaction between Dirac fields with propagation terms from the Proca Lagrangian). The form of the full classical Hamiltonian is $$\mathscr{H}(\mathbf{p},\mathbf{r},t)=\frac{1}{2m_e}\bigg(\mathbf{p} + \frac{e}{c}\mathbf{A}(\mathbf{r},t)\bigg)^2-e\phi(\mathbf{r},t)$$ where the canonical momentum now depends upon the old canonical momentum that we are used to plus a contribution from the vector potential. Expanding this out $$\mathscr{H}(\mathbf{p},\mathbf{r},t)=\frac{1}{2m_e}\mathbf{p}^2+\frac{e}{2m_e c}\mathbf{p}\cdot \mathbf{A}+ \frac{e}{2m_e c}\mathbf{A}\cdot \mathbf{p}+ \frac{e^2}{2m_e c^2}\mathbf{A}^2 - e\phi(\mathbf{r},t)$$ We drop the four term because it is extremely negligible compared to the rest of the terms. To figure out how to go further, we quantize our canonical variables and choose to work in position space $$\mathscr{H}(\mathbf{p},\mathbf{r},t) \rightarrow \hat{H}( \hat{\mathbf{p}},\hat{\mathbf{r}},t)=-\frac{\hbar^2}{2m_e}\nabla^2+\frac{-i\hbar e}{2m_e c}\nabla\cdot \hat{\mathbf{A}}+ \frac{-i \hbar e}{2m_e c}\hat{\mathbf{A}}\cdot \nabla+ - e\phi(\mathbf{r},t)$$ Moving into the Coulomb gauge (which will simplify our Hamiltonian, but rid our expression for the electromagnetic field operator of its Lorentz covariance), for which $\nabla \cdot \mathbf{A} = 0$ $$\hat{H}( \hat{\mathbf{p}},\hat{\mathbf{r}},t)=-\frac{\hbar^2}{2m_e}\nabla^2- \frac{i \hbar e}{2m_e c}\hat{\mathbf{A}}\cdot \nabla+ - e\phi(\mathbf{r},t)$$ $$=\frac{1}{2m_e}\hat{\mathbf{p}}^2- - e\phi(\mathbf{r},t)+ \frac{ e}{2m_e c}\hat{\mathbf{A}}\cdot \hat{\mathbf{p}}$$ \newpage which we now recognize as $\hat{H}=\hat{H}_0 + \hat{H}_r$ with $\hat{H}_r = \frac{ e}{2m_e c}\hat{\mathbf{A}}\cdot \hat{\mathbf{p}}$. Plugging in the expression for the quantized electromagnetic field $$\hat{H}_r = \frac{e}{m_e}\sum_{\mathbf{k}}\sum_{\lambda = 1}^2 \sqrt{\frac{2\pi\hbar}{\omega_k V}}\bigg[\hat{a}_{\mathbf{k},\lambda}e^{i\mathbf{k}\cdot \mathbf{r}}\epsilon_\lambda \cdot \mathbf{\hat{p}}e^{i\omega_\mathbf{k} t}+ \hat{a}_{\mathbf{k},\lambda}^\dagger e^{i\mathbf{k}\cdot \mathbf{r}}\epsilon_\lambda \cdot \mathbf{\hat{p}}e^{-i\omega_\mathbf{k} t}\bigg]$$ Sandwiching this between the initial and final states (the direct product states of the electron Hilbert space and photonic Fock space) to obtain the S-matrix element ($\mathbf{p}$ and $\mathbf{r}$ act on the electron's Hilbert space, while $\mathbf{a}$ and $\mathbf{a}^\dagger$ act on the photon's Fock space [I should probably mention that a Fock space is simply a direct sum of single-particle Hilbert spaces, so there is little discrepancy between the two]): $$\langle \Psi |\hat{H}_r|\Psi \rangle = \langle p_n^{(f)} | \langle 0 | \sum_{\mathbf{k}}\sum_{\lambda = 1}^2 \sqrt{\frac{2\pi\hbar}{\omega_k V}}\bigg[\hat{a}_{\mathbf{k},\lambda}e^{i\mathbf{k}\cdot \mathbf{r}}\epsilon_\lambda \cdot \mathbf{\hat{p}}e^{i\omega_\mathbf{k} t}...$$ $$+\hat{a}_{\mathbf{k},\lambda}^\dagger e^{i\mathbf{k}\cdot \mathbf{r}}\epsilon_\lambda \cdot \mathbf{\hat{p}}e^{-i\omega_\mathbf{k} t}\bigg]| 0 \rangle | p_n^{(i)}\rangle =\frac{e}{m_e}\sqrt{\frac{2\pi\hbar}{\omega_k V}} \langle p_n^{(f)} | e^{i\mathbf{k}\cdot \mathbf{r}} \epsilon \cdot \mathbf{\hat{p}}| p_n^{(i)}\rangle$$ Recognizing that this new interaction potential has the same form as the harmonic perturbation $$\Gamma_{i\rightarrow f}^{\textrm{emi.}}=\frac{4\pi^2e^2}{m_e^2 \omega_k V} |\langle p_n^{(f)} | e^{i\mathbf{k}\cdot \mathbf{r}} \epsilon \cdot \mathbf{\hat{p}}| p_n^{(i)}\rangle|^2 \delta(E_f-E_i+\hbar \omega_k)$$ and going with the dipole approximation $$\Gamma_{n\rightarrow m}^{\textrm{emi.}}=\frac{4\pi^2 e^2 \omega_{mn}^2}{\omega_k V}|\epsilon_\lambda \cdot \langle p_m(x) |\mathbf{r}| p_n(x)\rangle|^2\delta(E_f-E_i+\hbar \omega_k)$$ where $\omega_{mn}=\frac{E_m-E_n}{\hbar}$. We aim to find the total emission rate, which will be independent of the polarization of the photon. We can accomplish this goal by integrating the the transition rate over the total available phase space after the interaction. We can express the differential element of phase space as $$dN=\frac{Vd^3p}{(2\pi \hbar)^3}=\frac{V\omega^2}{(2\pi c)^3}d\Omega d\omega$$ Hence $$dW_{n\rightarrow m}^{\textrm{emi.}}=\frac{V}{(2\pi)^3c^3}d\Omega \int \omega^3 \Gamma_{n\rightarrow m}^{\textrm{emi.}}d\omega$$ $$=\frac{1}{2\pi \hbar c^3}|\epsilon_{\mathbf{k}}\cdot d_{nm}|^2d\Omega \int \omega_{fi}^2 \omega \delta(\omega_{fi}-\omega)d\omega$$ $$=\frac{\omega_{nm}^3}{2\pi \hbar c^3}|\epsilon_{\mathbf{k}}\cdot d_{fi}|^2 d\Omega$$ Finally, averaging over all of the polarizations $$\langle \epsilon_{\mathbf{k}}\rangle = \frac{\sum_{\lambda=1}^2 |\epsilon_\lambda \cdot d_{fi}|^2}{|\epsilon|^2}$$ we obtain $$dW_{n\rightarrow m}^{\textrm{emi.}}=\frac{q^2\omega_{nm}^3}{3\pi \hbar c^3}|\langle p_m(x) |\hat{x}| p_n(x)\rangle|^2 d\Omega$$

References:
Zettili, N. (2013). Quantum mechanics: concepts and applications. Chichester: Wiley.

