0
$\begingroup$

We know that the definition of the momentum operator $\hat{P_x}$ in an state space $\mathcal{E}$ is:

$$\hat{P_x}|\psi\rangle=P_x|\psi\rangle$$

where $P_x \in \mathbb{R}$. However we also know that the representation of $\hat{P_x}$ in the position basis $\{|\vec{r}\rangle\}$ is: $$ \hat{P_x}\to \frac{\hbar}{i}\frac{\partial}{\partial x}, \tag{1} $$

Now, my question is: Is $\frac{\hbar}{i}\frac{\partial}{\partial x}$ still an operator or is it an eigenvalue?

From what I gather I could also write (1) as:

$$ \frac{\hbar}{i}\frac{\partial}{\partial x}\psi(\vec{r})=P_x\psi(\vec{r}) $$

keeping $P_x \in \mathbb{R}$. My doubts get even worse when I define an operator like this:

$$ \hat{A}=\hat{X}+\hat{P}_x $$

Then I could write:

$$ \hat{A} |\psi \rangle=a|\psi \rangle $$

where $a$ is the eigenvalue of $\hat{A}$. But in this case $a=x+P_x$ where $x$ is the eigenvalue of the operator $\hat{X}$. Suppose I wanto to go to the position basis. Then I could write:

$$ \bigg(x+\frac{\hbar}{i}\frac{\partial}{\partial x}\bigg)\psi(\vec{r})=a\psi(\vec{r})=(x+P_x)\psi(\vec{r}) $$

from my reasoning the x on the left hand side is to be interpreted as an operator and on the right hand side as a number, but how can this be? This doesn't seem to make sense to me either way, that doesn't seem to be an "equation". What is this representation in position basis? Is it still an operator or is it the eigenvalue?

$\endgroup$
  • 1
    $\begingroup$ an eigenvalue is a number. I don’t think $d/dx$ is a number. $\endgroup$ – ZeroTheHero Nov 2 '19 at 23:58
1
$\begingroup$

Stick to one dimension to avoid superfluous complication.

The operator $\hat p$ has eigenvalues, but on specific eigenvectors, labelled as such, $$ \hat p | p\rangle = p|p\rangle, $$ But unless your state $|\psi\rangle$ were one such, your first equation will not hold. You may always compute $\langle \psi | \hat p | \psi\rangle$, some sort of momentum of that state, without assuming eigenvectors of the relevant operator, if that is what is confusing you.

In the position basis, the momentum operator resolves as $$ \hat p = \int\!\! dx ~~|x\rangle \frac {\hbar}{i} \partial_x \langle x| , $$ so, manifestly, the $|x\rangle$ s are not its eigenstates.

I'll tweak your next operator to Dirac's celebrated annihilation operator, $$ \hat a =\hat x + i \hat p , $$ and then its eigenvectors are the celebrated coherent states, $$ \hat a |\alpha \rangle = \alpha | \alpha \rangle \\ |\alpha\rangle \equiv e^{-|\alpha|^2 /2} \exp (\alpha ~~ \hat {a} ^\dagger) |0\rangle , $$ where $\hat a ^\dagger$ and $\hat |0\rangle =0$.

  • You might connect these Fock states to the x-basis in here, $$ |x\rangle=\frac{e^{x^2/2}}{\pi^{1/4}} e^{-(a^\dagger-\sqrt{2} x)^2/2} |0\rangle ~. $$

The takeaway is that you only get eigenvalue equations when acting on eigenstates, which usually define a convenient basis for your operator of choice.

$\endgroup$
  • $\begingroup$ "You may always compute ⟨ψ|p^|ψ⟩, some sort of momentum of that state, without assuming eigenvectors of the relevant operator, if that is what is confusing you." This sentence here did the trick, thanks $\endgroup$ – Bidon Nov 3 '19 at 2:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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