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

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.

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

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