## Sunday, October 8, 2017

### Derivation of the Non-Relativistic Path Integral

The path integral formulation of both relativistic and non-relativistic quantum mechanics is arguably one of the most beautiful and far-reaching ideas of physics post-1950's. It is, however, one of the least intuitive. By introducing the path integral in the non-relativistic case, where canonical quantization is done on position and momentum, we introduce the idea that a particle moving between any two points takes an infinite number of paths during its journey. In the relativistic case, where canonical quantization is done on fields, we introduce the idea that fields themselves occupy an infinite number of classical field configurations in a field's journey between any two states of the field (say, the ground state configuration to some excitation). The derivation for both is surprisingly similar, as simply switching $\hat{x}$ for $\hat{\phi}$ (where, in the most simple case, $\hat{\phi}$ is a scalar field) and $\hat{p}$ for $\partial_t \hat{\phi}$ supplemented with the correct completeness relations will yield the path integral for the non-relativistic case. In this short derivation, I will present the non-relativistic case simply because canonical quantization done on position and momentum is more familiar to most physicists who have not taken extensive courses in quantum theory and/or quantum field theory and hence we can confidently use tools that even most undergraduate physics majors have had extensive exposure to. Before I go ahead, I would like to make a couple notes about the path integral formulation.

a.) You have already been exposed the main ideas underlying the path integral. The path integral formulation takes the wave nature of fundamental particles and beefs it up ten-fold. Why might I say that you've already been exposed to it? Well, let's think about a plane wave (the wave described by an idealized free particle): by Huygen's principle, a plane wave is the sum of an infinite number of point sources of waves. Scaling down from infinity to two, we have a wave pattern that looks like the constructive/destructive interference pattern generated by a double slit that partitions an incident plane wave coming from left to right and the resulting wave that comes out of the two slits when the plane wave passes through it. In the case of quantum mechanics, this is identical to the wave amplitude of a particle going through both slits in the double slit experiment and interfering with itself. If we go to three slits, the same thing will happen - the probability amplitude of a particle arriving at a particular location on the other side of the partition containing the slits will be the modulus square of the sum of each wave amplitudes sourcing from all three slits. A particle propagating between any two points can be thought of as a generalization of the double slit experiment to an infinite number of slits and an infinite number of partitions, where each partition is an infinite set of possible position configurations. Because the double slit experiment tells us that no slit is preferred, a particle is allowed to go through each slit as it passes through each partition - e.g. it can travel an infinite number of paths. But hold on! If I wave my hand from left to right, it doesn't move an infinite number of paths! This simply can't be! This is where the beauty of the path integral really comes in: even though each path is taken, the amplitude of arriving at a particular location is weighted by a phase factor $$\delta = e^{i\frac{S}{\hbar}}$$ where S is the classical action, which is stationary for the classical path. If we were to add up all of the paths as they are weighted by this phase factor, then the total propagation amplitude could be calculated by the method of stationary phase, as $\delta = e^{iS / \hbar }$ oscillates slowly only when $S$ nears its stationary point, which occurs when the particle takes the classical trajectory that satisfies Newton's laws of motion. For particles that are sufficiently large, and hence have very small wavelengths, this is the only path that we see. However, when we look at very small objects like electrons with relatively large wavelengths, the fact that the electron takes an infinite number of paths cannot be discounted and the path integral formulation gives an extremely elegant way of utilizing this to explain a variety of phenomena, such as the results of the double slit experiment.

b.) The path integral formulation provides an alternative way to think about the same phenomena. The propagation amplitude that we obtain from the path integral satisfies Shrodinger's equation of motion. The results of the path integral formulation are identical to the canonical picture that we are most familiar with. What is crazy, however, is that the path integral formulation makes no reference to unitary operators acting on a Hilbert space. The non-commutative nature of quantum mechanics is almost hidden by the path integral formulation. The trade-off is that the path integral formulation is manifestly Lorentz invariant. It involves the Lagrangian, which is a Lorentz invariant quantity. The path integral doesn't care about what reference frame I choose. All it cares about is that one rids themselves of the notion that a particle has to take a classical trajectory between any two points, which is, admittedly, quite a disturbing thing to do, as we're so used to Newton's laws of motion - we've been using it ever since our early days of calculating the trajectories of boxes on inclines! If you've taken a course on quantum mechanics, then you may be much more inclined to throw this assumption away, as complete certainty of a particle's position and momentum is fundamentally barred by the non-commutativity of operators in quantum mechanics.

Okay, I'm done talking. Let's get to the actual derivation. Suppose I wish to calculate the probability that a particle in a position eigenstate, $x_a$, will propagate to a position eigenstate, $x_b$. What I would want to do to get this probability is take the modulus square of the following matrix element: $$S = \langle f | i \rangle = \bigg\langle f \bigg| \exp\bigg[ -i\frac{(t-t_1)\hat{H}(t)}{\hbar}\bigg] \bigg| i_0 \bigg\rangle$$ $$= \bigg\langle x_b \bigg| \exp\bigg[ -i\frac{(t-t_1)\hat{H}(t)}{\hbar}\bigg] \bigg| x_a \bigg\rangle$$ I could imagine splitting our time interval up into a discrete number of n time points, such that $t_k = t_1 + k\delta t$. Our propagation amplitude would then turn into $$\bigg\langle x_b \bigg| \exp\bigg[ -i\frac{\delta t\hat{H}(t_f)}{\hbar}\bigg] \cdots\exp\bigg[ -i\frac{\delta t\hat{H}(t_k)}{\hbar}\bigg] \cdots$$ $$\cdots\exp\bigg[ -i\frac{\delta t\hat{H}(t_1)}{\hbar}\bigg] \bigg| x_a \bigg\rangle$$ Let us introduce the position space completeness relation between each of our complex exponentials in the above expansion, such that $$\int dx_{n-1} \cdots dx_1 \bigg\langle x_b \bigg| \exp\bigg[ -i\frac{\delta t\hat{H}(t_f)}{\hbar}\bigg] \bigg|x_{n}\bigg\rangle \bigg\langle x_{n} \bigg|$$ $$\cdots \bigg\langle x_{k+1} \bigg| \exp\bigg[ -i\frac{\delta t\hat{H}(t_k)}{\hbar}\bigg] \bigg|x_k\bigg\rangle \cdots$$ $$\cdots\bigg| x_1 \bigg\rangle \bigg\langle x_1 \bigg| \exp\bigg[ -i\frac{\delta t\hat{H}(t_1)}{\hbar}\bigg] \bigg| x_a \bigg\rangle$$ Now we make this a tad bit more complex. Imaging that, between each of our propagator sandwiches of position eigenstates, we insert the momentum space completeness relation, such that $$\bigg\langle x_{k+1} \bigg| \exp\bigg[ -i\frac{\delta t\hat{H}(t_k)}{\hbar}\bigg] \bigg|x_k\bigg\rangle$$ $$=\int \frac{dp}{2 \pi \hbar}\bigg\langle x_{k+1} \bigg| p \bigg\rangle \bigg\langle p \bigg|\exp\bigg[ -i\frac{\delta t\hat{H}(t_k)}{\hbar}\bigg] \bigg|x_k\bigg\rangle$$ $$=\int \frac{dp}{2\pi \hbar } \exp\bigg[ -i\frac{\delta t\hat{H}(t_k)}{\hbar}\bigg] \exp\bigg[-i\frac{p(x_k - x_{k+1})}{\hbar}\bigg]$$ Since $V(x_{k+1},x_{k})$ obviously doesn't depend upon the momentum, the above equation goes to $$\exp\bigg[-i\frac{V(x_{k+1},x_k)}{\hbar}\bigg]\int \frac{dp}{2\pi \hbar} \exp\bigg[ -i\frac{\delta tp^2}{2m \hbar}\bigg] \exp\bigg[i\frac{p(x_{k+1}-x_{k})}{\hbar}\bigg]$$ Now all we need to do is a Gaussian integral ($\int_{-\infty}^\infty dl e^{-\frac{1}{2}cl^2+kl}=\sqrt{\frac{2\pi}{c}}\exp(k^2/2c)$)! Let $\frac{1}{2\pi \hbar}\sqrt{\frac{2\pi}{c}}=N$. $$=N\exp\bigg[-i\frac{V(x_{k+1},x_k)\delta t}{\hbar}\bigg]\exp\bigg[i\frac{m(x_{k+1}-x_{k})^2}{2 \hbar(\delta t)^2}\delta t\bigg]$$ $$=N\exp\bigg[\frac{i}{\hbar}\bigg(\frac{m(x_{k+1}-x_{k})^2}{2 (\delta t)^2} -V(x_{k+1},x_k)\bigg)\delta t\bigg]$$ Well look what we have here! The exponential is nothing more than the Lagrangian times a constant! Let us now put all of this work together $$\int dx_{n-1} \cdots dx_1 \bigg\langle x_b \bigg| \exp\bigg[ -i\frac{(t_f-t_1)\hat{H}(t_f)}{\hbar}\bigg] \bigg|x_{n}\bigg\rangle \bigg\langle x_{n} \bigg|$$ $$\cdots \bigg\langle x_{k+1} \bigg| \exp\bigg[ -i\frac{(k\delta t)\hat{H}(t_k)}{\hbar}\bigg] \bigg|x_k\bigg\rangle \cdots$$ $$\cdots\bigg| x_1 \bigg\rangle \bigg\langle x_1 \bigg| \exp\bigg[ -i\frac{(t_f-t_1)\hat{H}(t_1)}{\hbar}\bigg] \bigg| x_a \bigg\rangle$$ $$=N^n \int dx_{n-1} \cdots dx_1 \exp\bigg[\sum_{k=0}^{n-1} \frac{i}{\hbar}\bigg(\frac{m(x_{k+1}-x_{k})^2}{2 (\delta t)^2} -V(x_{k+1},x_k)\bigg)\delta t\bigg]$$ Let's take a pause before state the final result. What do you notice here? For each time, we integrate over an infinite number of positions. Each position, however, is weighted by a phase factor $$\exp\bigg[\sum_{k=0}^{n-1} \frac{i}{\hbar}\bigg(\frac{m(x_{k+1}-x_{k})^2}{2 (\delta t)^2} -V(x_{k+1},x_k)\bigg)\delta t\bigg]$$ This is what we mean by a particle taking an infinite number of paths between any two points! Our matrix element (which is actually a Green's function that allows us to propagate free particles between positions in free space) tells us that at each time, the particle hits a "partition". At each of these partitions, we integrate over all of the positions ("slits") that the particle can occupy ("go through"). Taking the limit as $\delta t \rightarrow 0$, this infinite set of integrals turns into a functional integral with a new measure $$\lim_{\delta t \rightarrow 0}\int dx_{n-1} \cdots dx_1=\int Dx$$ and the argument of our exponential turns into $$\lim_{\delta t \rightarrow 0}\sum_{k=0}^{n-1} \frac{i}{\hbar}\bigg(\frac{m(x_{k+1}-x_{k})^2}{2 (\delta t)^2} -V(x_{k+1},x_k)\bigg)\delta t$$ $$=\frac{i}{\hbar}\int \bigg(\frac{m}{2}\dot{x}^2-V(x_{k+1},x_k)\bigg)dt=\frac{i}{\hbar}S$$ Hence, our path integral is $$\langle f | i \rangle = N^n \int Dx e^{i\frac{S}{\hbar}}$$ Gorgeous.

