# Can a ladder operator have imaginary expectation value?

I was recently taught about the expectation values of observables (Hermitian operators), namely that they are always real. I am wondering about non-Hermitian operators that have complex expectation values (and I know that that notion does not make "physical sense", but never mind). For example, there are the ladder operators $$\hat a$$ and $$\hat a^\dagger$$ which are not Hermitian, the raising and lowering operator respectively. If we have a superposition of eigenstates of the harmonic oscillator, say $$\psi=\sum_{n=1}^k\psi_n$$, then $$\hat a|\psi\rangle=\sum\hat a |\psi_n\rangle=\sum\sqrt{n+1}|\psi_{n+1}\rangle$$. Then $$\langle\psi|\hat a|\psi\rangle=\sum\sum\sqrt{n+1}\langle\psi_m|\psi_{n+1}\rangle=\sum_{n=2}^k\sqrt{n+1}$$, if I am not mistaken. But this is a very "real" quantity, just a sum of square roots of some integers.
Does that mean the raising operator observed, even if it isn't Hermitian? I assume this is not the case, since I have seen other posts on this website mention that the expectation value can in fact be imaginary. But based on what I did above, I don't see how this is possible. Did I make a mistake somewhere?

Any observable quantity is a real number, which means that it corresponds to a real eigenvalue. This is why they are described by Hermitian operators.

Real number here means that it is not an imaginary number, not that it is "real" in some other philosophical or mathematical sense.

• Yes, I believe I understand that part. What I'm wondering is whether the fact that the expectation is a real value means that in this case, it can be associated to a physical quantity that is measurable. Sep 28, 2020 at 11:39
• A measurable quantity must be a real number, but it doesn't mean that every real number has to be measurable and/or correspond to an actual physical quantity. Sep 28, 2020 at 11:51
• Again, I understand that it's not an if and only if relation. This doesn't really answer my question, which is: is it possible for it to be a measurable quantity, even if it's not Hermitian? You're only telling me that it's not necessarily a measurable quantity. Sep 28, 2020 at 11:54
• No, because the measurable quantities are described by Hermitian operators. However, $a^ \dagger + a$ or $i(a^\dagger - a)$ may correspond to measurable quantities. Sep 28, 2020 at 11:58
• If you accept an answer you can also upvote it. Sep 28, 2020 at 12:56

In your example, if you had considered the non-Hermitian operator $$i\hat a$$, you would have obtained imaginary expectation value: $$\left<\psi\right|i\hat a\left|\psi\right> = i\sum_{j=2}^k\sqrt{j+1}$$

• I agree, but at that point it's no longer the ladder operator I was talking about. Am I missing something here? Sep 28, 2020 at 11:40
• @QuaereVerum $i\hat a$ is still a ladder operator. Sep 28, 2020 at 12:39
• You said that you don't see how it's possible to have an imaginary expectation value, I just wanted to give you a simple example
– Emmy
Sep 28, 2020 at 18:39

Yes it is completely possible for ladder operator to have complex expectation values. The coherent state \begin{align} \vert\alpha\rangle = e^{-\vert\alpha\vert^2/2} \sum_{n=0}^{\infty}\frac{\alpha^n}{\sqrt{n!}}\vert 0\rangle \end{align} is in fact an eigenstate of the ladder operator $$\hat a$$ with eigenvalue $$\alpha$$, which can be any complex number. Thus, \begin{align} \langle \alpha \vert \hat a \vert \alpha\rangle =\alpha\, ,\qquad \qquad \alpha \in \mathbb{C}\, . \end{align}

Ladder operators as usually not hermitian (or self-adjoint) so none of the usual properties of the hermitian operators are valid. They are thus not observables. In particular, they can have complex expectations values and even complex eigenvalues, and their eigenstates (when they exist) aren’t orthogonal: \begin{align} \langle \beta \vert\alpha\rangle =e^{-(\vert \beta\beta^2+\vert\alpha\vert^2 -2\beta^*\alpha)/2}\ne \delta(\alpha-\beta)\, . \end{align}