I am reading through the chapter on second quantization in Advanced Quantum Mechanics by Schwabl. In §1.5.1, the book suggests that because the two orthonormal basis $\{|\lambda\rangle\}_\lambda$ and $\{|i\rangle\}_i$ of the single particle Hilbert space are related by the unitary transformation $$|\lambda\rangle=\sum_i |i\rangle\langle i|\lambda\rangle$$ This means that the basis transformation for the corresponding creation operators of the two basis on the Fock space is $$\hat a_\lambda^+=\sum_i \hat a_i^+\langle i|\lambda\rangle$$ While this intuitively makes sense, I am looking for a rigorous proof of this transformation law from the definition of the creation operators in terms of increasing the occupation number of their corresponding state. I have also seen the following reasoning from Quantum Mechanics by Merzbacher:

Take the single particle state transformation law and rewrite it in terms of the creation operators acting on the vacuum state (which is the same in both basis): $$\hat a_\lambda^+|0\rangle=\sum_i\langle i|\lambda\rangle \hat a_i^+|0\rangle$$ This suggests that the transformation laws for the creation operators is the one mentioned above. However this is far from rigorous because the creation operators are operators on the Fock space and to show that $\hat a_\lambda^+$ and $\sum_i\langle i|\lambda\rangle \hat a_i^+$ are equal, we need to prove that their action on a complete basis of the vector space is the same, not just the vacuum state. Merzbacher addresses this issue in the following way:

We therefore assume the transformation equations to hold beyond the no-particle and one-particle subspaces as operator equations in the entire state vector space of the system of any number of identical particles. This assumption is technical rather than fundamental and determines the form of the ensuing theory and not its physical content.

I don't find this very convincing either. Is there any way to prove the above transformation relations for the creation operators?


4 Answers 4


This is a messy question, because the Fock states over the different bases are different: that is, the state $$ |n_1,n_2,n_3,\ldots\rangle_\lambda $$ means a state with $n_1$ particles in state $|\lambda_1\rangle$, $n_2$ particles in state $|\lambda_2\rangle$, and so on, and this is pretty messy if you try to expand all of those $|\lambda_j\rangle$ states into the $|i\rangle$ basis. However, the structure is a bit easier to understand if you go back to the initial definition, as $$ |\{n\}\rangle_\lambda = \frac{1}{\sqrt{N!}}\sum_{P\in S_N}(\pm1)^P P |\lambda_1\rangle^{\otimes n_1} |\lambda_2\rangle^{\otimes n_2} |\lambda_3\rangle^{\otimes n_3}\cdots|\lambda_k\rangle^{\otimes n_k}, $$ and to the action of $a_{\lambda_1}^+$ on that basis, $$ a_{\lambda_1}^+ |\{n\}\rangle_\lambda = \frac{1}{\sqrt{(N+1)!}}\sum_{P\in S_{N+1}}(\pm1)^P P |\lambda_1\rangle^{\otimes (n_1+1)} |\lambda_2\rangle^{\otimes n_2} |\lambda_3\rangle^{\otimes n_3}\cdots|\lambda_k\rangle^{\otimes n_k}. $$

Is there something we can do to simplify how this transforms to the $|i\rangle$ basis? Well, the only thing that's changed from the previous state, so one thing we should really try to go for is just the identity $$ a_{\lambda_1}^+|\{n\}\rangle_\lambda = \sum_i \langle i|\lambda_1\rangle a_{i}^+|\{n\}\rangle_\lambda, $$ which is a lot more reasonable. To put this into action, we just expand that single added state into the new basis: $$ a_{\lambda_1}^+ |\{n\}\rangle_\lambda = \sum_i \langle i|\lambda_1\rangle \frac{1}{\sqrt{(N+1)!}}\sum_{P\in S_{N+1}}(\pm1)^P P |i\rangle |\lambda_1\rangle^{\otimes n_1} |\lambda_2\rangle^{\otimes n_2} |\lambda_3\rangle^{\otimes n_3}\cdots|\lambda_k\rangle^{\otimes n_k}, $$ so we've done a good deal of the job.