REFERENCES:
Feynman, Richard Phillips, et al. Quantum mechanics and path integrals. Dover Publications, 2014.
Schwartz, Matthew Dean. Quantum field theory and the standard model. Cambridge University Press, 2016.
Shankar, Ramamurti. Principles of quantum mechanics. Springer, 2014.

## Saturday, September 9, 2017

### Emmy Noether's Theorem and Energy/Momentum Conservation

Perhaps one of the most important consequences of classical field theory is a definitive answer to the question "why is momentum and energy conserved"? Indeed, even during your first days of being a wee little physics student, you took energy and momentum conservation as a given - an axiom of the theories underlying physics. Momentum and energy conservation, however, come from a much more deep connection between symmetries and physical laws. In this article, we will try to give the reader an understanding of where these symmetries come from and how they birth energy and momentum conservation. Let us first introduce the definition of the Lagrangian density $$S=\int d^4 x L(\phi, \partial_\mu \phi)$$ where $S$ is the action and the argument of the integral is the "Lagrangian density", which depends upon the scalar field, $\phi$, and its spatio-temporal derivatives, $\partial_\mu \phi$. The integration goes over time and three spatial dimensions. Minimizing the action yields the famous Euler-Lagrange equations of motion for fields: $$\frac{\partial L}{\partial \phi}-\partial_\mu \bigg(\frac{\partial L}{\partial (\partial_\mu \partial \phi)}\bigg)=0$$ Let us now discuss the beef of our derivation. Let us take the variation of the Lagrangian: $$\delta L = \sum_n\bigg(\bigg[\frac{\partial L }{\partial\phi_n}-\partial_\mu \frac{\partial L}{\partial(\partial_\mu \phi_n)}\bigg]\delta \phi_n+\partial_\mu \bigg[ \frac{\partial L}{\partial (\partial_\mu \phi_n)}\delta \phi_n\bigg]\bigg)$$ Probing just a bit deeper, let us take these variations with respect to some parameter, $\beta$: $$\frac{\delta L}{\delta \beta} = \sum_n\bigg(\bigg[\frac{\partial L }{\partial\phi_n}-\partial_\mu \frac{\partial L}{\partial(\partial_\mu \phi_n)}\bigg] \frac{\delta \phi_n}{\partial \beta}+\partial_\mu \bigg[ \frac{\partial L}{\partial (\partial_\mu \phi_n)}\frac{\delta \phi_n}{\delta \beta}\bigg]\bigg)$$ If $\beta$ is a symmetry of the Lagrangian, the first order term (and all other higher order terms) of the series expansion about the original Lagrangian should drop out. Because the functional derivative is a total derivative with respect to the varied parameter (see my post on the derivation of the Euler-Lagrange equations of motion for more information on functional derivatives), the above equation should go to zero. Moreover, because the Euler-Lagrange equations of motion already go to zero, the second term must independently go to zero (it is important to keep in mind that this is true when the particles we are talking about are "on shell", meaning the the equations of motion for $\phi$ are indeed satisfied and the particles therefore obey the mass-shell condition (also called the Einstein energy-momentum relation)). The four integral of this last term is what we call a "Noether current" because its four divergence goes to zero. $$J^\mu=\sum_n \frac{\partial L}{\partial (\partial_\mu \phi_n)}\frac{\delta \phi_n}{\delta \beta}$$ Because $J^\mu$'s four divergence goes to zero, by the continuity equation, there exists a conserved "charge" $$\partial_\mu J^\mu = \frac{\partial \rho}{\partial t}\implies Q=\int d^3 x J^0\implies \frac{\partial Q}{\partial t}=0$$ Let us now get to the full beauty of our discussion. Suppose that we vary the Lagrangian via some sort of space-time translation (that is, suppose that we propagate the Lagrangian to a different point in space or we propagate it in time). The translated Lagrangian will have the following form $$L\rightarrow L'=L+x^\nu \partial_\nu \phi$$ We can write the last term in the following manner $$\partial_\nu L=\partial_\mu \sum_n \frac{\partial L}{\partial (\partial_\mu \phi_n)}\partial_\nu \phi$$ where $\frac{\delta L}{\delta x^\nu}=\partial_\nu L$ and $\frac{\delta \phi}{\delta x^\nu}=\partial_\nu \phi$. Utilizing the metric tensor as an operator that turns contravariant tensors into covariant tensors and vice-versa, we can rewrite this Lagrangian as $$\partial_\mu\bigg( \sum_n \frac{\partial L}{\partial (\partial_\mu \phi_n)}\partial_\nu \phi-\eta{}^\nu_\mu L\bigg)=0$$ Because the four divergence of the term in parenthesis goes to zero, we can think of it as a conserved current. It is actually what we call the "energy-momentum" tensor and it has four currents associated with it, such that $$T^{\mu \nu}=\sum_n \frac{\partial L}{\partial (\partial_\mu \phi_n)}\partial_\nu \phi-\eta{}^\nu_\mu L$$ The $T^{00}$ term reproduces the Legendre transformation of the Lagrangian density, which, if you looked at my post on deriving Hamilton's equations of motion from the Lagrangian via the Legendre transformation, you should recognize this as the Hamiltonian density. Because the four divergence of the energy-momentum tensor goes to zero, we should have an associated charge conservation $$H=\int d^3 x T^{00}$$ which is, *drum roll*, the total energy! This term is what corresponds to time translation invariance of the Lagrangian (and, actually, more generally, time translational invariance of the action itself). What about the the three spatial indices, $T^{0i}$? These correspond to momentum densities in the three Cartesian coordinate momenta! Again, because the four divergence of the energy-momentum tensor goes to zero, there is a conserved charge for each one of these densities, which then correspond to, *drum roll*, the three Cartesian coordinate momenta! Willie nillie, we found ourselves a definitive answer to why energy and momentum are conserved (from space-time symmetries of the action) using our understanding of charge conservation (and, really, the continuity equation)!

