I'm self-learning many body theory and right now I'm trying to solve Problem 1.3 from Quantum Theory of Many-Particle Systems by Fetter and Walecka.


Given a homogeneous system of a spin-zero particles interacting through a potential V show that the expectation value of the hamiltonian in the noninteracting ground state is $$ \frac{E^{(1)}}{N} = \frac{(N − 1)V(\vec{0})}{2V}$$ where $$ V(\vec q)=\int d_3x V(\vec{x})e^{-i\vec{q}\cdot\vec{x}} $$ Then show that the second-order contribution to the ground-state energy is $$ \frac{E^{(2)}}{N} = -\frac{(N − 1)}{2V} \int \frac{d^3q}{(2\pi)^3} \frac{m|V(\vec q)|^2}{\hbar^2\vec q^2}$$ using second order perturbation theory $$E^{(2)}=\sum_{n\neq0}\frac{\langle0|H_1|n\rangle\langle n|H_1|0\rangle}{E_0-E_n}$$ where $|n\rangle$ is the excited state and $|0\rangle$ is the ground state.

My try: The kinetic operator is $$ T=\sum_{m \vec k} \frac{ \hbar ^2k^2 }{2m} n_{m \vec k} = \sum_{\vec k} \frac{ \hbar ^2k^2 }{2m} n_{\vec k} $$

The sum over m goes away since spinless particles only have 1 possible m. Moreover since they are bosons at $T=0K$ they all are in the lowest energy state.

I managed to write the interaction hamiltonian as $$ H_1 = \sum_{m m' \vec k \vec p\vec q} \frac{ 1}{V} V(\vec q) c_{\vec p m}^\dagger c_{\vec k m'}^\dagger c_{\vec k + \vec q m'}c_{\vec p - \vec q m} $$

First order perturbation theory says that $$ E^{(1)} = \langle 0 |H_1|0 \rangle = \langle 0 |\sum_{m m' \vec k \vec p\vec q} \frac{ 1}{V} V(\vec q) c_{\vec p m}^\dagger c_{\vec k m'}^\dagger c_{\vec k + \vec q m'}c_{\vec p - \vec q m} |0\rangle$$

Now I act right with the destructors and left with the creators, which will act as destructors: I need to create holes where there is an occupied state, so I can only do it at the lowest energy. That means that $ m = m', \vec p = \vec k = \vec 0 $ and $\vec q = 0$. I get: $$\langle 0|\sum_{ \vec 0} \frac{ 1}{V} V(\vec 0) c_{\vec 0}^\dagger c_{\vec 0}^\dagger c_{\vec 0} c_{\vec 0} |0\rangle$$

I use commutation rules to pair up the operators to get $N^2 - N$, finding finally $$ \frac{E^{(1)}}{N} = \frac{(N − 1)V(\vec{0})}{V}$$

which lacks the $1/2$, where does that come from? Is there another double counting?

Now regarding the second order energy I need to evaluate the matrix element $$ \langle 0|H_1|n \rangle = \langle 0 |\sum_{m m' \vec k \vec p\vec q} \frac{ 1}{V} V(\vec q) c_{\vec p m}^\dagger c_{\vec k m'}^\dagger c_{\vec k + \vec q m'}c_{\vec p - \vec q m} |n\rangle $$

I know I need to transform the sum into an integral but before that I should probably find a way to eliminate the sum over $\vec k$ and $\vec p$ through some considerations on the allowed creations/destructions, but I can't manage to do so.

Edit I try to work it all out following the suggestion in the answer, because I can't manage to obtain the right result.

So I evaluate this matrix element

$$ \langle n|H_1|0 \rangle = \langle n|\sum_{m m' \vec k \vec p\vec q} \frac{ 1}{V} V(\vec q) c_{\vec p m}^\dagger c_{\vec k m'}^\dagger c_{\vec k + \vec q m'}c_{\vec p - \vec q m} |0\rangle $$

first operating with the destructors one the right, getting $\delta_{\vec k + \vec q, 0}\delta_{\vec p - \vec q, 0}\delta_{m, m'}\sqrt{N(N -1)}$, then with the creators on the right, which now have became $c_{\vec q m}^\dagger c_{-\vec q m}^\dagger$, getting $ \sqrt{N(N -1)} $. Thus

$$ \langle n|H_1|0 \rangle = \sum_{\vec q} \frac{ 1}{V} V(\vec q) N(N-1)$$

I switch to the integral $$ = \int \frac{d_3q V}{(2\pi)^3} \frac{ 1}{V} V(\vec q) N(N-1)$$ Now I need to do the modulus squared, divide it by the energy difference and put it in the sum. I don't see how this can lead me to the final result, where for example $ (N-1)$ isn't squared. What am I missing?


