# Expectation value of a ladder operator

I am going back over old Q.M simple harmonic motion material and, as I can't see an answer on the web, I would like to confirm the validity of an assumption.

Using the ladder operators:

{\displaystyle {\begin{aligned}a&={\sqrt {m\omega \over 2\hbar }}\left({\hat {x}}+{i \over m\omega }{\hat {p}}\right)\\a^{\dagger }&={\sqrt {m\omega \over 2\hbar }}\left({\hat {x}}-{i \over m\omega }{\hat {p}}\right)\end{aligned}}}

My early reading was on the discrete energy levels of potential wells, and the expectation values of, for example $x$, $x^n$ , $p^2$ etc. that can be calculated using these orthoganal energy eigenstates.

I know that I can easily rearrange the above to get $x$ and $p$ in terms of $a$ and $a^{\dagger }$ and that should give me all the aspects of expectation values which I am used to using in 1 D expectation values.

I also know that you can find the expectation value if ${\displaystyle A}$ has a complete set of eigenvectors ${\displaystyle \phi _{j}}$, with eigenvalues ${\displaystyle a_{j}}$.

My question is: does trying to find the expectation value of $\langle \Psi | a |\Psi \rangle$ or $\langle \Psi | a^{\dagger } |\Psi \rangle$ implicity assume that you get $x$ and $p$ in terms of $a$ and $a^{\dagger }$ and then use those expressions in calculation of the expectation values?

Apologies, this is basic stuff but it's been a while and the answer might help someone else. I can see related questions regarding the number operator but if there is a duplicate I will remove this.

EDIT Do these expressions make any physical sense? Thanks to ACuriousMind for this answer below.

As mathematical expressions the "expectation values" of $a$ and $a^†$ are perfectly fine, but they are physically non-sensical since the operators are not self-adjoint and therefore are not observables - you're not computing expectation values because there's no measurement you could expect those values for.

END EDIT

• The position and momentum operators have purely continuous spectrum, and therefore no eigenvectors. I don't see the problem in taking the expectation value as well; as long as you are in the (form) domain of the (closed) operator, the expectation value makes perfect sense as a number that is complex for non-self-adjoint and real for self-adjoint operators. – yuggib Nov 22 '16 at 13:57
• @yuggib Thank you, my wording is probably still bad, but hopefully not as bad as the first draft. – user108787 Nov 22 '16 at 14:20
• I'm...not exactly sure what the question is. As mathematical expressions the "expectation values" of $a$ and $a^\dagger$ are perfectly fine, but they are physically non-sensical since the operators are not self-adjoint and therefore are not observables - you're not computing expectation values because there's no measurement you could expect those values for. – ACuriousMind Nov 22 '16 at 14:27
• – Qmechanic Nov 22 '16 at 15:07
• @bleuofblue If you know $\langle a\rangle$ and $\langle a^\dagger\rangle$ it's straightforward to calculate $\langle x\rangle$ and $\langle p\rangle$. Moreover, since $\langle a^\dagger\rangle=\langle a\rangle^*$, you only need to compute a single expectation value. So it's certainly not a stupid thing to calculate. It just depends on how you like to organize your algebra. – Jahan Claes Nov 22 '16 at 17:21

## 1 Answer

The operators $a$ and $a^\dagger$ are not hermitian. From the definition of the operators above it is not hard to see that $$x~=~\sqrt{\frac{2\hbar}{m\omega}}(a~+~a^\dagger)$$ and $$p~=~i\sqrt{2\hbar m\omega}(a~-~a^\dagger)$$ are however hermtian. The role of the raising and lowering operators is $$a|n\rangle~=~\sqrt{n}|n-1\rangle,~a^\dagger|n\rangle~=~\sqrt{n+1}|n+1\rangle.$$ Clearly these operators are not hermitian, which would imply $a~=~a^\dagger$

From a matrix perspective if is clear that these operators have a representation according to off diagonal entries. The lowering operator for instance would have $a_{n+1,n}~=~\sqrt{n+1}$ and the raising operator would have $a_{n,n+1}~=~\sqrt{n}$, with all other entries zero. The eigenvalues of any operator with matrix representation $M$ are computed by $det|M~-~\lambda\mathbb I|$ $=~0$. If the diagonal entries of a matrix are zero there are no eigenvalues. Think of $M~=~e^\mu$ with $det~M~=~e^{tr\mu}$, where the exponential of the trace is zero corresponding to zero diagonal entries. Therefore there are no eigenvalues.

• "If the diagonal entries of a matrix are zero there are no eigenvalues" is false: On the one hand, it's trivially false because things like $\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$ clearly have 0 as an eigenvalue, and on the other hand, matrices like $\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$ have vanishing trace but non-vanishing determinant (your formula should have $\det(\exp(M))$ on the l.h.s). This matrix is diagonalizable with eigenvalues $-1,1$. Lastly, the eigenvectors of $a$ are well-known and used - they are the coherent states! – ACuriousMind Nov 22 '16 at 16:54