## Saturday, July 8, 2017

### How Do Magnets Work? Spin, Paramegnetism, Ferromagnetism and the Ising Model

 Photo Credit: Newton Henry Black, Harvey N. Davis (1913) Practical Physics, The MacMillan Co., USA, p. 242, fig. 200.

Magnets are everywhere - they play a crucial role in correcting the trajectories of charged particles in particle accelerators, they act as very great emergency breaking systems, they encode your credit card information whenever you swipe to make a purchase, and you may even have a few sitting on your refrigerator. Unfortunately, however, even many undergraduate physics majors never get to fully appreciate the origin of this strange phenomena that we call "permanent magnetism". Indeed, not every metal can become a permanent magnet. As it turns out, magnetic materials tend to require that the individual atoms that make up the metal be sufficiently close to each other to experience an "interaction potential" whose origin is fundamentally quantum mechanical. I will attempt to make this discussion tailored toward the intermediate undergraduate physics major or the even the advanced high school student. Why so? Well, the full beauty of permanent magnetism is best described in its dirty quantum mechanical nature. We could talk all day about the classical origin of magnetism, treating every atom as a miniature dipole (which they are), but this watered-down description does not give the reader an understanding of where permanent magnetism really comes from; rather, it gives the reader a classical approximation to what permanent magnetism is. We will begin with a short warm-up with spin and build our way (in surprisingly short time) to the classical, macroscopic nature of magnetism through the mean field Ising model. Let's do it!

### I. A QUICK REFRESHER ON SPIN

So what is spin? Angular momentum! But it is a special kind of angular momentum that, for particles with a spin greater than zero, allows us to write down the full rotation operator (a group element of SU(2)) as it acts upon spinor states (representations of SU(2)) via treating it as the generator of finite rotations: $$T(\theta)=\exp\bigg(\frac{-i \mathbf{\theta} \cdot (\mathbf{\hat{L}}+\mathbf{\hat{S}})}{\hbar}\bigg)$$ Okay, so what about the stationary states of $\mathbf{\hat{S}}$? Well, it is angular momentum, so the eigenvalues for the stationary states for the total angular momentum operator should at least be: $$\hat{S}^2| \psi \rangle= \hbar^2 s(s+1)|\psi \rangle$$ Moreover, the eigenvalue of the z-projection operator should then also be: $$\hat{S}_z|\psi \rangle = \hbar m|\psi\rangle$$ Now, this is where we start jumping out of the realm of normal orbital angular momentum, $\mathbf{\hat{L}}$, and into the strange nature of spin. It is this very nature that causes spin to have no classical analogue - it is purely a quantum mechanical object with macroscopic classical effects. For fermions, the values of s are only half-integral, while for bosons, the values of s are only integral. Moreover, the stationary states of the spin operator are NOT functions - they are spinors (two-dimensional representations of the SU(2) group). For electrons, the value of s is $$s=\frac{1}{2}$$ so the z-projection is either $$\lambda_z=\frac{\hbar}{2}$$ which is called 'spin-up' or $$\lambda_z=-\frac{\hbar}{2}$$ which is called 'spin-down'. Therefore, the stationary states for an electron are 2-dimensional spinors, which either represent spin-up, spin-down, or a linear combination of the two (a superposition of spin-up and spin-down). It turns out that much of the information presented here (including our quick blurb about the rotation operator) is just about all you need to have as an elementary basis in starting your journey to understanding quantum computing! As a final side note on our digression on the fundamentals of spin, recall that we can write the spin operators of an electron (or any spin-1/2 fermion) in terms of Pauli matrices, $\sigma_i$, where i=1,2,3, which correspond to the x, y and z coordinates, respectively. That is: $$\hat{S}_x=\frac{\hbar}{2}\sigma_1=\frac{\hbar}{2}\begin{bmatrix}0 & 1\\1 & 0\end{bmatrix}$$ $$\hat{S}_y=\frac{\hbar}{2}\sigma_2=\frac{\hbar}{2}\begin{bmatrix}0 & -i\\i & 0\end{bmatrix}$$ $$\hat{S}_z=\frac{\hbar}{2}\sigma_3=\frac{\hbar}{2}\begin{bmatrix}1 & 0\\0 & -1\end{bmatrix}$$ The eigenvalues of the Pauli spin matrices are $+1$ for spin-up and $-1$ for spin-down. Let us then ask: what happens when a spin-1/2 particle that is subject to an external magnetic field? We will answer this question in a semi-classical analysis of the situation (i.e. we are not quantizing the magnetic field) - the results are a hallmark of 20th century physics.

The Microscopic Effects of a Uniform Magnetic Field

Let us think of the most simple situation possible: an electron in a uniform magnetic field. What happens? If we think of spin as, perhaps, some sort of current (it's not) that creates a dipole moment that depends on the angular momentum of what is spinning, then the dipole moment is: $$\mathbf{\mu}=C\mathbf{L}$$ where C is some constant that keeps the dimensions consistent and L is the angular momentum of the spinning current. This model is not without its merit for the case of a single electron, as the intrinsic spin does actually generate a magnetic moment even though no current is actually orbiting the electron. We can write down the dipole moment produced by the spin of the electron as: $$\mathbf{\mu}=\frac{g e}{2mc}\mathbf{\hat{S}}$$ where e is the elementary charge of the electron, m is the mass, and c is the speed of light. If you have studied very much classical electrodynamics, you may be wondering where the constant 'g' came in. This constant is referred to as the 'LandÃ© g-factor' or, simply, the 'g-factor' and was first discovered experimentally, then derived from the framework of relativistic quantum mechanics (for more information, see [1]). If a stationary dipole is subject to a uniform magnetic field, its total (non-relativistic) energy is: $$E=-\mathbf{\mu} \cdot \mathbf{B}$$ In an elementary E&M course, one may hear the heuristic argument that dipoles prefer to align with the magnetic field because it minimizes the energy. We will see why this is true within the framework of statistical mechanics in the next section. Via direct analogy, we can write the Hamiltonian of an electron subject to a uniform magnetic field as: $$\hat{H}=-\frac{g e}{2mc} \mathbf{B} \cdot \mathbf{\hat{S}}$$ Suppose that the magnetic field is pointed up in the z-direction. The Hamiltonian is then: $$\hat{H}=-\frac{g e}{2mc}B\hat{S}_z$$ Because we are only dealing with the spin operator, the eigenvalues of this Hamiltonian are: $$E=-\frac{g \hbar e}{4mc} B \text{ or } E=\frac{g \hbar e}{4mc}B$$ corresponding to spin-up or spin-down, respectively. The eigenvectors therefore diagonalize $\hat{S}_z$. Okay, that's some great information, but what is actually happening? Let us evolve the state via the time-evolution operator: $$\hat{U}(t-t_0)\lvert\psi_0\rangle=a\exp\bigg(\frac{ige}{4mc} B(t-t_0)\bigg)\lvert+\rangle+b\exp\bigg(-\frac{ige}{4mc} B(t-t_0)\bigg)\lvert-\rangle$$ where $\lvert+\rangle$ denotes a spin-up state, $\lvert-\rangle$ denotes a spin-down state, and a and b are normalization constants. If a=0 (the electron is initially in a spin-up state), then: $$\hat{U}(t-t_0)\lvert\psi_0\rangle=a\exp\bigg(\frac{ige}{4mc} B(t-t_0)\bigg)\lvert+\rangle$$ Let us think of the action of the time evolution operator in a different fashion. By identification with the rotation operator, we can see that the effect of the time evolution operator on the spin-up state is to rotate it in Hilbert space by an angle of: $$\theta = \frac{ge\hbar}{4 mc} B(t-t_0)$$ Therefore, the state rotates at a constant rate of: $$\omega=\frac{ge\hbar}{4 mc} B$$ This is called the 'Larmor' frequency. We can see that the effect on a spin-down state is to rotate in the opposite direction. A linear combination of spin-up and spin-down then yields a superposition of rotations in both directions at the Larmor frequency, which means that which state we observe to electron to be in is time-dependent in the case of being a linear combination of the two, for the combined state is no longer stationary. Hence, if the initial state is indeed a linear combination of spin-up and spin-down, we will observe a time-dependent probability of observing it in either a spin-up state or spin-down state, but we will NEVER observe it in some other state. Let us now move on to shortly analyzing the effect of spin in physical space, not just Hilbert space. Suppose the original orientation of the spin vector is in some intermediate direction between spin-up and spin-down, then the expectation value of the components are[2]: $$\langle \hat{S}_x \rangle=\frac{\hbar}{2}\sin(\beta)\cos\bigg(\frac{ge\hbar}{4 mc} Bt\bigg)$$ $$\langle \hat{S}_y \rangle=-\frac{\hbar}{2}\sin(\beta)\sin\bigg(\frac{ge\hbar}{4mc} Bt\bigg)$$ $$\langle \hat{S}_z \rangle=\frac{\hbar}{2}cos(\beta)$$ where $\beta$ is the angle of the average spin vector from the z-axis. Therefore, if the spin vector is in some intermediate value between spin-up and spin-down and the uniform magnetic field is in the z-direction, the expectation value of the spin vector will rotate about the z-axis at the Larmor frequency. What is the gist of this? Electrons in uniform magnetic fields precess about the magnetic field lines at a particular frequency called the Larmor frequency! The situation gets much more interesting when we introduce a non-uniform magnetic field, for the electron will not only rotate at the Larmor frequency, but it will also gain momentum in either the up or down direction depending upon the original orientation of the electron as it passes through the magnetic field. We will not go into detail about this, but I do highly encourage the reader to read through page 399 of [1] or 182 of [2] for more information about the particular behavior of silver atoms (which have a total spin of 1/2) as they pass through a non-uniform magnetic fields in a famous classical experiment called the Stern-Gerlach experiment.

