Why is the position operator acting on a state equal to position times the wave function in the position basis?

Question

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.

Answer 1

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.

Answer 2

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 $\psi(x)$.