3
$\begingroup$

Usually, in second quantization formalism, we are dealing with hamiltonians that are certain functions of the creation and annihilation operators.

I have been given a particular state and I want to build an Hamiltonian, or more generally an operator, that is diagonal in said basis. So I want an Hamiltonian of the type \begin{equation} \hat{H}=\sum_i c_i|v_i\rangle\langle v_i| \end{equation} and I want to see how it looks like as a function of the creation/annihilation operators. I know how my states $|v_i\rangle$ look like (in the Fock basis representation they are $|v_i\rangle=\hat{a}_i \sum_{n_1,n_1,\dots} f(n_1)f(n_2)\dots|2n_1,2n_2,\dots\rangle$) and I can substitute them into the Hamiltonian to get the representation of the Hamiltonian in the Fock basis. \begin{equation} \hat{H}=\sum_ic_i\hat{a}_i \sum_{n_1,n_1,\dots,m_1,m_2,\dots} f(n_1)f(n_2)\dots f(n_1)f(n_2)\dots|2m_1,2m_2,\dots\rangle\langle2m_1,2m_2,\dots|\hat{a}^\dagger_i \end{equation} and then I could write $|2n_1,2n_2,\dots\rangle$ as $(a_1^\dagger)^{n_1}(a_2^\dagger)^{n_2}\dots|0,0,\dots\rangle$.

It may be a stupid question but how do I ''get rid'' of the vacuum states $|0,0,\dots\rangle\langle 0,0,\dots|$ or can I even do that? I know that when we have a first quantization Hamiltonian and we want to ''upgrade'' it to the second quantization form we sum over all the particles and over all the states and then we use a completeness relation to get rid of the bra\kets but it doesn't look like I can do it here so I was wondering if I can even ger rid of it.

$\endgroup$
3
  • 1
    $\begingroup$ The question looks to me a bit too general to give a clear answer. Perhaps, you could say more about the properties of states in question. One thing one could do is choose one of these states as a new vacuum and count all the excitations from it. $\endgroup$ – Roger Vadim Feb 22 at 12:35
  • $\begingroup$ So my states are $|v_i \rangle = \hat{a}_i |\psi|rangle$. They are shown to be orthogonal states, i.e. $\langle v_i| v_j \rangle = \delta_{i,j}$. Are you suggesting to choose $|\psi \rangle$ as a new vacuum? (I had previously done that btw but I thought that creating an excitation $v_i$ out of it was not consistent with the definition of creation\annihilation operators) $\endgroup$ – Luthien Feb 22 at 22:56
  • $\begingroup$ Are they all just non-interacting one particle states? Why writing the Hamiltonian as a sum of number operators doesn't work? $\endgroup$ – Roger Vadim Feb 23 at 5:43
4
+50
$\begingroup$

Let's consider a much simple case: a single harmonic oscillator, where the Fock states are simply $|n \rangle$ where $n \geq 0$. Your original Hamiltonian has the form $$H_{\text{int}} = \sum_n c_n |n \rangle \langle n |.$$ We can of course rewrite it in the form $$H_{\text{int}} = \sum_n \frac{c_n}{n!} (a^\dagger)^n |0 \rangle \langle 0| a^n$$ where I'm using a slightly different normalization from you, the standard one for the harmonic oscillator. But this isn't what we want; in second quantized notation the Hamiltonian should be written purely in terms of creation and annihilation operators. For example, if we want a normal ordered Hamiltonian, $$H_{\text{int}} = \sum_n b_n (a^\dagger)^n a^n.$$ We can't just "erase the $|0 \rangle \langle 0|$" to conclude $b_n = c_n / n!$, because that would not give the same Hamiltonian. However, the coefficients $a_n$ can be related to the $c_n$ in a more complex way. We have $$c_n = \langle n | H_{\text{int}} |n \rangle = \sum_m b_m \langle n | (a^\dagger)^m a^m | n \rangle = \sum_{m \leq n} \binom{n}{m} b_m.$$ So the second quantized coefficients $b_n$ are the binomial transform of the first quantized coefficients $c_n$, $$b_n = \sum_{m \leq n} \binom{n}{m} (-1)^{n-m} c_m.$$ This expression, suitably generalized for multiple harmonic oscillators, gives the second-quantized coefficients you want.

The reason you don't often see this in textbooks is because it's usually not needed. Most of the time in field theory, we are directly specifying the $b_n$, so there's no need to get it from the $c_n$. At other times, we specify the $c_n$ but only for low $n$, say up to $n = 1$ or $n = 2$, because higher-body interactions are negligible. This fixes the $b_n$ for low $n$, and then we set the higher $b_n$ to zero.


Here's a simple example of this in action. A simple harmonic oscillator has $$c_n = n \hbar \omega.$$ This implies that $$b_n = (-1)^n \hbar \omega \sum_{m \leq n} \binom{n}{m} (-1)^{m} m.$$ Using the defining property of the binomial coefficients, $$\binom{n}{m} = \binom{n-1}{m-1} + \binom{n-1}{m}$$ the sum becomes $$b_n = (-1)^n \hbar \omega \sum_{m \leq n} \left(\binom{n-1}{m-1} + \binom{n-1}{m}\right) (-1)^{m} m = (-1)^n \hbar \omega \sum_{k \leq n-1} \binom{n-1}{k} (-1)^{k+1}.$$ Using the binomial theorem, we have $$b_n = (-1)^{n+1} \hbar \omega (1 - 1)^{n-1} = (-1)^{n+1} \hbar \omega \delta_{n, 1} = \hbar \omega \delta_{n, 1}.$$ As expected, we have $b_1 = \hbar \omega$ with all other coefficients equal to zero.

$\endgroup$
3
  • $\begingroup$ Thank you for your answer knzhou. From what you said, it emerges that for the harmonic oscillator we have $H= \sum_n \sum_{m\leq n} \binom{n}{m}(-1)^{n-m} c_m \hat{a}^{\dagger n} \hat{a}^n$. If I'm not mistaken then we should be able to "work through" the binomial coefficients sum to get the "typical" Hamiltonian of the quantum harmonic oscillator $\hbar \omega (\hat{a}^\dagger \hat{a}+1/2)$. Am I correct? $\endgroup$ – Luthien Feb 22 at 22:35
  • $\begingroup$ @Luthien That's right, I added an explicit demonstration of this. $\endgroup$ – knzhou Feb 23 at 5:45
  • $\begingroup$ Great! Thank you, I had an extensive search before asking the question here and this particular approach is not really addressed in the typical derivation of the second quantized Hamiltonian. $\endgroup$ – Luthien Feb 23 at 11:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.