0
$\begingroup$

In the position basis, we define the basis kets $\vert x\rangle$ like this: $$\hat{x}\vert x \rangle = x \vert x \rangle \quad \forall x\in\mathbb{R}.$$ We also have that for any state $\vert \psi \rangle$, $\langle x \vert \psi \rangle = \psi(x)$, the wave function in the x-basis. My lecturer then claims that it is 'obvious' that $\hat{x} \vert \psi \rangle = x\psi(x).$

Note we are working in only one dimension at the moment for simplicity.

I tried using the completeness condition: $$\hat{x} \vert \psi \rangle = \int_{-\infty}^{\infty}dx \;\hat{x} \vert x \rangle \langle x \vert \psi \rangle.$$ Then, using the definition of the $\vert x \rangle$ kets and the fact that $\langle x \vert \psi \rangle = \psi(x)$: $$\hat{x} \vert \psi \rangle = \int_{-\infty}^{\infty} dx \; x \vert x \rangle \psi(x)$$ I don't see where to go from here.

$\endgroup$
2
  • 2
    $\begingroup$ You are fine. A state is not a function. Your lecturer slipped up. $\endgroup$ Commented Feb 10, 2022 at 3:43
  • $\begingroup$ @CosmasZachos I understand the difference between a state and a function, but I don't see where he slipped up. Are you saying that $\hat{x} \vert \psi \rangle \neq x\psi(x)$? $\endgroup$ Commented Feb 10, 2022 at 4:08

2 Answers 2

3
$\begingroup$

Your lecturer is mistaken. The correct equation is $$\langle x\rvert\hat x\lvert\psi\rangle = x\psi(x).$$ We can easily see that the given equation is wrong from the fact that $\hat x\lvert\psi\rangle$ is a ket, i.e. a vector in the state space, while $x\psi(x)$ is a complex number, i.e. a scalar, so they cannot possibly be equal. To correct this, we note that $x\psi(x)$ is equal to the "component" of $\hat x\lvert\psi\rangle$ in the "$\lvert x\rangle$ direction," so the expression on the left needs to multiplied by the bra $\langle x\rvert$ to get the appropriate scalar.

$\endgroup$
2
$\begingroup$

Yes, your lecturer has slipped up.

As you've likely encountered, you can think of $|x\rangle$ as a vector and $\hat{x}$ as a matrix. This means that $\hat{x} |\psi\rangle$ is a vector, not a number as your lecturer has given $x \psi(x)$.

The final equation that you have given \begin{equation} \hat{x} |\psi \rangle = \int dx \, x \psi(x)|x \rangle \end{equation} captures what I imagine your lecturer was intending to teach. That is, in a position basis one replaces $\hat{x}$ with $x$ and $| \psi \rangle$ with $\psi(x)$.

$\endgroup$
1
  • $\begingroup$ This answer and the answer by @Sandejo really helped me understand, thanks a lot! $\endgroup$ Commented Feb 10, 2022 at 5:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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