1 Answer 1


$H_1$ is Hermitian, so it's easier to operate first with the destruction operators on $|0\rangle$. Notice that unless their momenta is zero, they will give zero. So you will find you must destroy two zero momenta particles, and create a pair, one with $q$ and one with $-q$. These are the only $\langle n|$ states that contribute. The energy difference is then $\hbar^2 q^2/m$.

Edit with details

The second order energy term from elementary quantum mechanics is \begin{equation} E^{2} = \sum_{n \neq 0} \frac{\langle 0|H_1|n\rangle \langle n|H_1|0\rangle}{E_0-E_n} = \sum_{n \neq 0} \frac{|\langle n|H_1|0\rangle|^2}{E_0-E_n} \end{equation} The only states $|n\rangle$ that have non zero matrix elements are the normalized states with \begin{equation} |n(\vec q)\rangle = c^\dagger_{\vec q} c^\dagger_{-\vec q} |0'\rangle \end{equation} where $|0'\rangle$ is the normalized state with $N-2$ particles in the zero momentum orbitals. and $\vec q$ not equal to zero. Notice that since $c^\dagger_{-\vec q} c^\dagger_{\vec q}= c^\dagger_{\vec q} c^\dagger_{-\vec q}$, to sum over the $|n(\vec q)\rangle$ states you should either include just one of $|n(\vec q)\rangle$ and $|n(-\vec q)\rangle$ or sum over all $\vec q \neq 0$ and divide by $2$.

The energies are $E_0 = 0$, $E_{n(\vec q)} = 2\left [ \frac{\hbar^2 q^2}{2m}\right ] = \frac{\hbar^2 q^2}{m}$, so the perturbation theory result is \begin{equation} E^{(2)} = -\frac{1}{2}\sum_{\vec q\ne 0} \frac{|\langle n(\vec q)|H_1|0\rangle|^2}{\hbar^2 q^2/m} \end{equation} The matrix element is, as I discussed in the comments, \begin{equation} \langle n(\vec q)|H_1|0\rangle = \frac{\sqrt{N(N-1)}}{V}V(\vec q) \,, \end{equation} where the factor of 2 is canceled since there are two ways to generate the same $|n(q)\rangle$. Putting these together gives \begin{equation} E^{(2)} = -\frac{N(N-1)}{2V^2} \sum_{\vec q \ne 0} \frac{|V(\vec q)|^2}{\hbar^2 q^2/m} = -\frac{N(N-1)}{2V}\int \frac{d^3q}{(2 \pi)^3} \frac{|V(\vec q)|^2}{\hbar^2 q^2/m} \,. \end{equation}

  • $\begingroup$ Not sure I understood, can you show me what are the deltas that come out? $\endgroup$
    – user336294
    Commented Jun 14, 2022 at 11:21
  • $\begingroup$ For $|0\rangle$ the ground state of $N$ particles, $c_{\vec p-\vec q}c_{\vec k+\vec q}|0\rangle =\sqrt{N(N-1)}\delta_{0,\vec k+\vec q}\delta_{0,\vec p-\vec q} |0'\rangle$, where $|0'\rangle$ is the ground state of N-2 particles. The creation operators are now $c^\dagger_{\vec q}$ and $c^\dagger_{-\vec q}$ giving the intermediate state described in the answer. $\endgroup$
    – user200143
    Commented Jun 14, 2022 at 15:15
  • $\begingroup$ I tried to do it with your hint (see edit) but I can't see how to get to the result, what am I missing? $\endgroup$
    – user336294
    Commented Jun 14, 2022 at 18:49
  • $\begingroup$ Only if you create with $q=0$ do you get $\sqrt{N(N-1)}$ from the creation operators. That state is not included in the second order sum. For the unoccupied states the factor is $1$ from the creation operators. Squaring the result gives the $N(N-1)$ factor and $|V(q)|^2$. Careful counting will give the correct result. $\endgroup$
    – user200143
    Commented Jun 15, 2022 at 3:30
  • $\begingroup$ yes right, but what about the modulus square of the integral? And how should I insert the $-\hbar^2q^2/2m $ inside the integral? I'm sorry for being pedantic but this is the first time I do a similar exercise. $\endgroup$
    – user336294
    Commented Jun 15, 2022 at 12:43

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