Homework Set 2

Problem 1: Pair correlation function of noninteracting spinless bosons

The pair correlation function gives the relative probability of finding a particle at position ${\mathbf{𝐫}}^{\prime }$ if we know that there is one at position $\mathbf{𝐫}$. It is defined by:

$G(\mathbf{𝐫}-{\mathbf{𝐫}}^{\prime })=⟨\mathrm{\Phi }(\mathbf{𝐫})|{\hat{\mathrm{\Psi }}}^{†}({\mathbf{𝐫}}^{\prime })\hat{\mathrm{\Psi }}({\mathbf{𝐫}}^{\prime })|\mathrm{\Phi }(\mathbf{𝐫})⟩=⟨{\mathrm{\Phi }}_0|{\hat{\mathrm{\Psi }}}^{†}(\mathbf{𝐫}){\hat{\mathrm{\Psi }}}^{†}({\mathbf{𝐫}}^{\prime })\hat{\mathrm{\Psi }}({\mathbf{𝐫}}^{\prime })\hat{\mathrm{\Psi }}(\mathbf{𝐫})|{\mathrm{\Phi }}_0⟩$

Here $|\mathrm{\Phi }(\mathbf{𝐫})⟩=\hat{\mathrm{\Psi }}(\mathbf{𝐫})|{\mathrm{\Phi }}_0⟩$ is a state with $N-1$ particles, obtained after removal of one particle at position $\mathbf{𝐫}$, so that pair correlation function is computed as the expectation value of the density operator in this new quantum state. Compute the pair correlation function for a system of translationally invariant noninteracting spinless bosons in many-body state

$|{\mathrm{\Phi }}_0⟩=|n_{{\mathbf{𝐤}}_0},n_{{\mathbf{𝐤}}_1},\dots ⟩$

by transforming field operators to creation and annihilation operators in momentum representation while assuming that bosons are enclosed in a box of volume $V$ with periodic boundary conditions. Your final result should be a function of $\mathbf{𝐫},{\mathbf{𝐫}}^{\prime }$ and sums over $n_{\mathbf{𝐤}}$, where you can also use that

$⟨{\mathrm{\Phi }}_0|{\hat{\mathrm{\Psi }}}^{†}(\mathbf{𝐫})\hat{\mathrm{\Psi }}(\mathbf{𝐫})|{\mathrm{\Phi }}_0⟩=\frac{1}{V}\sum _{\mathbf{𝐤}}{n}_{\mathbf{𝐤}}=N$.

Problem 2: Magnons in one-dimensional Heisenberg model

Consider the low-energy excitations (magnons) above the ground state of a one-dimensional spin-S ferromagnet described by the isotropic Heisenberg model ($J>0$):

$\hat{H}=-\frac{J}{{\hslash }^2}\sum _{n=1}^N{\widehat{\mathbf{𝐒}}}_n\cdot {\widehat{\mathbf{𝐒}}}_{n+1}=-\frac{J}{2{\hslash }^2}\sum _{n=1}^N({\hat{S}}_n^{+}\cdot {\hat{S}}_{n+1}^{-}+{\hat{S}}_n^{-}\cdot {\hat{S}}_{n+1}^{+}+2{\hat{S}}_n^z\cdot {\hat{S}}_{n+1}^z)$

The periodic boundary conditions, ${\widehat{\mathbf{𝐒}}}_{n+1}={\widehat{\mathbf{𝐒}}}_1$ and ${\widehat{\mathbf{𝐒}}}_N={\widehat{\mathbf{𝐒}}}_0$, are imposed on spin operators.

(a) Apply the Holstein-Primakoff transformation, in the approximation where the density of magnons is small so that

${\hat{S}}_n^{+}\approx \hslash \sqrt{2S}{\hat{a}}_n^{†}$

${\hat{S}}_n^{-}\approx \hslash \sqrt{2S}{\hat{a}}_n$

${\hat{S}}_n^z=\hslash (S-{\hat{a}}_n^{†}{\hat{a}}_n)$

and then expand the Hamiltonian above to the quadratic order in boson operators.

(b) Using the following Fourier transform

${\hat{a}}_n=\frac{1}{\sqrt{N}}\sum _k{e}^{ikna}{\hat{a}}_k$

diagonalize the approximative Hamiltonian you obtained in (a) to find the magnon energy-momentum dispersion $\hslash {\omega }_k$ in terms of $J,S,a$ parameters. Here $k=2\pi n/Na$ with $a$ as the lattice constant and $n=0,\pm 1,\dots ,(N-1)/2,\pm N/2$.

