Expectation value of an operator (not its corresponding observable value)

v0id
1. Problem statement
This isn't a homework question itself, but is related to one. More specifically, I'm computing the time-derivative of $$\langle x \rangle$$ using the correspondence principle. One side simplifies to $$\left\langle \frac{\hat{p}}{m} \right\rangle$$, but what is the physical meaning of this? How does one compute the expectation value of an operator? The concept is alien to me.

Homework Equations

$$\langle Q \rangle = \int_{-\infty}^{\infty}\Psi^* \hat{Q} \Psi \; dx$$
$$\langle \hat{Q} \rangle = ?$$

Homework Helper
The expectation value of an operator A in a pure quantum state $|\psi\rangle$ is the complex number

$$\langle A\rangle_{|\psi\rangle} =\langle \psi|A|\psi\rangle$$

v0id
The expectation value of an operator A in a pure quantum state $|\psi\rangle$ is the complex number

$$\langle A\rangle_{|\psi\rangle} =\langle \psi|A|\psi\rangle$$

Well yes, but that's simply the generalized form of the first equation I posted. So what you're saying is that they are the same? $$\left\langle \frac{\hat{p}}{m} \right\rangle = m^{-1}\int_{-\infty}^{\infty}\Psi^*\left(-i\hbar\frac{\partial}{\partial x}\right)\Psi \; dx$$?

Homework Helper
That is correct. The generalization doesn't single out the (rigged) Hilbert space. That's the advantage of using abstract bra-ket notation.

v0id
That makes sense. I mistakenly thought there is a difference between $$\langle Q \rangle$$ and $$\langle \hat{Q} \rangle$$, but that may have resulted from an abuse of notation.

Thanks a lot, dextercioby.

omephy
Actually, what is difference between <Q> and <Q_hat> ?

Homework Helper
Gold Member
Read the three posts above your post, again. The OP and dextercioby came to the conclusion that there was no difference between the two. The OP's confusion resulted from an abuse or misuse of the notation involved.

Also, this thread is almost 5 years old! If you still have questions on this topic, please start a new thread.