Suppose we are given the vector $$|n_1, n_2, \ldots \rangle \in H^{\otimes n}_s$$ where $H^{\otimes n}_s$ is the $n$-fold symmetric tensor product of a Hilbert space $H$, and $|n_1, n_2, \ldots \rangle$ represents the state where $n_i$ particles are in the $i$th state one (of course we require that $\sum n_k$ = n).

The creation and annihilation operators, $A^\dagger$ and $A$, increase/decrease the total number of particles by one. More explicitly, $$A: H^{\otimes n}_s \rightarrow H^{\otimes (n-1)}_s\\ A^\dagger: H^{\otimes n}_s \rightarrow H^{\otimes (n+1)}_s.$$

It is my current understanding that if the set $\{e_i\}$ form a basis for $H$, then for any $k$ we have (omitting any normalization constants): $$A(e_k)| n_1, n_2, \ldots n_k,\ldots\rangle = | n_1, n_2, \ldots n_k-1,\ldots\rangle\\A^\dagger(e_k)| n_1, n_2, \ldots n_k,\ldots\rangle = | n_1, n_2, \ldots n_k+1,\ldots\rangle$$ In other words, when applied to the $k$th basis vector these operators create/annihilate a single particle in the $k$th state.

I am now curious what happens if I apply this operator to any $\psi \in H$, and not just a basis vector. How do the total number of particles still increase/decrease by 1 if $\psi = \sum \psi_i e_i$ is now a linear combination of states? What is the physical interpretation of this?


1 Answer 1


Omitting normalization is not a good idea in this case. The reason is because change in the normalization is a function of the state fed to the operator, so when you operate on a state that is not an eigenstate of the number operators then the balance among the states gets shifted.

The first step is to expand your generic state in the number operator basis: $$|\psi\rangle = \sum_{\{n\}} \psi_{\{n\}} | n_1, n_2, \ldots n_k,\ldots\rangle .$$ We assume that this basis is, for our purposes, complete. If it is incomplete then that means that some of the number states are actually degenerate multiplets, but that's not relevant for this. Then the effect of raising and lowering operators should be obvious \begin{align} A(e_k)|\psi\rangle = \sum_{\{n\}} \sqrt{n_k} \psi_{\{n\}} | n_1, n_2, \ldots n_k - 1,\ldots\rangle \ \text{and}\\ A^\dagger(e_k)|\psi\rangle = \sum_{\{n\}} \sqrt{n_k+1} \psi_{\{n\}} | n_1, n_2, \ldots n_k + 1,\ldots\rangle. \end{align} Because $\psi$ isn't an eigenstate of the number operator, we cannot say that it has a definite number of particles that was increased by 1. We can, however, say that the expectation value of the number operator was increased by at least 1. The reason why the expected number can increase by more than one is because the raising operator also up-weights the high number states.

Here's some of the math behind that. First, we need to compute the normalization constant: $$\mathcal{N}^2 = \langle \psi|A(e_k) A^\dagger(e_k)|\psi\rangle = \sum_{\{n\}} (n_k+1) \psi_{\{n\}}^* \psi_{\{n\}}.$$ Next, we insert a number operator $N_j = A^\dagger(e_j) A^\dagger(e_j)$ to get the expectation: $$\langle \psi|A(e_k) A^\dagger(e_j) A(e_j) A^\dagger(e_k)|\psi\rangle = \mathcal{N}^{-1}\sum_{\{n\}} (n_k+1) (n_j + \delta_{jk}) \psi_{\{n\}}^* \psi_{\{n\}},$$ with $\delta_{jk}$ the Kronecker delta (it equals 1 if $j=k$, 0 otherwise).