(c) What is the total number of such non-interacting magnons at temperature ? You should simply write the integral expression without fully evaluating it.

NOTE: Useful formula from Fourier analysis: $\frac{1}{N}\sum _{n=1}^N{e}^{i(k^{\prime }-k)na}={\delta }_{k{k}^{\prime }}$.

Problem 3: Roton excitation in a superfluid

A real superfluid is more complicated than a weakly interacting boson gas because the interaction is strong and can extend over a finite range instead of being just a delta function assumed in the class. Suppose that interaction takes the form of a square function

$V(\mathbf{𝐫}-{\mathbf{𝐫}}^{\prime })=U\mathrm{r}\mathrm{e}\mathrm{c}\mathrm{t}\left(\frac{x-x^{\prime }}{2r_0}\right)\left(\frac{y-y^{\prime }}{2r_0}\right)\left(\frac{z-z^{\prime }}{2r_0}\right)$

where $\mathrm{r}\mathrm{e}\mathrm{c}\mathrm{t}(t)=1$ when $|t|<1/2$ and $\mathrm{r}\mathrm{e}\mathrm{c}\mathrm{t}(t)=0$ when $|t|>1/2$. The chosen interaction is highly anisotropic to make the calculations below tractable, but the qualitative features of the result will remain the same if we choose a more realistic isotropic form of the interaction potential energy.

(a) Write the second quantized Hamiltonian in terms of operator anihilating bosons in the eigenstates of momentum operator

${\hat{b}}_{\mathbf{𝐤}}=\frac{1}{\sqrt{V}}\int d^{3}r{e}^{-i\mathbf{𝐤}\cdot \mathbf{𝐫}}\hat{\mathrm{\Psi }}(\mathbf{𝐫})$

and its Hermitian conjugate ${\hat{b}}_{\mathbf{𝐤}}^{†}$. Use the relation

$\underset{-\mathrm{\infty }}{\overset{+\mathrm{\infty }}{\int }}\mathrm{r}\mathrm{e}\mathrm{c}\mathrm{t}(t)e^{-i2\pi ft}dt=\frac{\sin (\pi f)}{\pi f}=\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{c}(\pi f)$

(b) Close to the ground state the interaction terms is dominated by contributions involving ${\hat{b}}_{\mathbf{𝐤}=0}$. Suppose that in the ground state ${\hat{b}}_{\mathbf{𝐤}=0}=\gamma$. As long as $|\gamma |\gg 1$, you can treat ${\hat{b}}_{\mathbf{𝐤}=0}$ and ${\hat{b}}_{\mathbf{𝐤}=0}^{†}$ as numbers $\gamma$ and ${\gamma }^{*}$, respectively, while neglecting interaction contributions containing four nonzero momenta since they are much smaller. This approximation will allow you to reduce the Hamiltonian to a quadratic form, as done in the class.

(c) Employ Bogoliubov transformation

${\hat{\alpha }}_{\mathbf{𝐤}}=u_{\mathbf{𝐤}}{\hat{b}}_{\mathbf{𝐤}}+v_{\mathbf{𝐤}}{\hat{b}}_{-\mathbf{𝐤}}^{†}$

where each ${\hat{\alpha }}_{\mathbf{𝐤}}$ represents a new boson mode so that

$[{\hat{\alpha }}_{\mathbf{𝐤}},{\hat{\alpha }}_{{\mathbf{𝐤}}^{\prime }}^{†}]={\delta }_{\mathbf{𝐤},{\mathbf{𝐤}}^{\prime }};[{\hat{\alpha }}_{\mathbf{𝐤}},{\hat{\alpha }}_{{\mathbf{𝐤}}^{\prime }}]=0$

This transformation should allow you to diagonalize the Hamiltonian from (b) by writing it in the form

$\hat{H}=\sum _{\mathbf{𝐤}\ne 0}{E}_{\mathbf{𝐤}}{\hat{\alpha }}_{\mathbf{𝐤}}^{†}{\hat{\alpha }}_{\mathbf{𝐤}}+E_0$. How does eigenenergy $E_{\mathbf{𝐤}}$ depend on $|\mathbf{𝐤}|$?

(d) Plot $E_{\mathbf{𝐤}}$ along the $k_x$ direction for small $|k_x|$ and show that schematically it looks like the Figure below:

The excitations near $\mathbf{𝐤}=0$ have a linear dispersion and are called phonons in the superfluid, while excitations near $|\mathbf{𝐤}|=k_0$ have quadratic dispersion and are called rotons in the superfluid.