What remains to be done? We want that right-hand side to really match the action of our transformed creation operator, $$ a_{i}^+|\{n\}\rangle_\lambda = \frac{1}{\sqrt{(N+1)!}}\sum_{P\in S_{N+1}}(\pm1)^P P |i\rangle |\lambda_1\rangle^{\otimes n_1} |\lambda_2\rangle^{\otimes n_2} |\lambda_3\rangle^{\otimes n_3}\cdots|\lambda_k\rangle^{\otimes n_k}, $$ and here the easy way out is just to say "hey, we've just put in an extra $|i\rangle$, what else do you need?, which does work but which you might be a bit unhappy with. If you really want to do the grueling details, you need to expand all the states in that tensor product, giving you an extremely messy sum: \begin{align} \frac{1}{\sqrt{(N+1)!}} & \sum_{P\in S_{N+1}}(\pm1)^P P |i\rangle |\lambda_1\rangle^{\otimes n_1} |\lambda_2\rangle^{\otimes n_2} |\lambda_3\rangle^{\otimes n_3}\cdots|\lambda_k\rangle^{\otimes n_k} \\ & = \frac{1}{\sqrt{(N+1)!}}\sum_{P\in S_{N+1}}(\pm1)^P P |i\rangle \left(\sum_{j_1} |i_{j_1}\rangle\langle i_{j_1} |\lambda_1\rangle \right)^{ \otimes n_1} \left(\sum_{j_2} |i_{j_2}\rangle\langle i_{j_2}|\lambda_2\rangle\right)^{ \otimes n_2} \\ & \qquad \otimes \left(\sum_{j_3} |i_{j_3}\rangle\langle i_{j_3}|\lambda_3\rangle\right)^{ \otimes n_3} \cdots \left(\sum_{j_k} |i_{j_k}\rangle\langle i_{j_k}|\lambda_k\rangle\right)^{ \otimes n_k}, \end{align} and if you expand all of those coefficients out you'll run out of ink. However, there's a common thread in that mess, and it's the fact that all the stuff you're adding up is now Fock states over the new basis: \begin{align} \frac{1}{\sqrt{(N+1)!}} & \sum_{P\in S_{N+1}}(\pm1)^P P |i\rangle |\lambda_1\rangle^{\otimes n_1} |\lambda_2\rangle^{\otimes n_2} |\lambda_3\rangle^{\otimes n_3}\cdots|\lambda_k\rangle^{\otimes n_k} \\ & = \sum_\text{mess} \prod_{i',\lambda} \langle i' |\lambda\rangle \times \frac{1}{\sqrt{(N+1)!}}\sum_{P\in S_{N+1}}(\pm1)^P P |i\rangle |i_{1}\rangle^{ \otimes n_1'} |i_{2}\rangle^{ \otimes n_2'} |i_{3}\rangle^{ \otimes n_3'} \cdots |i_{k'}\rangle^{ \otimes n'_{k'}}, \end{align} and in each of those, you've just got $a_i^+$ acting on the new Fock state, \begin{align} \frac{1}{\sqrt{(N+1)!}} & \sum_{P\in S_{N+1}}(\pm1)^P P |i\rangle |\lambda_1\rangle^{\otimes n_1} |\lambda_2\rangle^{\otimes n_2} |\lambda_3\rangle^{\otimes n_3}\cdots|\lambda_k\rangle^{\otimes n_k} \\ & = a_i^+ \sum_\text{mess} \prod_{i',\lambda} \langle i' |\lambda\rangle \times \frac{1}{\sqrt{N!}}\sum_{P\in S_{N+1}}(\pm1)^P P |i_{1}\rangle^{ \otimes n_1'} |i_{2}\rangle^{ \otimes n_2'} |i_{3}\rangle^{ \otimes n_3'} \cdots |i_{k'}\rangle^{ \otimes n'_{k'}}, \end{align} at which stage you can just replace the mess back to how it was, which, because nothing has changed, just collapse back to the product $|\lambda_1\rangle^{\otimes n_1} |\lambda_2\rangle^{\otimes n_2} |\lambda_3\rangle^{\otimes n_3}\cdots|\lambda_k\rangle^{\otimes n_k}$ and therefore to our initial $\lambda$-basis Fock state, \begin{align} \frac{1}{\sqrt{(N+1)!}} & \sum_{P\in S_{N+1}}(\pm1)^P P |i\rangle |\lambda_1\rangle^{\otimes n_1} |\lambda_2\rangle^{\otimes n_2} |\lambda_3\rangle^{\otimes n_3}\cdots|\lambda_k\rangle^{\otimes n_k} \\ & = a_i^+ |\{n\}\rangle_\lambda, \end{align} which is exactly what we wanted to prove. And that also concludes the bigger proof. $$\tag{$\square$}$$

