I have a query about the Dirac notation in quantum mechanics. I am reading notes that say

$\left \langle x|\psi \right \rangle=\psi(x)$


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

But is there a mathematical reason for the latter expression? I get that you could argue

$\Big(\int dx|x\rangle\langle x|\Big)|y\rangle=\int dx|x\rangle\langle x|y\rangle =\int dx|x\rangle\delta(x,y)=|y\rangle$

But what about the more general case?

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

Why should the right hand side equal $|\psi\rangle$ ?

Basically, is there a proof of $1=\int dx|x\rangle\langle x|$ ?

Also, I am a little confused about what kind of object $|\psi\rangle $ is; I am assuming it is a function? Are there any good concrete examples that I could use to get my head around these expressions and what exactly the objects $|x\rangle $ and $|\psi\rangle $are ?

  • 2
    $\begingroup$ You can use the commands \langle and \rangle to make left and right angle brackets - these look much nicer. $\lvert \psi \rangle$ is a vector, not a function. $\endgroup$
    – jacob1729
    Commented May 30, 2019 at 19:40
  • 3
    $\begingroup$ related: physics.stackexchange.com/q/364208 $\endgroup$ Commented May 31, 2019 at 0:19
  • $\begingroup$ @ZeroTheHero This sure looks like an exact duplicate, but the OP should make up his mind on linear algebra concepts first. I think he should experiment with a 2-state system. $\endgroup$ Commented May 31, 2019 at 0:47

5 Answers 5


The equation $$ |\psi\rangle=\int dx |x\rangle\psi(x) $$ is correct. It is the expansion of the state $|\psi\rangle$ in terms of the eigenstates $|x\rangle$ of the $\hat x$ operator.

The statement that $$ \mathbb I= \int dx |x\rangle\langle x| $$ is a completeness relation that it goes along with the orthogonality relation $$ \langle x|x'\rangle =\delta(x-x'). $$

A completeness relation is the statement that a given set of vectors (here the $|x\rangle$) is complete in that it is sufficient to expand out any vector. Any self adjoint operator such as $\hat x$ has a complete set of mutually orthogonal eigenvectors.

$ |\psi\rangle$ is not a function. It is an element of the abstract Hibert space. The wavefunction is $\langle x|\psi \rangle$.

  1. Given the position basis $|x\rangle$ and an element $|\psi\rangle \in \mathcal{H}$, the wave function is defined as $$ \psi(x) = \langle x|\psi\rangle $$

  2. $|\psi\rangle$ is not a function, it is a generic element of a Hilbert space.

  3. A Hilbert space $\mathcal{H}$ is complete, namely every Cauchy sequence of vectors admits a limit in the space itself. This implies that every element can be written onto a basis $|\phi_n\rangle$ like $$ |\psi\rangle = \sum_n c_n |\phi_n\rangle $$ with $c_n = \langle \phi_n|\psi\rangle$. Rearranging this back into the previous equation one gets: $$ |\psi\rangle = \sum_n \langle \phi_n|\psi\rangle\, |\phi_n\rangle = \sum_n |\phi_n\rangle \otimes\langle\phi_n|\psi\rangle = \left(\sum_n |\phi_n\rangle \otimes\langle\phi_n|\right)|\psi\rangle $$ therefore $$ \left(\sum_n |\phi_n\rangle \otimes\langle\phi_n|\right) = 1 $$

  • $\begingroup$ Yes that is fine for when we are talking about a countable Hilbert space, but the case I am talking about seems to be for an uncountable basis. What then? $\endgroup$ Commented May 30, 2019 at 20:04
  • $\begingroup$ @RoryCornish Well, it really makes no difference at all: you just replace the sum with an integral and that's it; conceptually the statement holds the same. $\endgroup$
    – gented
    Commented May 30, 2019 at 20:11

First of all, $\vert \psi \rangle$ and $\vert x\rangle$ are vectors in a Hilbert space. If you are begining to learn Quantum Mechanics, I think a long exposition about what this mean would be harmful rather than helpful. Let's just say that $\vert \psi \rangle$ represents the state of a system. For example, $\vert x\rangle$ represens the state of a system whose position is exactly $x$.

You also have to know that $\int \vert x\rangle \langle x \vert $ is not $1$ as in the sense of the number one, but as in the identity operator 1 or I. This is called 'completeness' and I'm not able to give you a proof rigth now but you can search for it under that name.

Answering your question $\vert \psi \rangle$ and $\vert x\rangle$ are abstract object that follow certain mathematical rules, while $\langle x \vert \psi \rangle$ = $\psi(x)$ is a regular fucntion: a complex value for every $x$. The expression $\int \vert x\rangle \langle x \vert \psi \rangle $ = $\int \vert x\rangle \psi(x)$ is another state, one formed by the sum (integral) of the states $\vert x\rangle$, each of them multiplied by a complex value given by $\psi(x)$.