Problem 4: Gross-Pitaevski equation

The dynamics of an ultracold dilute gas of bosonic atoms in an optical lattice can be described by a Bose-Hubbard model where the system parameters are controlled by laser light. The Bose-Hubbard Hamiltonian is given by

$\hat{H}=-t\sum _{⟨\mathbf{𝐫},{\mathbf{𝐫}}^{\prime }⟩}({\hat{b}}_{\mathbf{𝐫}}^{†}{\hat{b}}_{{\mathbf{𝐫}}^{\prime }}+{\hat{b}}_{{\mathbf{𝐫}}^{\prime }}^{†}{\hat{b}}_{\mathbf{𝐫}})+\frac{U}{2}\sum _{\mathbf{𝐫}}{\hat{n}}_{\mathbf{𝐫}}({\hat{n}}_{\mathbf{𝐫}}-1)-\mu \sum _{\mathbf{𝐫}}{\hat{n}}_{\mathbf{𝐫}}$

where $⟨\mathbf{𝐫},{\mathbf{𝐫}}^{\prime }⟩$ denotes nearest neighbor (NN) pairs. The first term is a kinetic energy term which describes hopping of bosonic atoms between NN sites; the second term describes the interaction energy between bosons on the same site (note that energy is zero when there is no or only one boson); and the third term describes chemical potential. The action of the first and second terms are illustrated in the Figure below:

Thus, the kinetic energy term competes with the interaction term and the ground state can be different in different phases depending on the choice of parameters. In the case of superfluid ground state background with a uniform order parameter, we can study excitations by using the so-called Gross-Pitaevskii (GP) equation.

(a) Using the Heisenberg equation of motion for the annihilation operator

$i\hslash \frac{\partial }{\partial t}{\hat{b}}_{\mathbf{𝐫}}(t)=[{\hat{b}}_{\mathbf{𝐫}}(t),\hat{H}]$

evaluate its right hand side.

(b) Write the Heisenberg equation of motion above, with evaluated right hand side, in the continuum limit. In the superfluid phase near ground state, ${\hat{b}}_{\mathbf{𝐫}}(t)\approx ⟨{\hat{b}}_{\mathbf{𝐫}}(t)⟩=\mathrm{\Phi }(\mathbf{𝐫},t)$ can be treated as numbers, where $\mathrm{\Phi }(\mathbf{𝐫},t)$ is classical position- and time-dependent field. Show that equation of motion for $\mathrm{\Phi }(\mathbf{𝐫},t)$ is the GP equation ($a$ is the lattice spacing):

$i\hslash \frac{\partial }{\partial t}\mathrm{\Phi }(\mathbf{𝐫},t)=[-a^{2}t(\frac{{\partial }^2}{\partial x^2}+\frac{{\partial }^2}{\partial y^2})-(\mu +U/2+4t)+U|\mathrm{\Phi }(\mathbf{𝐫},t){|}^2]\mathrm{\Phi }(\mathbf{𝐫},t)$

which resembles Schrodinger equation, but this is deceptive since GP equation is nonlinear.

(c) If we are only interested in the low energy excitation, which corresponds to smooth small variations on top of a uniform background, we can assume that the solution takes the form of the GP equation takes the form:

$\mathrm{\Phi }(\mathbf{𝐫},t)={\mathrm{\Phi }}_0+u(\mathbf{𝐫})e^{-i\omega t}+v(\mathbf{𝐫})e^{i\omega t}$

where the variation on the top of the uniform background takes a plane wave form and the variation is much smaller than the background, $|u|,|v|\ll |{\mathrm{\Phi }}_0|$. Note that if GP equation had been linear, each plane wave component would be independent of each other. However, due to the nonlinear term in the GP equation, components with opposite frequencies, $\omega$ and $-\omega$, mix with each other so they need to be combined when solving for eigenmodes. Plug this ansatz into the GP equation and by retaining only terms to zeroth order in $u$ and $v$ find the mean-field solution for $|{\mathrm{\Phi }}_0{|}^2$.

(d) By retaining the first order terms in $u$ and $v$ in (c), and by assuming that $u$ and $v$ contain only momentum $\mathbf{𝐤}$ and $-\mathbf{𝐤}$ as plane wave solutions, compute the dispersion $\omega (\mathbf{𝐤})$.