Given a particle whose position can be anywhere on the $x$-axis we may write the following:

$$\hat x|x'\rangle=x'|x'\rangle$$

and so we may write the expansion postulate as:

$$|\psi\rangle=\int\langle x'|\psi \rangle |x'\rangle dx' $$

From which we make the association:

$$\psi(x')=\langle x'|\psi \rangle$$

This I understand just fine. However in our notes the next few lines puzzle me slightly. They proceed as follows:

By taking the scalar product of $|\psi\rangle $ & $|x\rangle$ we get:

$$\langle x|\psi \rangle =\int \langle x|x'\rangle \langle x'|\psi \rangle dx'$$

And so we see that the integral form of the resolution of the identity is:

$$\int |x\rangle \langle x|dx =1$$

The above is a word for word transcription of our notes. The way that I am thinking about this is that the eigenvectors $|x'\rangle$ essentially represent delta functions at the point $x'$, and so $\psi (x')$ as is written above tells us the value of $\psi (x)$ AT $x'$.

However I don't understand the part where we take inner product $\langle x|\psi \rangle$. I see the motivation, which I believe is to obtain an expression which we can identify as $\psi (x)$, but I don't understand how we get from the integral to the resolution of the identity expression. How have the bra's and ket's been manipulated in this way? Is it to do with the fact that $\langle x|$ & $| \psi\rangle$ have no explicit dependency on $x'$?

Any help greatly appreciated - I understand this is a basic question so my apologies if it has been done to death!

  • 1
    $\begingroup$ Do you know that $\langle x|x'\rangle=\delta(x\!-\!x')~?$ $\endgroup$
    – Qmechanic
    Oct 14, 2016 at 10:19
  • $\begingroup$ The notes actually use this to justify that - they say that because of the properties above we can see that is the case. I agree, and it makes good sense, but I just don't really understand how the bras and kets have been manipulated. $\endgroup$
    – arcturus7
    Oct 14, 2016 at 10:28

1 Answer 1


$$\langle x|\psi \rangle =\int \langle x|x'\rangle \langle x'|\psi \rangle dx'$$ $$\phantom{\langle x|\psi \rangle} = \langle x|\left( \int dx' |x'\rangle \langle x'|\right) |\psi \rangle $$ because as you have stated, $\langle x|$ and $|\psi \rangle$ are independent of $x'$ and the scalar product is linear. And now we immediately see that the brackets need to equal $1$.

If I have misunderstood your question, please tell me in the comment and I'll happily delete this answer.

  • $\begingroup$ Oh that is so much ore obvious when you write it like that... Can't believe I didn't spot that. Thanks! $\endgroup$
    – arcturus7
    Oct 14, 2016 at 12:10

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