To be honest, I've probably been a bit shoddy with signs and constants at a few points during the proof ─ but because of the structural features, it has to work. The details are left as an exercise for the reader ;-).

  • $\begingroup$ Had the same idea. $\endgroup$
    – WoofDoggy
    Nov 9, 2017 at 20:10

The operator $a_f^\dagger$ adds the (one particle) vector $f$ to whatever state, with the correct symmetry. Completely symmetric, anti-symmetric, for bosons, fermions, respectively. All you have to do is to check that for vectors $f,g$ and scalars $\alpha, \beta$ one has

$$ a^\dagger_{\alpha f + \beta g}= \alpha a^\dagger_{f } + \beta a^\dagger_{g} $$

This can easily be checked using the fifth equation of Emilio Pisanty's answer.

Hence $a^\dagger_f$ is linear in $f$. Interestingly, since $a_f$ is the adjoint of $a_f^\dagger$ it follows that $a_f$ is actually antilinear in $f$.


The following is basically a refinement of Emilio Pisanty's answer.

Denote the single-particle Hilbert space by $\mathcal{H}$, which has an orthonormal basis $\lbrace \left|i\right>\rbrace=\lbrace \left|1\right>,\left|2\right>,\cdots \rbrace$. We know that $\lbrace\left|i\right>\rbrace$ induces an orthonormal basis of the Fock space, namely $$\left| n_1 n_2 \cdots \right>=\sqrt{\frac{N!}{\prod_i n_i!}}\hat\Pi_\zeta\Big(\underbrace{\left|1\right>\cdots\left|1\right>}_{n_1}\underbrace{\left|2\right>\cdots\left|2\right>}_{n_2}\cdots\Big),$$ where $$\hat\Pi_\zeta\big(\left|i_1\right>\cdots\left|i_N\right>\big)=\frac1{N!}\sum_{p\in S_N}\zeta^p\left|i_{p(1)}\right>\cdots\left|i_{p(N)}\right>$$ is the projection operator which maps a vector in $\mathcal{H}^{\otimes N}$ to its symmetric (asymmetric) part in the eigensubspace of exchange operators with eigenvalue $\zeta\pod{\zeta=\pm 1}$.