The Microscopic Effects of Other Electrons

We will now discuss a different type of potential that is only non-negligible when there exists a high degree of overlap between the wave functions of two electrons. This potential is called the 'exchange interaction' (it will play a crucial role in our analysis of ferromagnetism). We can write the Hamiltonian down for this interaction as: $$\hat{H}=-k_{1,2}\mathbf{\hat{S}}_1\cdot\mathbf{\hat{S}}_2$$ where $k_{1,2}$ is some arbitrary constant that depends upon the degree of overlap between the wave functions (we shall elaborate in section III). The stationary state solutions to this Hamiltonian are direct product states of the two electrons, such that the three following states are degenerate for the energy eigenvalue corresponding to a total spin of $s=1$ ($E=\frac{k\hbar^2}{4}$), called 'triplet states': $$|\psi_1\rangle=|+\rangle \otimes |+\rangle$$ $$|\psi_2\rangle=|+\rangle\otimes|-\rangle+|-\rangle\otimes|+\rangle$$ $$|\psi_3\rangle=|-\rangle\otimes|-\rangle$$ while the energy eigenvector for the energy eigenvalue ($E=\frac{-3k\hbar^2}{4}$) corresponding s=0 is just the 'singlet state': $$|\psi_3\rangle=|-\rangle\otimes|+\rangle-|+\rangle\otimes|-\rangle$$ By the Pauli exclusion principle, coupled electrons are not allowed to be in the same state (i.e. cannot be simultaneously in the same spin state and spatial wave function), for they occupy the anti-symmetric singlet states. However, in details that we shall explain in due time, combined systems of spin-1/2 can occupy triplet states depending upon signage of $k_{1,2}$, which will play a crucial role in our analysis of ferromagnetism. Now that we have discussed the two important potentials at play, let us get into the nitty-gritty details of the Ising model by discussing first a few simple situations that will allow us to appreciate the full scope of ferromagnetism.

### II. THE EFFECT OF AN INCIDENT MAGNETIC FIELD (PARAMAGNETISM)

Consider a system composed of weakly interacting spin-1/2 atoms (i.e. the exchange interactions of each atom can be neglected) at some temperature T. Then we can consider the probability of a single atom occupying a particular energy state to be described by the Boltzmann distribution: $$p(E_r,T)=\frac{\exp(-\beta E_r)}{\sum_r \exp(-\beta E_r)}$$ which we can obtain from a Taylor expansion of the natural log of the number of accessible states of the entire system of atoms while using the fundamental relations $\frac{\partial S}{\partial E}=\frac{1}{kT}=\beta$, where k is the Boltzmann constant, T is the temperature, and S is the entropy. $E_r$ denotes the energy of the atom in a state $r$, which is either spin-up or spin-down. We could also just get it from considering the partition function (which is the denominator of this expression). The Hamiltonian of this system is: $$H=-\mu B$$ Generally, this is an inner product, but we are assuming the $\mathbf{B}$ is uniform in magnitude a direction, so $\mu$ points along the same direction as $\mathbf{B}$. Since the atom either occupies a spin-up state or spin-down state, the two possible energies are either positive or negative, corresponding to the magnetic moment being anti-aligned with the external field or aligned with the external field, respectively. Hence, the probability of being spin-up is: $$p_{+}(E_r,T)=\frac{\exp(\beta \mu B)}{\exp(-\mu B)+\exp(\mu B)}$$ while spin-down is: $$p_{-}(E_r,T)=\frac{\exp(-\beta \mu B)}{\exp(-\mu B)+\exp(\mu B)}$$ In the regime of relatively low temperatures: $$\bigg(\frac{\exp(\beta \mu B)}{\exp(-\mu B)+\exp(\mu B)}\bigg )>\bigg( \frac{\exp(-\beta \mu B)}{\exp(-\mu B)+\exp(\mu B)}\bigg)$$ and thus: $$p_{+} > p_{-}$$ so the probability of the magnetic moment being aligned with the magnetic field is greater than being anti-aligned. However, if the temperature is high, then $\beta \implies 0$, so $$p_{+} \approx p_{-}$$ This behavior of the probabilities in the high temperature regime can be attributed to the random thermal motion of the atoms, for the higher the temperature, the higher the average kinetic energy of each atom and hence the higher the resistance of the atoms to being aligned in a particular direction of spin as dictated the direction of the incident uniform magnetic field. Therefore, the preference of the magnetic moment to align with the magnetic field is fundamentally an effect of statistical mechanics! We can, however, do even better than simply calculating the probabilities of a particular atom occupying some particular state - we can go as far as calculating the magnetization of the material from the mean magnetic moment. From the definition of the mean: $$\bar{\mu} = \sum_r p_r \mu_r=\mu \frac{\exp(\mu B)-\exp(-\mu B)}{\exp(\mu B)+\exp(-\mu B)}=\mu \tanh\bigg(\frac{\mu B}{kT}\bigg)$$ from the definition of the magnetization (dipole per unit volume), we can let N define the number of atoms per volume. Hence: $$M=N\mu \tanh\bigg(\frac{\mu B}{kT}\bigg)$$ So in the regime of high temperature ($\tanh$ approaches zero): $$M\approx \frac{N\mu^2}{kT}B=\chi B$$ where $\chi$ is the magnetic susceptibility. However, by the shape of hyperbolic tangent, in the regime of low temperatures and/or strong magnetic field, $M$ approaches $M=N \mu$. Very beautiful!

