# Physical meaning of $\left \langle \psi_{nml} | x | \psi_{n'm'l'} \right \rangle$ for hydrogen atom?

I know how to calculate the matrix element $\left\langle \psi_{nml} | x | \psi_{n'm'l'} \right\rangle$, but what is the physical meaning of it? In general, what does the following mean: $$\left \langle \psi_{nml} | A | \psi_{n'm'l'} \right \rangle$$ where $A$ is a Hermitian operator (or does $A$ have to be Hermitian?)?

• As worded, this question is very broad and confusing. First it asks for the meaning of the expectation value of $x$, which has a pretty simple physical meaning. Then it asks for the meaning of an expectation value for an arbitrary operator. Could you please edit the post so that it asks one specific question? Otherwise, this post will likely be closed as being too broad, or even for being unclear what you're asking. – DanielSank Nov 27 '15 at 18:56
• @DanielSank they're both matrix elements between different eigenstates, not expectation values. – fqq Nov 27 '15 at 19:00
• Have you read the Wikipedia entry on expectation values in QM? – Kyle Kanos Nov 27 '15 at 19:00
• @fqq Ah, I see I missed the primes. In any case, it's still unclear whether the author wants to know specifically about $x$ or more generally about any operator. – DanielSank Nov 27 '15 at 19:27

$\lvert \psi_{n'm'l'} \rangle$ is the state you start out with. $A {\mid} \psi_{n'm'l'} \rangle$ is the new state you get when you apply $A$ to the original state. $\langle \psi_{nml} {\mid} A {\mid} \psi_{n'm'l'} \rangle$ is the projection of this new state onto $\lvert \psi_{nml} \rangle$; that is, it measures the overlap between the unprimed state and the result of operating on the primed state. Loosely, it measures how much the operator mixes the two states.
That's really all that can be said for a general operator. However, if the operator is describing the time evolution of the system, the matrix element describes the rate at which the primed state is turned into the unprimed state. This is captured by Fermi's golden rule, which says the transition rate is proportional to $\lvert \langle \psi_{nml} {\mid} A {\mid} \psi_{n'm'l'} \rangle \rvert^2$.
That is the transition dipole moment integral. It is basically the probability that an electric dipole (i.e. a photon) can cause a transition between the states $\psi_{nml}$ and $\psi_{n'm'l'}$.
The quantity $A|\psi\rangle$, for general Hermitian operators $A$, is mostly meaningless. The eigenvalues and eigenvectors of $A$ are meaningful, as is the quantity $\langle \psi | A | \psi \rangle$, but $A|\psi\rangle$ by itself is not. This is a sort of confusing point when first learning QM because it feels like the most important thing about operators should be, well, how they operate on state vectors.
The exception is when the Hamiltonian contains a term that's proportional to $A$, such as when $A = \mathbf{r}$ and the Hamiltonian contains an electric dipole moment term. Then since $d\psi/dt \propto H\psi$, the rate of change of $\psi$ contains $A\psi$, in which case $A |\psi \rangle$ really does tell you something physical: the rate at which $\psi$ evolves into other states. Projecting out a specific state and squaring to get a probability gives the transition rate, $|\langle \psi' | A | \psi \rangle|^2$.