Now we consider another orthonormal basis $\lbrace\left|\lambda_i\right>\rbrace=\lbrace\left|\lambda_1\right>,\left|\lambda_2\right>,\cdots\rbrace$, choose an arbitrary $\left|n_{\lambda_1}n_{\lambda_2}\cdots\right>$, and denote $c_{n_1n_2\cdots}=\left<n_1n_2\cdots\middle|n_{\lambda_1}n_{\lambda_2}\cdots\right>$. By the definition of $a^\dagger_{\lambda_1}$, we have $$\begin{aligned} a^\dagger_{\lambda_1}\left|n_{\lambda_1}n_{\lambda_2}\cdots\right> &=\sqrt{n_{\lambda_1}+1}\left|n_{\lambda_1}+1,n_{\lambda_2},\cdots\right>\\ &=\sqrt{n_{\lambda_1}+1}\sqrt{\frac{(N+1)!}{(n_{\lambda_1}+1)!n_{\lambda_2}!\cdots}} \hat{\Pi}_\zeta \Big(\left| {\lambda_1} \right>\underbrace{\left|{\lambda_1}\right>\cdots\left|{\lambda_1}\right>}_{n_{\lambda_1}} \underbrace{\left|{\lambda_2}\right>\cdots\left|{\lambda_2}\right>}_{n_{\lambda_2}}\cdots\Big)\\ &=\sqrt{n_{\lambda_1}+1}\sqrt{\frac{(N+1)!}{(n_{\lambda_1}+1)!n_{\lambda_2}!\cdots}}\hat{\Pi}_\zeta \Big(\sum_{i}\left< i|\lambda_1 \right>\left| i \right>\hat{\Pi}_\zeta\big(\underbrace{\left|{\lambda_1}\right>\cdots\left|{\lambda_1}\right>}_{n_{\lambda_1}} \underbrace{\left|{\lambda_2}\right>\cdots\left|{\lambda_2}\right>}_{n_{\lambda_2}}\cdots\big)\Big)\\ &=\sqrt{N+1}\sum_{i}\left< i|\lambda_1 \right>\hat{\Pi}_\zeta \big(\left| i \right>\left| n_{\lambda_1}n_{\lambda_2}\cdots \right>\big)\\ &=\sqrt{N+1}\sum_{i}\left< i|\lambda_1 \right>\sum_{n_1,n_2,\cdots}c_{n_1n_2\cdots}\,\hat{\Pi}_\zeta \big(\left| i \right>\left| n_1 n_2 \cdots \right>\big)\\ &=\sqrt{N+1}\sum_{i}\left< i|\lambda_1 \right>\sum_{n_1,n_2,\cdots}c_{n_1n_2\cdots}\sqrt{\frac{N!}{n_1!n_2!\cdots}} \hat{\Pi}_\zeta \Big(\left| i \right> \hat{\Pi}_\zeta\big(\underbrace{\left|{1}\right>\cdots\left|{1}\right>}_{n_{1}} \underbrace{\left|{2}\right>\cdots\left|{2}\right>}_{n_{2}}\cdots\big)\Big)\\ &=\sum_{i}\left< i|\lambda_1 \right>\sum_{n_1,n_2,\cdots}c_{n_1n_2\cdots}\zeta^{\sum_{j=1}^{i-1}n_i}\sqrt{n_i+1}\left| n_1\cdots,n_i+1,\cdots \right>\\ &=\sum_{i}\left< i|\lambda_1 \right>\sum_{n_1,n_2,\cdots}c_{n_1n_2\cdots}a_i^\dagger \left| n_1\cdots n_i\cdots \right>\\ &=\sum_{i}\left< i|\lambda_1 \right>a_i^\dagger \left| n_{\lambda_1}n_{\lambda_2}\cdots \right> ,\end{aligned} $$ and the proof is done. Note that in the third equality we have used the property: $$\hat\Pi_\zeta\Big(\left|\psi\right>\left|\phi\right>\Big)=\hat\Pi_\zeta\Big(\left|\psi\right>\hat\Pi_\zeta\left|\phi\right>\Big).$$


All you need is to show the change of basis formula for the creation operator, which follows by linearity. Linearity can be seen using this definition in wikipedia (note that the insertion operator is well defined, i.e. if you choose any nonzero state $|v\rangle$, not necessarily normalized, we can define an insertion operator as in wikipedia; this insertion operator and the resulting creation operator is also clearly independent of the way you may extend $|v\rangle$ into a basis):


The equivalence can be easily checked in both bosonic and fermionic cases.


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