### III. THE EFFECT OF EXCHANGE INTERACTIONS

Now that we have played around with the case of only an incident magnetic field with no coupling, let us now consider a 1-D line of spin-1/2 atoms with no external magnetic field, but with a non-negligible spin-spin coupling. Let the this line not fold into itself (i.e. let us not impose periodic boundary conditions on the atoms). The discussion we gave in section I on the effect of the exchange interaction was quite elementary. We should now go into a bit more detail on exchange interactions to give the reader a good idea of what they are. As mentioned earlier, the constant, k, in $$\hat{H}=-k_{1,2}\hat{\mathbf{S}}\cdot\hat{\mathbf{S}}$$ determines whether the material favors triplet states or singlet states. We can write k in terms of a new constant, J, such that $k=J$ and J is[3]: $$J=2\frac{J_{ex}-CB^2}{1-B^4}$$ where $$J_{ex}=\int \psi_a^*(\mathbf{r_1}) \psi_b^*(\mathbf{r_2}) \bigg(\frac{1}{R_{ab}}+\frac{1}{r_{12}}-\frac{1}{r_{a1}}-\frac{1}{r_{a2}}\bigg)\psi_a(\mathbf{r_1})\psi_b(\mathbf{r_2})d^3\mathbf{r_1}d^3\mathbf{r_2}$$ which is called the 'exchange integral'. $\psi_a$ is the atomic state of atom a and $\psi_b$ is the atomic state of atom b. The quantity $R_{ab}$ is the distance between the protons a and b of each atom, $r_{12}$ is the distance between electrons 1 and 2 of each atom, $r_{a1}$ is the distance between electron 1 and proton a, and $r_{a2}$ is the distance between electron 2 and proton b. The quantity, C, which is called the 'Coulomb integral', is: $$C=\int [\psi_a(\mathbf{r_1})]^2\bigg(\frac{1}{R_{ab}}+\frac{1}{r_{12}}-\frac{1}{r_{a1}}-\frac{1}{r_{a2}}\bigg)[\psi_b(\mathbf{r_2})]^2d^3\mathbf{r_1}d^3\mathbf{r_2}$$ and finally, the quantity B is called the 'overlap integral', which measures the degree of overlap between the two atomic wave functions. $$B=\int \psi_a(\mathbf{r_1})\psi_b(\mathbf{r_2})d^3\mathbf{r_2}$$ For much more complicated atoms, calculating these integrals is a daunting task, as one must resort to numerical methods for calculation of these integrals, given the vast majority of wave functions (recall that the only atomic wave function that we have an analytical solution for, thanks to the Dirac equation or the Shrodinger equation with perturbative corrections, is that of the hydrogen atom) have no straight-forward (as in, non-approximate) solution. However, we can, and quite extensively have, obtain the value of J experimentally. Materials that are anti-ferromagnetic have a value of $J>0$, thus favoring anti-alignment of their spins (i.e., they favor singlet states of the two electron system) because it is the lowest energy state. Materials that are ferromagnetic have a value of $J>0$ and hence the lowest energy states are that of the degenerate triplet states of the two electron. Within the Ising model, we make assumption that, instead of the pure form that we explored for the two electron system, we can write the approximate Hamiltonian down for this system as: $$H=-\sum_{i=1}^{n-1} J_{i}\sigma_i\sigma_{i+1}=\sum_{i=1}^{n-1}\langle \mathbf{\hat{S}}_i\cdot\mathbf{\hat{S}}_{i+1}\rangle$$ where $\sigma_i$ and $\sigma_{i+1}$ are the eigenvalues for the Pauli spin matrices explored in section I (+1 for spin-up and -1 for spin-down). As a simplification, we let the value of J be $J=J_i$ for all $i\in\{1,2,...,n-1\}$. What does this mean? Simply that the interaction potential cares only about the nearest neighbors (i.e. that each atom does not 'sense' the presence of electrons located more than one atom away) and that each atom is equally spaced (i.e. that the integrals that J depends upon are the same for every i). This is a feasible approximation, for the value of J drops off extremely fast. The value of $i$ ends at $n-1$ because at the end of the line of atoms, there is no $i+1$ atom. Let us, just for convenience, let $\sigma_i\sigma_{i+1}=\mu_i$ since the product is either +1 or -1. If we just focus our attention upon the spin of just a single atom and consider the rest of the system as a heat bath, we can calculate all of the necessary thermodynamic quantities straight from the partition function: $$Z=2\sum_{\mu_i=+1,\mu_i=-1}\exp\bigg(\beta J\sum_{i=1}^{n-1}\mu_i \bigg)$$ where, again, $\beta=(kT)^{-1}$, k is the Boltzmann constant, and T is the temperature. The sum goes over each value of the product of $\sigma_i$ and $\sigma_{i+1}$. The factor of 2 is introduced because, even though there are two possible values of the product of $\sigma_i\sigma_{i+1}$, there are four possible combinations of spin orientations (++,+-,-+,--) and the partition function must go over every possible state (and hence every possible combination). Moreover, because a sum in the argument of the exponential simply multiplies each exponential over the sum: $$2\sum_{\mu_i=+1,\mu_i=-1}\exp\bigg(\beta J\sum_{i=0}^{n-1}\mu_i \bigg)=2\prod_{i=1}^{n-1}\sum_{\mu_i=+1,\mu_i=-1}\exp(\beta J \mu_i)$$ $$=2\bigg(\sum_{\mu_1=+1,\mu_1=-1}\exp(\beta J \mu_1)\sum_{\mu_2=+1,\mu_2=-1}\exp(\beta J \mu_2)\text{ }\dotsi\sum_{\mu_{n-1}=+1,\mu_{n-1}=-1}\exp(\beta J \mu_{n-1})\bigg)$$ $$=2\bigg([\exp(\beta J)+\exp(-\beta J)]_1\text{ }\dotsi[\exp(\beta J)+\exp(-\beta J)]_{n-1}\bigg)$$ $$=2\bigg(\exp(\beta J)+\exp(-\beta J)\bigg)^{n-1}$$ Finally, we can simply use the identity $\exp(\theta)+\exp(-\theta)=2\cosh(\theta)$ to obtain a full expression for the partition function: $$Z=2\bigg(2\cosh(\beta J)\bigg)^{n-1}$$ How nice! Don't get too comfortable with the beautiful simplicity, however, as simply increasing the dimensionality of the problem to $D\gt 2$ steps out of the bounds of an analytical solution for the partition function. Moreover, adding an incident magnetic field changes the dynamics of the problem in a manner that we shall explore later. Let us now, just as we did in the last section, calculate the particular spin orientation that this Hamiltonian prefers. This amounts to calculating the average value of $\mu_i$, as the average will tell us whether the spins prefer to align or anti-align. The average can be expressed in terms of the partition function in the following manner: $$\langle \mu \rangle = 2Z^{-1}\sum_{\mu_i=+1,\mu_i=-1}\exp\bigg(\beta J\sum_{i=0}^{n-1}\mu_i \bigg)\mu_i$$ Which, following along the same lines as we had in the last section, is evaluated as: $$2Z^{-1}\sum_{\mu_i=+1,\mu_i=-1}\exp\bigg(\beta J\sum_{i=0}^{n-1}\mu_i \bigg)\mu_i$$ $$=\bigg(\bigg(2\cosh(\beta J)\bigg)^{n-1}\bigg)^{-1}\sum_{\mu_i=+1,\mu_i=-1}\exp\bigg(\beta J\sum_{i=0}^{n-1}\mu_i \bigg)\mu_i$$ $$=\frac{\bigg(\mu\exp(\beta J)-\mu\exp(-\beta J)\bigg)^{n-1}}{\bigg(2\cosh(\beta J)\bigg)^{n-1}}=\frac{\bigg(2\mu\sinh(\beta J)\bigg)^{n-1}}{\bigg(2\cosh(\beta J)\bigg)^{n-1}}=\bigg(\mu\tanh(\beta J)\bigg)^{n-1}$$ where we have used the identity $\exp(\theta)-\exp(-\theta)=2\sinh(\theta)$. The steps leading to the evaluation of the numerator are identical to those in which we used to obtain the partition function; however, the sum over the exponentials is now weighted by $\mu$, producing the $\sinh$ function. Hence: $$\langle \mu \rangle=\bigg(\tanh(\beta J)\bigg)^{n-1}$$ What does this tell us? Well, $\tanh$ is either always negative or always positive depending upon the parity of its argument, but the only parameter that is allowed to vary in the argument of the $\tanh$ function is T, which can never be negative. We therefore surmise that the preferred orientation of the individual spins in this line of atoms is in the same as the neighboring atoms. The lower the temperature, the easier it becomes to align all of the spins into a single direction (given that $\beta = (kT)^{-1}$ and thus lower temperatures increase the magnitude of $\tanh$'s argument). We are now ready to develop a theory for the existence of ferromagnetism, where exchange interactions play a crucial role permanent magnetization of a material, while an incident magnetic field plays a crucial role in generating magnetization in the first place.