If it helps, try to see it as a discrete sum. $\int \vert x\rangle \psi(x)$ = $\vert \phi \rangle$ = $\psi(x_0)\vert x\rangle \Delta x + \psi(x_0 + \Delta x)\vert x + \Delta x\rangle \Delta x+ ... $ as $\Delta x$ approaches $0$

  • $\begingroup$ No I am not new to quantum mechanics. I completed Quantum mechanics I and II at University but skipped III, which delt with Dirac notation. I am also used to Hilbert spaces from mathematical analysis and Fourier transforms, the works, etc. That's why I would really like some concrete examples of the above to work on some examples. $|\phi>$ . For instance, particle in a box, hydrogen atom, the eigen vectors are functions. In spin, they are matrices. Some examples would be nice. Also, there must be a proof of $1=\int dx|x><x|$. Can you point me where? $\endgroup$ Commented May 30, 2019 at 20:15

Actually, the answer is quite simple. We basically need to show that when the expression

$I=\int dx|x><x|$

acts on either type of vector $|x>$ or $|\phi>$ that it returns that same vector. So first consider former and take the vector $|y>$. In this case we want to show that:


To prove this, all we need to do is act with the vector $<\phi|$ on $I|y>$ and check that it returns $\phi^{*}(y)$. So lets do it

$<\phi|I|y>=<\phi|\int dx|x><x|\cdot|y>=\int dx<\phi|x><x|y>=\int dx\phi^{*}(x)\delta(x-y)=\phi^{*}(y)$

But, by definition, this means that we must have:


as required. Now we need to consider what happens when I acts on a vector of the type $|\phi>$. We need to show that:


But, to show this, all we need to do is act with the vector $<y|$ on $I|\phi>$ and check that the result equal $\phi(y)$

So lets do it:

$<y|I|\phi>=\left \langle y|\int dx |x \right \rangle<x|\phi>=\int dx\left \langle y|x \right \rangle<x|\phi>=\int dx \delta(y-x)\phi(x)=\phi(y) $

So we do indeed get $\phi(y)$, so, by definition we must have:


So finally we have proven that when it acts on both types of vector it can act on, e.g. $|x>$ and $\phi>$ you get back that same vector. Thus we have shown that:

$\int dx |x><x|=1$

Of course, I am making the implicit assumption of completeness since we are dealing with a Hilbert space.


Also I asked for concrete examples of $|x>$ and $\phi>$ but was given none. I think the reason is that it is not easy express these quantities in concrete terms in the Physics picture. What I was really trying to do was relate the physicists notation to the mathematicians notation. Obviously there has to be a one-to-one correspondence between the way physicists and mathematicians deal with the same space.

Below I identify this one-to-one correspondence. So please look away if it might confuse you. This is only for physicists who might be interested in how mathematicians would view the same Hilbert space.

In pure mathematics the equivalent of $|\phi>$ is essentially just the function $\phi$. In pure mathematics, the function $\phi$ is considered to be the vector in the Hilbert space.

In pure mathematics the equivalent to $<\phi|$ is the element of the dual space which equals the map $\Phi$ defined by:

$\Phi(\cdot ):(\cdot ) \mapsto (\phi,.)$

where $(\phi,\psi):=\int dx\phi^{*}(x)\psi(x)$ is the inner product of pure mathematics.

Also, in then mathematics view, $<x|$ and $|x>$ are essentially equivalent to the functionals:

$<x|$ $:\phi \mapsto \phi(x)$

$|x>$ $:\Phi \mapsto \phi(x)^{*}$

So as for a concrete example. If $\phi(x)=Ae^{2\pi i(x)}$ were a wave function

$|\phi>$ would in mathematics simply be the function $\phi(\cdot )=Ae^{2\pi i(\cdot )}$ which is a vector in a Hilbert space

and $<y|$ would be the functional that maps $\phi()=Ae^{2\pi i()}$ to the real number $r=Ae^{2\pi i(y)}$

Thus, in the mathematicians way of looking at this Hilbert space, it is easy to write down in concrete terms what the vectors of the space are.

When $<x|$ and $|y>$ act on each other, they behave as a different sort of function and produce $\delta(x-y)$. This reminds me a bit of function overloading in computing.

It is all a pretty clever way of working so that:

$\left \langle \phi|\psi \right \rangle=\int dx<\phi|x><x|\psi>=\int dx\phi(x)^{*}\psi(x)=:\left ( \phi,\psi \right )$

So, using the bra and ket vectors in this way you end up with precisely the inner product $\left ( \phi,\psi \right )$ of pure mathematics.

And what I have described is effectively the ono-to-one correspondence between the Physics way of looking at it, and the Mathematics way of looking at it. Physicists use the former and mathematicians use the latter; except mathematicians don't bother with the $<x|$ or $|x>$ vectors.

I just wrote this because I think it is always good to be able to look at a problem in a different way. Or at least to be aware that there are other representations of the same thing. Physicists and mathematicians could probably benefit from each others work.


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