1. From the first postulate of quantum mechanics we known that the vector $|\psi\rangle$ is the mathematical entity that says, intuitively, "in a time $t$, the (state of a) system is a vector".

  2. Then, by the virtue of the second postulate, the next fundamental ingredient of a physical theory are expressed by operators $\mathcal{O}$ acting on Hilbert space.

with 1. we have a formal way to say how to represent our physical phenomena; with 2. we have a formal entity that carries a way to extract important information about our physical phenomena: position $\mathcal{X}$, linear momentum $\mathcal{P}$, etc...

Moreover, about 2., I read the second postulate (using now the position operator) as:

$"$I want say something about $|\psi\rangle$ currently position. Therefore, I must to apply the position operator on $|\psi\rangle$ as:

$$\mathcal{X}|\psi\rangle." \tag{1}$$

  1. From the third postulate, the way formal way that we extract information (we want to extract a numerical result) about $(1)$ is by virtue of the help of an eigenproblem such as:

$$\mathcal{X}|\psi\rangle = x|\psi\rangle \tag{2}$$

Expression $(2)$ is the root of my confusion. I'm going to explain it in the following with physics jargon.

$$---------$$ Suppose we have a system (a electron say) $| \psi \rangle$. Now, I want to discover the position of this electron. Therefore, I must to solve: $\mathcal{X}|\psi\rangle = x|\psi\rangle$.

On the one hand, we know, from postulate 1 of quantum mechanics, that the system state is given by:

$$|\psi\rangle = a_{1}|\psi_{1}\rangle+...+a_{n}|\psi_{n}\rangle.\tag{3}$$


$\mathcal{X}(a_{1}|\psi_{1}\rangle+...+a_{n}|\psi_{n}\rangle) = x(a_{1}|\psi_{1}\rangle+...+a_{n}|\psi_{n}\rangle) = xa_{1}|\psi_{1}\rangle+...+xa_{n}|\psi_{n}\rangle. \tag{4}$

On the other hand, we know, that the mere existence of a eigenvalue $x$ and it position operator $\mathcal{X}$ states the existe of eigenvectors as $|\psi_{x}\rangle \equiv|x\rangle$:

$$\mathcal{X}|\psi_{x}\rangle = x|\psi_{x}\rangle \equiv \mathcal{X}|x\rangle = x|x\rangle \tag{5}$$.

Together with the fact that $\mathcal{X} = \mathcal{X}^{\dagger}$, we know that $|\psi_{x}\rangle \equiv|x\rangle$ forms an (infinite) basis for the Hilbert space and the system (the electron say) can be written as:

$$| \psi \rangle = \int c(x)|x\rangle dx \equiv \int c(x)|\psi_{x}\rangle dx. \tag{6}$$

The thing is, when I wrote $(3)$, I kept in my mind the following reasoning:

The whole superpositon $|\psi\rangle$ are expanded in terms of system states $|\psi_{n}\rangle$; the basis kets are the same "electron". Then in $(4)$ I have the position information $x$, the coefficients $a_{i}$ and the same electron state $|\psi_{i}\rangle$ (that will be possibly collapsed after the measurement).

but when it comes equation $(5)$ I kept in my mind the following reasoning:

I'm using the whole linear algebra to construct a particular basis, the position basis (since I'm using the position operator), and then I write the whole state in terms of these position vectors $| x \rangle$.

I'm very, very confused.

So what is the difference between position vectors $|x \rangle \equiv |\psi_{x}\rangle$ and the basis states of $(3)-(4)$, the $|\psi_{n}\rangle$?.

Maybe It would be possible to state the question as:

What is the difference between: $\mathcal{X}|x\rangle = x|x\rangle$ and $\mathcal{X}|\psi\rangle = x|\psi\rangle ?$

where $|x\rangle$ is the position vector and $|\psi\rangle$ a generic state vector.


2 Answers 2


It is true that $X (\psi(x))=x\psi(x)$ (in the position basis). But the $x$ in front is a function. The function $\psi(x)$ gets pointwise multiplied by the function $x$.