### IV. FERROMAGNETISM (The Ising Model)

Let us reiterate what we learned in the last two sections before we really jump into the theory of ferromagnetism. I would like the reader to be capable of appreciating the feet that the advent of quantum mechanics and its use in modern statistical mechanics is in explaining a phenomena that, in the scope of modern history, has only been well-understood in very recent times (there is actually still a lot of research done in developing a better understanding of ferromagnetism, but the essentials of ferromagnetism, which are what we are exploring, are well-understood). Perhaps the particular case of ferromagnetism that you are most familiar with (and of which are not actually ferrous at all) is that of the convenience store neodymium magnets. If one was capable of probing a ferromagnetic material (we are) and observing the orientation of all of the spins of the individual atoms that make up the material, they would likely see something of the following:

The arrows indicate the direction of the spins. In each of the domains, called ferromagnetic domains, most of the spins are aligned in the same direction. The existence of these domains perplexed scientists for quite some time, as no then-current theory of electromagnetism was capable of explaining why these domains existed in some materials and not others. We now understand that the existence of these domains is a byproduct of what we just explored in the last section, such that each of the spin-1/2 atoms tends to prefer to align its spin vector in the same direction. We could bias all of these domains to align their spins in the same direction if we utilize the existence of an external field to align them along the same direction, as we explored in the second section of this article. The particular curve that the magnetization follows for different magnitudes of magnetic fields was first discovered experimentally and is called a 'hysteresis curve', as given by the image below.

The magnitude of the magnetization depends upon the history of the already existing magnetic field, such that, if one starts at any point on this curve and adds a particular amount of magnetic field, H, the path that the magnetization will follow depends upon that particular starting point. One of the easiest ways one could carry out such a task is to simply wrap a ferromagnetic metal bar with copper wire, then add a constant current, which creates an approximately constant magnetic field that depends upon the direction of the current down the axis of the metal bar. This magnetic field will bias the spins to align along a particular direction and produce a net magnetization and hence a net magnetic field (for a net magnetization produces a magnetic field). Because the material is ferromagnetic, the preferred orientation of any individual spin depends upon the spin of its neighbors, so the material will remain magnetized. What one has done is create a permanent magnet out of simply wrapping wire around a rod and running current through it! Permanent magnetism is hence a result of quantum mechanics - to think that it was hiding under our noses the whole time! Wonderful, now let us get into the Ising model in its full detail by considering a lattice of ferromagnetic atoms and adding the two interactions that we discussed in the last two sections together for each atom in the following manner. $$H=-\sum_{\langle ij \rangle}J_{ij}\sigma_i\sigma_j-\mu B\sum_i\sigma_i$$ In this Hamiltonian, $i$ denotes each atom in the lattice, $j$ denotes the set of atoms nearest to atom $i$. Hence the notation in first sum, $\langle ij \rangle$ means to sum over i first, then each of the nearest atoms. The potential on the right is the same as the one we considered in section II, but now we are summing over each of the atoms in the lattice. Let us now make our first approximation before we get into the big approximation (the mean field approximation) - let's assume that every atom is surrounded by six nearest neighbors, two on each Cartesian axis. Moreover, assume that each of these atoms are equally spaced from atom $i$ and that $J>0$ for each atom. This looks like the following picture.

Now, unfortunately, this 3-dimensional model of ferromagnetism does not have an analytical solution for the partition function. That is, unless we make one more approximation. This 'mean field' approximation, however, still allows us to probe the existence of second-order phase transitions in the model and many other properties that are important to our understanding of ferromagnetic materials. We have actually already made one key assumption and that is to assume that each of the atoms are equally spaced with a constant J value that is greater than zero. The second key assumption is that the average spin of any particular atom is equal to some constant order parameter, m and that the deviation from this average is some small quantity, $\delta \sigma_{i \text{ or } j}$. We can then do a cute little trick by adding zero to each value of spin: $$\sigma_i=m+(\sigma_i-m)=m+\delta \sigma_i$$ We can then write the Hamiltonian as: $$H=-\sum_{\langle ij \rangle}J_{ij}(\sigma_i+\delta \sigma_i)(\sigma_j+\delta \sigma_j)-\mu B\sum_i\sigma_i$$ $$=-\sum_{\langle ij \rangle}J_{ij}(m^2+m(\delta\sigma_i+\delta\sigma_j+\delta\sigma_i\delta\sigma_j))-\mu B\sum_i\sigma_i$$ Dropping terms of order $O(2)$ $$H=-\sum_{\langle ij \rangle}J_{ij}(m^2+m(\delta\sigma_i+\delta\sigma_j))-\mu B\sum_i\sigma_i$$ $$=-\sum_{\langle ij \rangle}J_{ij}(-m^2+m(\sigma_i+\sigma_j))-\mu B\sum_i\sigma_i$$ where in the last part, we re-inserted the expression for $\sigma_i$ in terms of m and itself to re-obtain the sigmas. For the next part, a lot of resources tend to leave out the details, seemingly writing the next step as some trivial operation on the Hamiltonian. It is true that it is trivial, but without some illumination, it can be somewhat frustrating to figure out what is going on. Consider the first sum: $$-\sum_{\langle ij \rangle}J_{ij}(-m^2)$$ We already assumed the necessary assumptions to treat J as the same quantity for every neighboring atom to lattice site, i. Let us approximate this sum, then, to simply be a multiple of N atoms by z neighbors (this approximation will only be valid for large numbers of atoms). We must, however, introduce a factor of 1/2, for each lattice site shares each of its six atoms with the neighboring lattice sites and hence the 1/2 prevents us from double counting the j's. $$-\sum_{\langle ij \rangle}J_{ij}(-m^2)=\frac{1}{2}NzJm^2$$ We can approximate the second sum by arguing that, since the sum contains exactly one $\sigma_i$ for each of the bonds, we can let $$-\frac{1}{2}\sum_{\langle ij \rangle}m(\sigma_i+\sigma_j)=\sum_i \sigma_i$$ because there are z bonds for each i. Wonderful, so now our Hamiltonian is: $$H=\frac{1}{2}JzNm^2-(JzNm+\mu B)\sum_i \sigma_i$$ In calculating the partition function from this, what we have done may not seem like much of an improvement, but we have actually already calculated the partition function for this in section III, as: $$Z=\sum_{\sigma_i=1,\sigma_i=-1}\exp\bigg(-\frac{1}{2}\beta JzNm^2+\beta(Jzm+\mu B)\sum_i \sigma_i\bigg)$$ $$=\exp(-\frac{1}{2}\beta JzNm^2)\sum_{\sigma_i=1,\sigma_i=-1}\exp\bigg(\beta(Jzm+\mu B)\sum_i \sigma_i\bigg)$$ which we have already calculated, but with multiples of sigma's instead of single sigma's alone. Therefore: $$Z=\exp\bigg(-\frac{1}{2}\beta JzNm^2\bigg)\bigg[2\cosh\bigg(\beta(Jzm+\mu B)\bigg)\bigg]^N$$ Wonderful! Now comes the important part of our derivation. We have made a lot of assumptions. Let's see if we can obtain a self-consistency equation for m that will justify all of the assumptions that we have made in getting this far. We do this by minimizing the Helmholtz free energy, which we can calculate using the partition function. $$F=-\beta^{-1}\ln(Z)=-\beta^{-1}\ln\bigg(\exp\bigg(-\frac{1}{2}\beta JzNm^2\bigg)\bigg[2\cosh\bigg(\beta(Jzm+\mu B)\bigg)\bigg]^N\bigg)$$ Hence: $$F=\frac{1}{2}JzNm^2-\beta^{-1}N\ln\bigg(\bigg[2\cosh\bigg(\beta(Jzm+\mu B)\bigg)\bigg]\bigg)$$ Via elementary calculus, we can minimize this free energy to obtain the following self-consistency equation for the value of m (the average value of $\sigma$): $$\frac{\partial}{\partial m}F=0 \implies m=\tanh\bigg(\beta(Jzm+\mu B)\bigg)$$ There is a lot of detail in this equation, as it codes much of the information that we 'expect' about how permanent magnets behave. First of all, phase transitions. Are there any phase transitions evident in this? This may seem like somewhat of a cute analysis, but it will answer the question in a simple way. Let us define a new variable: $$x=\beta(Jzm+\mu B)$$ Hence, our equation becomes: $$\frac{kT}{Jzm+\mu B}x=\tanh(x)$$ If we look at the simple form of $\tanh$ using an online graphing calculator, we note the following form:

desmos.com
The solutions to our consistency equation should lie at the intersection of a line with a slope of $\frac{kT}{Jzm+\mu B}$ and the $\tanh$ function. Hence, for slopes that are large (that is, for high temperatures or small contributions from the already established value of m [hysteresis hinting itself in the denominator] and B), there will exist one solution at $\tanh(x)=0$, i.e. $m=0$. Hence, the average spin is zero. This is precisely what we should expect, as the random thermal motions caused by large temperatures should randomly orient the spins. Only with a stronger magnetic field can we begin to re-order the spins. Our phase transition occurs when we decrease the slope. For a matter of fact, it occurs right at: $$\frac{kT}{Jzm+\mu B}=1$$ This is because the slope of $\tanh$ approaches 1 as its argument approaches zero. What do we expect then? Three solutions!

