I'm reading Shankar QM and in chapter 11.2 Eq. 11.2.7, he uses the projection operator (summed over infinity) to project the $|\psi\rangle$ state in the $x$-basis as follows:

$$ \int |x\rangle \langle x|\psi\rangle dx\,.$$

My confusion is perhaps from my misuse of the Dirac notation in the following assumptions:

On one hand the

$$ \int |x\rangle \langle x|\psi\rangle dx = I |\psi\rangle = |\psi\rangle\,,\tag{1}\label{1}$$

if we integrate over $dx$ first. But I think that this leaves ambiguity in the basis of $|\psi\rangle$ because the integral can also be reduced as:

$$ \int |x\rangle \langle x|\psi\rangle dx = \int \psi(x)|x\rangle dx\,,\tag{2}\label{2}$$

which leaves an integral over $dx$ basis vectors with coefficients of $\psi(x)$. I interpret this as $\psi(x)$ as well but that disagrees with Shankar's definition.

So, my question is: If $\langle x|\psi\rangle$ is $\psi(x)$, then what is $|x\rangle \langle x|\psi\rangle$? What is wrong with the assumptions/thinkings \eqref{1} and \eqref{2}?

  • $\begingroup$ Hi MannyC: I salvaged your rejected suggested edit. $\endgroup$
    – Qmechanic
    Commented Apr 12, 2019 at 3:01
  • $\begingroup$ Ah thanks. What did you change? It looks exactly as I intended it...Well doesn't matter anyway :) $\endgroup$
    – MannyC
    Commented Apr 12, 2019 at 3:05
  • $\begingroup$ I interpret this as $\psi(x)$ as well... But why? That is incorrect, and you haven't explained at all why you think this should be the case. 1 and 2 are correct $\endgroup$ Commented Apr 12, 2019 at 3:45
  • $\begingroup$ I just remembered: related. $\endgroup$
    – MannyC
    Commented Apr 12, 2019 at 4:04
  • $\begingroup$ @MannyC: Your suggested edit was rejected by the system because it was submitted simultaneously with another edit. $\endgroup$
    – Qmechanic
    Commented Apr 12, 2019 at 6:39

3 Answers 3


Let me offer an analogy with a simple example: consider a vector in three dimensions, $\vec{v} = v_x \hat x + v_y \hat y + v_z \hat z$. $\vec{v}$ analogous to the abstract state $|\psi\rangle$, while writing the vector $\vec v$ in coordinate basis is analogous to writing $$|\psi\rangle = \int dx \langle x|\psi\rangle |x\rangle.$$ Now suppose I ask you what is the $x$-component of the vector $\vec{v}$? That is $v_x$! Similarly, $\langle x|\psi\rangle \equiv \psi(x)$ is the component of $|\psi\rangle$ in the $|x\rangle$ 'direction'.

Now if I asked you to project $\vec v$ in $z$-direction then you will drop the $x$ and $y$ components and give me a vector $v_z \hat z$. That is precisely analogous to writing $\langle x|\psi\rangle |x\rangle$.

For someone new it might be hard to fathom a delta function normalized basis of $|x\rangle$, and as @MannyC pointed out, one can use any basis of $L^2(\mathbb R)$ functions. I hope this example makes things slightly clearer to the OP, who I gain is beginning to learn the subject.


The results $(1)$ and $(2)$ of OP(v3) are identical. Namely $$ |\psi\rangle = \int dx\,\psi(x)|x\rangle\,.\tag{3}\label{3} $$ This can be seen by projecting over an arbitrary vector, which can be chosen to be drawn from the basis $|x\rangle$ as well. $$ \langle x' | \psi\rangle = \int dx\,\psi(x)\,\langle x'| x\rangle\,. $$ And this holds because $$ \langle x' | \psi\rangle = \psi(x')\,,\qquad \langle x' | x\rangle = \delta(x-x')\,. $$ Recall that indeed the states $|x\rangle$ are delta-function normalized. If you are unhappy with this definition it's also possible to pick an arbitrary $L^2(\mathbb{R})$ function as a state: $|\chi\rangle$ and then check that \eqref{3} projected on $\langle\chi|$ is consistent: $$ \langle \chi | \psi \rangle = \int dx\,\psi(x) \langle\chi|x\rangle = \int dx\,\psi(x) \,\chi^*(x)\,, $$ which is the definition of the $L^2(\mathbb{R})$ inner product, so \eqref{3} holds for all states.


Both (1) and (2) are fine. The problem comes when you say

[...] I interpret this as 𝜓(𝑥) as well

You're claiming that

$$\int\psi(x) | x\rangle dx = \psi(x)$$

which doesn't make sense. Firstly, an integral over $x$ doesn't subsequently depend on $x$. $x$ is a dummy variable, so this expression makes precisely as much sense as $$\sum_{n=1}^3 n^2 = 5n$$ where the left hand side is simply a number ($1^2+2^2+3^2=14$) while the right hand side is a function of some variable $n$, so the symbol $n$ has been used to mean two different things on either side of the equals sign.

Furthermore, inasamuch as $|x\rangle$ is a basis vector of the Hilbert space, the object on the left is a linear combination of Hilbert space vectors, and is therefore a Hilbert space vector itself. On the other hand, the right hand side is just a complex number.

It is always worth remembering that in a strict sense, $|x\rangle$ is not a "real" basis for the Hilbert space (because $|x\rangle$ is not a normalizable state) and the expression $\int \psi(x) |x\rangle dx$ only makes sense when you insert it into an inner product. However, I didn't judge this to be the most important issue in your question, so I'm just mentioning it as a side note to be addressed when you're comfortable with everything else.


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