Note that this isn't the eigenvector equation. It only looks like one. This is merely defining the action of $X$ in the position basis. It's analogous to $P \psi(x) =-i\frac{d}{dx} \psi(x) $. Furthermore, it's only true in the position basis.

In the eigenvector equation, $X|x_0\rangle=x_0|x_0\rangle$, the $x_0$ is a constant eigenvalue.

  • $\begingroup$ Surely something’s not right with your answer. the $x$ is not a function:it’s an eigenvalue. Now, $X\vert \psi\rangle=x\vert \psi\rangle$ only if $\vert\psi\rangle$ is the eigenket of $X$ with eigenvalue $x$, i.e. only if $\vert\psi\rangle=\vert x\rangle$. What is confusing maybe is that there is a continuous discrete set of eigenvalues so it is not labelled by a subscript like $E_n$ is, but the $x$ “in front” is nevertheless a constant eigenvalue: it is the eigenvalue $x$ only if $\vert \psi\rangle=\vert x\rangle$. $\endgroup$ Commented Nov 13, 2022 at 6:54
  • $\begingroup$ @ZeroTheHero Hmmm, is it not true that, in the position basis, the action of $X$ on any function $f(x) $ gives $xf(x) $? For example, $X(e^{ipx}) =xe^{ipx}$? $\endgroup$
    – Ryder Rude
    Commented Nov 13, 2022 at 6:57
  • $\begingroup$ yes but that’s because $f(x)=\langle x\vert f\rangle$ and $x f(x):=\langle x\vert X\vert f\rangle= x\langle x\vert f\rangle$. There’s a distinction in working with ket vectors like $\vert f\rangle$ and its component at $x$, which is $f(x)$. To get to multiplication by $x$ you have used the action of $X$ on $\langle x\vert$. It is certainly not true for instance that, if $\vert\psi\rangle=a_0\vert E_0\rangle+a_1\vert E_1\rangle$, then $H\vert\psi\rangle = E\vert\psi\rangle$ since $\vert\psi\rangle$ is not an eigenstate of $H$. $\endgroup$ Commented Nov 13, 2022 at 7:03
  • 1
    $\begingroup$ In other words, when using the position basis you are implicitly using $\langle x\vert X=x\langle x\vert$ constantly. Alternatively, you statement $X\vert\psi\rangle=x\vert\psi\rangle$ is not yet a statement in the position basis. $\endgroup$ Commented Nov 13, 2022 at 7:07
  • 1
    $\begingroup$ @RyderRude Even in the equation of your last comment, it is not a function. $X$ maps $\psi$ to $X\psi$. If you evaluate this new function at $x$, you obtain $x\psi(x)$. But of course, if you meant the identity function (by an abuse of notation), then you're right. Put differently, we have $X\psi = \mathrm{Id}\, \psi$, where $\mathrm{Id}(x) = x$ and the multiplication is defined pointwise. $\endgroup$ Commented Nov 13, 2022 at 9:10

First of all, it is true that $\chi|x\rangle=x|x\rangle$, but what is not true, is that $\chi|\psi\rangle=x|\psi\rangle$.

When we write $|x\rangle$, we are talking about the x-representation, that is a continous base of the Hilbert space. This representation helps you to expand a general state in this base. Using the completeness relation,

$$|\psi\rangle =\mathbb{1}\cdot |\psi\rangle =\int_{\mathbb{R}^3}d^3x |\vec{x}\rangle\langle\vec{x}|\psi\rangle = \int_{\mathbb{R}^3}d^3x \cdot c(x)|\vec{x}\rangle$$

You can verify from here, that $$\hat{\vec{x}}|\psi\rangle\neq \vec{x}|\psi\rangle$$

this is because, $$\vec{x}|\psi\rangle =\vec{x}\int_{\mathbb{R}^3}d^3x´ \cdot c(x´)|\vec{x}´\rangle=\int_{\mathbb{R}^3}d^3x´ \cdot c(x´)\vec{x}´|\vec{x}´\rangle$$

So, the main difference is that $\chi|x\rangle=x|x\rangle$ is the definition of the position eigenkets, and the other expression is false.


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