desmos.com

There hence exists one solution at m=a, m=-a, and m=0, where a is an arbitrary number set by the slope; however, the m=0 solution corresponds to an unstable solution, so the order parameter will tend to stick around either m=a or m=-a, i.e. either highly biased toward being spin-up or highly biased toward being spin-down. We can see the beauty of these results in the following 2-D Monte Carlo mean field Ising model simulations performed by the University of Texas. First, we have simply a panel of atoms at a temperature of $T=40T_c$, where $T_c$ is the critical temperature in the zero magnetic field regime (even though we assumed that there existed a magnetic field in our entire derivation, we can set B=0 either at the beginning or end of the derivation and retrieve the same result).

http://farside.ph.utexas.edu/teaching/329/lectures/node110.html

The black denotes spin-up and the white denotes spin-down. Notice that the spins are fairly well distributed at this higher temperature and hence we expect that the order parameter be inching $m=0$; however, if we take a look at the same simulation at $T=3.6T_c$

http://farside.ph.utexas.edu/teaching/329/lectures/node110.html

The spins become heavily biased toward spin-up! Very beautiful! We expect the same behavior if we were to bias the simulation by increasing the magnetic field while keeping the temperature constant, for our self-consistency equation tells a story of a fine balance between temperature and the introduction of an incident magnetic field.

Wonderful! We have now thoroughly scratched the surface of ferromagnetic materials using the theoretical tools of statistical mechanics! I cannot lie to my reader, however, and assert that we have probed the full beauty of ferromagnetism. For a matter of fact, we have only considered a semi-classical approximation to the properties of ferromagnets. A much more rigorous treatment of ferromagnetism must come from the operator formalism of quantum mechanics and the broader subject of quantum field theory. This is, however, still subject to much research and hence I leave the reader with a suggestion of going further and exploring more complicated models of ferromagnetism now that you have a subtle basis of what is one of the less complicated models out there (I must note, however, that the Ising model is still subject to a high degree of research, as, though it is simple in the mean field theory, it still does not have any non-approximate analytical solutions in higher dimensions).

REFERENCES
[1] Shankar, R. (2014). Principles of Quantum Mechanics. New York, NY: Springer.
[2] Griffiths, D. J. (2017). Introduction to Quantum Mechanics. Cambridge: Cambridge University Press.
[3] Van Vleck, J. H.: Electric and Magnetic Susceptibilities, Oxford, Clarendon Press, p. 318 (1932).
[4] Kopietz, P., Bartosch, L., & Schütz, F. (2013). Introduction to the Functional Renormalization Group. Berlin: Springer Berlin.
[5] Kittel, C., & Kroemer, H. (1998). Thermal physics. New York: W.H. Freeman and Company.
[6] Griffiths, D. J. (1999). Introduction to Electrodynamics. Upper Saddle River: Prentice Hall.
[7] Schwartz, M. (2005). Principles of Electrodynamics. New York: Dover Publications.
[8] Reif, Frederick. Fundamentals of Statistical and Thermal Physics. Calcuta: Levant, 2010. Print.

## Sunday, June 11, 2017

### An Explanation of Paramagnetism from the Framework of Statistical Mechanics

Here, we discuss a beautifully straight-forward way of calculating the effects of an incident magnetic field on a substance composed of atoms with a net spin of $s=\frac{1}{2}$ from the framework of statistical mechanics. As a side note on notation, most derivations will use the notation for the magnetic field being as being $\mathbf{H}$; however, I am not fond of this misuse of notation, for the "true" magnetic field is $\mathbf{B}$, $\mathbf{H}$ being merely a measure of the magnetic field caused by free currents, called the auxiliary field (which does not have the same units of $\mathbf{B}$). Now, consider a system composed of a of weakly interacting atoms spin-1/2 atoms (i.e. the spin-spin interactions of each atom can be neglected) at some temperature T. Then we can consider the probability of a single atom occupying a particular energy state to be described by the Boltzmann distribution: $$p(E_r,T)=\frac{\exp(-\beta E_r)}{\sum_r \exp(-\beta E_r)}$$ which we can obtain from a Taylor expansion of the natural log of the number of accessible states of the entire system of atoms while using the fundamental relations $\frac{\partial S}{\partial E}=\frac{1}{kT}=\beta$, where k is the Boltzmann constant, T is the temperature, and S is the entropy. $E_r$ denotes the energy of the atom in a state $r$, which is either spin-up or spin-down. Recall from classical electrodynamics that the energy (or Hamiltonian in our notation) of a dipole interacting with an external field is: $$H=-\mu B$$ where $\mu$ is the magnetic moment. Generally, this is an inner product, but we are assuming the $\mathbf{B}$ is uniform in magnitude a direction, so $\mu$ points along the same direction as $\mathbf{B}$. Since the atom either occupies a spin-up state or spin-down state, the two possible energies are either positive or negative, corresponding to the magnetic moment being anti-aligned with the external field or aligned with the external field, respectively. Hence, the probability of being spin-up is: $$p_{+}(E_r,T)=\frac{\exp(\beta \mu B)}{\exp(-\mu B)+\exp(\mu B)}$$ while spin-down is: $$p_{-}(E_r,T)=\frac{\exp(-\beta \mu B)}{\exp(-\mu B)+\exp(\mu B)}$$ In the regime of relatively low temperatures: $$\bigg(\frac{\exp(\beta \mu B)}{\exp(-\mu B)+\exp(\mu B)}\bigg )>\bigg( \frac{\exp(-\beta \mu B)}{\exp(-\mu B)+\exp(\mu B)}\bigg)$$ and thus: $$p_{+} > p_{-}$$ so the probability of the magnetic moment being aligned with the magnetic field is greater than being anti-aligned. However, if the temperature is high, then $\beta \implies 0$, so $$p_{+} \approx p_{-}$$ This behavior of the probabilities in the high temperature regime can be attributed to the random thermal motion of the atoms, for the higher the temperature, the higher the average kinetic energy of each atom and hence the higher the resistance of the atoms to being aligned in a particular direction of spin as dictated the direction of the incident uniform magnetic field. Therefore, the preference of the magnetic moment to align with the magnetic field is fundamentally an effect of statistical mechanics! We can, however, do even better than simply calculating the probabilities of a particular atom occupying some particular state - we can go as far as calculating the magnetization of the material from the mean magnetic moment. From the definition of the mean: $$\bar{\mu} = \sum_r p_r \mu_r=\mu \frac{\exp(\mu B)-\exp(-\mu B)}{\exp(\mu B)+\exp(-\mu B)}=\mu \tanh\bigg(\frac{\mu B}{kT}\bigg)$$ from the definition of the magnetization (dipole per unit volume), we can let N define the number of atoms per volume. Hence: $$M=N\mu \tanh\bigg(\frac{\mu B}{kT}\bigg)$$ So in the regime of high temperature ($\tanh$ approaches zero): $$M\approx \frac{N\mu^2}{kT}B=\chi B$$ where $\chi$ is the magnetic susceptibility. However, by the shape of hyperbolic tangent, in the regime of low temperatures and/or strong magnetic field, $M$ approaches $M=N \mu$. Very beautiful!

References:
Reif, Frederick. Fundamentals of Statistical and Thermal Physics. Calcuta: Levant, 2010. Print.