Let's now consider a concrete example of a space with two excitation types. Specifically, let's assume that $\psi = \frac{1}{2} e^{-\ln(2)(n_1+n_2) / 2}$, so that each excited state is twice as common as the next higher state. First, let's compute the expected number of particles in each state: \begin{align} \langle n_i\rangle &= \sum_{n_i=0}^\infty n_i \left(\frac{1}{2}\right)^{n_i+1} \\ &= \left[\frac{1}{2}\frac{\partial}{\partial r} \sum_{n_i=0}^\infty r^{n_i}\right]_{r=1/2} \\ &= \left[\frac{1}{2}\frac{\partial}{\partial r} \frac{1}{1-r} \right]_{r=1/2} \\ &= 2 \end{align} Next, our normalization constant when we raise the first number type: \begin{align} \mathcal{N} &= \frac{1}{2}\sqrt{\sum_{\{n\}} (n_k+1) \psi_{\{n\}}^* \psi_{\{n\}}} \\ &= \frac{1}{2}\sqrt{\sum_{n_1=0}^\infty\sum_{n_2=0}^\infty (n_1+1) \left(\frac{1}{2}\right)^{n_1+n_2}} \\ &= \frac{1}{2}\sqrt{\sum_{n_1=0}^\infty (n_1+1) \left(\frac{1}{2}\right)^{n_1}} 2 \\ &= \frac{1}{2}\sqrt{4 + \sum_{n_1=0}^\infty \left.r\frac{\partial r^{n_1}}{\partial r}\right|_{r=1/2} 2} \\ &= \frac{1}{2}\sqrt{4 + \left.r\frac{\partial}{\partial r}\left[\frac{1}{1-r}\right]\right|_{r=1/2} 2} \\ &=\sqrt{2}. \end{align} Next, the expected number of type 1 is \begin{align} \langle \psi|A(e_1) A^\dagger(e_1) A^\dagger(e_1) A^\dagger(e_1)|\psi\rangle &= \frac{1}{\sqrt{8}}\sum_{n_1=0}^\infty\sum_{n_2=0}^\infty (n_1+1)^2 \left(\frac{1}{2}\right)^{n_1+n_2} \\ &= \frac{1}{\sqrt{2}} \left[\frac{\partial}{\partial r} \left(r\frac{\partial}{\partial r}\right) \sum_{n_1=0}^\infty r^{n_1+1} \right]_{r=1/2} \\ &= \frac{12}{\sqrt{2}}, \end{align} which is an increase greater than 1.

Because the states in our example factor, the $n_i$ are independent. If we had chosen a state where they don't factor, for example $\psi \propto \frac{1}{\cosh(2 \ln(2)(n_1+n_2))}$, $\frac{1}{(n_1 + n_2)^4}$, or $\left(n_1 + n_2^{-1}\right) \left(\frac{1}{2}\right)^{(n_1+n_2)/2}$, then the raising operator on type 1 would also affect the expected number of type 2. In fact, raising type 1 could raise or lower type 2, depending on how they correlate in the original state.

Lowering can have even more counter-intuitive results because it destroys the ground state. Consider $|\psi\rangle = \frac{4}{5}|0\rangle + \frac{3}{5}|10\rangle$. The expected number in the base state is $\langle N\rangle = \frac{18}{5}=3.6$. After applying the lowering operator and normalizing, our state becomes \begin{align} \frac{A |\psi\rangle}{\sqrt{\langle \psi|A^\dagger A|\psi\rangle}} &= |9\rangle. \end{align} In other words, the lowering operator increased the expected number for $\psi$ because it annihilated the contribution from the ground state, which had a higher weight in the original sate.

I leave a general proof that raising increases the expected single particle number by 1 or more, and whether anything can be said about the general effects of the lowering operator, as exercises for the reader.

  • $\begingroup$ Thank you very much for your answer! So informally speaking, when the creation and annihilation operators are applied to an arbitrary $|\psi \rangle \in H$ we can say that statistically a particle has been created/destroyed, but we cannot say deterministically which state was increased by 1. Have I understood that correctly? $\endgroup$
    – CBBAM
    Commented Sep 28, 2022 at 5:12
  • $\begingroup$ @CBBAM All states get shifted by 1, but the balance among the states get shifted, too. I may actually be wrong about the expectation value being shifted by exactly 1 because of that, because the high number states see a larger upweighting than the low number states. $\endgroup$ Commented Sep 28, 2022 at 5:14
  • $\begingroup$ Now I am a little confused. If either of these operators creates or destroys more than one particle wouldn't this violate how we defined the operator (namely the codomain is no longer correct). $\endgroup$
    – CBBAM
    Commented Sep 28, 2022 at 5:17
  • $\begingroup$ I suppose another way to frame my question is if the creation operator creates a single particle in the $k$th state when applied to the basis vector $e_k$, in what state is the particle created when the creation operator is applied to a non-basis vector? I hope I have framed that well, sorry if there was any confusion. $\endgroup$
    – CBBAM
    Commented Sep 28, 2022 at 5:19
  • $\begingroup$ @CBBAM the problem you're running into is that you're thinking of the operators as just creating or just annihilating and doing nothing else. The reweighting of states has an effect. It's more complete to thing of $A$ and $A^\dagger$ as a factorization of the number operator that also happens to shift the state by 1. The only reason we can say that the creation operator definitely increases the expected number is because there is no maximum number state. Because there's a minimum number state, the annihilation operator can increase or decrease the expected number. $\endgroup$ Commented Sep 28, 2022 at 6:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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