In quantum mechanics, why is $\int (dp/2\pi) |p \rangle\langle p| = 1 $ where $|p \rangle$ represents momentum eigenstate?

  • $\begingroup$ Are you sure about the two pi? A family of orthonormal vectors spans the whole space (is 'complete') is the sum of all the projectors onto each single basis vector gives the identity operator. A projector onto the normalized state |p> is |p><p|, the sum over all projectors $\int dp |p><p|$, and the identity operator is just 1 $\endgroup$
    – A.O.Tell
    Oct 17, 2012 at 16:33
  • $\begingroup$ It's from A.Zee's Quantum Field Theory page 11.. So it is wrong? $\endgroup$
    – RRRR
    Oct 17, 2012 at 17:51
  • $\begingroup$ It's not wrong, it's just a way to normalize states that is a bit unconventional. Maybe it has advantages elsewhere. I'll check Zee later $\endgroup$
    – A.O.Tell
    Oct 17, 2012 at 18:10

3 Answers 3


The factor of $2\pi$ is a convention Zee uses which normalizes the $|p\rangle$ states so that they are not quite orthonormal:

$$ \langle p | p' \rangle = 2\pi \delta(p-p') $$

which means that Zee's $p$ states are different from Ahmed's $p$ states by a factor of $\sqrt{2\pi}$. Let me call these unit normalized $p$ states $||p\rangle$, then

$$ \sqrt{2\pi} || p \rangle = |p\rangle $$

This is something Dirac does also, and it becomes essential for relativistic states, when you want

$$ \int {d^3 p \over (2\pi)^3 2\omega_p} |p\rangle\langle p| = 1 , $$

because then all the parts are manifestly invariant andso the integral over $p$ is an integral over the mass-shell hyperboloid:

$$ \int {d^4 p \over (2\pi)^4} (2\pi) \delta(p_\mu p^\mu - m ^2) = \int {d^3 p\over (2\pi)^3 (2\omega_p)} , $$

where the delta function projects you onto the mass shell in a relativistically invariant way. But for this to be true, you need

$$ |p\rangle = (\sqrt{2\pi})^3 \sqrt{2\omega_p} || p \rangle $$

where $\omega_p = \sqrt{p^2 + m^2} $. This is called a relativistically normalized state. The relativistically normalized states are natural, and define relativistically normalized creation operators,

$$ \alpha(k) |0\rangle = |k\rangle , $$

which are related to the usual nonrelativistically normalized creation operators as follows:

$$ \alpha(k) = (\sqrt{2\pi})^3 \sqrt{2\omega_k} a_k . $$

In terms of the $\alpha(k)$'s and their conjugates, the field Fourier expansion is manifestly invariant:

$$ \phi(x) = \int e^{ikx} \alpha(k) + e^{-ikx} \alpha^\dagger(x) {d^3p \over (2\pi)^3 2\omega_k} . $$

This looks trivial, but it makes everything in scattering theory and mode expansions relativistically consistent. So you can make trivial all the first chapters of any old quantum field theory books with mode expansions which look opaque, because they are noncovariant.


I don't think this question has been answered. The original poster asked: why is \begin{align} \int dp |p\rangle \langle p| = 1, \end{align} up to unimportant normalization issues that have been discussed extensively (but are off the point).

The above statement is equivalent to the question of completeness of a set of states on an Hilbert space (which ahmed addressed but did not answer the question why the LHS of the original poster's equation is equal to the RHS).

Completeness is, heuristically, the statement that a set of vectors (functions on a Hilbert space) spans the space of possible (piecewise continuous and some degree of smoothness (differentiable)) functions on that space. In other words, just about any "well-behaved" function can be expanded in a basis of states that is complete.

A proof of completeness of eigenfunctions of differential equations is given in Chapter VI, Section 3, p. 424 of Courant & Hilbert's Methods of Mathematical Physics, Vol. 1, first English edition. It's too lengthy to reproduce here.

But motivation (not a "proof") for the completeness relation can be given as follows. We consider the expansion of the Dirac $\delta$ function in a particular basis, say the position basis. We assume that it can be written as an expansion over momentum eigenstates in the position basis, $e^{-ipx'}$: \begin{align} \delta(x-x') &= \int_{-\infty}^\infty\!dp\,c_p(x)\frac{e^{-ipx'}}{\sqrt{2\pi}}. \end{align} Clearly the coefficient, $c_p$ in the expansion must depend on $x$ since the $\delta$ function on the LHS does.

The completeness relation is determined as follows. Mulitply the above equation by $e^{ip'x'}/\sqrt{2\pi}$, interchange the order of the two integrals, and integrate over $x'$ on the interval $(-\infty,\infty)$. Use orthonormality of the plane waves, $\int_{-\infty}^\infty \frac{dx'}{2\pi} e^{i(p'-p)x'}=\delta(p'-p)$ to get \begin{align} c_{p'}(x) &= \frac{1}{\sqrt{2\pi}} e^{ip'x}. \end{align} Substitution back into the expression for $\delta(x-x')$ gives \begin{align} \delta(x-x') = \int_{-\infty}^\infty \frac{dp}{2\pi}e^{ip(x-x')}. \end{align} Rewriting this \begin{align} \delta(x-x') &= \langle x|x' \rangle = \langle x|1|x' \rangle \\ &= \int_{-\infty}^\infty\! dp\, \langle x|p\rangle\langle p|x'\rangle \\ \langle x|1|x' \rangle &= \langle x| \cdot\Big[\int_{-\infty}^\infty\! dp\, |p\rangle\langle p|\Big] \cdot |x'\rangle. \end{align} And completeness follows from the arbitrariness of the values of $x,x'$.

Nota Bene: This is not a proof as we have made several questionable steps (like reversing the orders of integration $dx' \leftrightarrow dp$) and essentially assumed the result we wanted (by using orthonormality).

What this does, though, is to make plausible the idea that if you need to expand a function in a complete set of states, you need a relation like $\int\! dp\, |p\rangle\langle p| =1$ to hold.

  • 1
    $\begingroup$ -1: OP was asking about why the $2\pi$ factors appear in some books and not others, not about why basis states are supposed to be orthonormal. $\endgroup$
    – Ron Maimon
    Mar 28, 2013 at 3:15
  • $\begingroup$ You need to read the question again, Maimon: "In quantum mechanics, why is ∫(dp/2π)|p⟩⟨p|=1 where |p⟩ represents momentum eigenstate?" This is asking about completeness not normalization. -1 for being off-topic. $\endgroup$
    – MarkWayne
    Mar 28, 2013 at 4:49
  • 1
    $\begingroup$ I was here when OP asked the question, the issue was the 2pi factor, which confused OP, because it is not orthonormal, but a different convention from other books. I understand what was being asked, really, I am not offtopic. I shouldn't have downvoted you, I suppose, you didn't say wrong things, just things OP already knows. $\endgroup$
    – Ron Maimon
    Mar 28, 2013 at 19:20

The composition of any state in terms of the momentum eigenstates is $\lvert\psi\rangle = k\int \mathrm{d}^{3}p\lvert p\rangle\langle\psi \vert p\rangle$. The set of momentum eigenstates form a complete set of orthogonal state vectors in Hilbert space (this set is uncountably infinite). The components of this decomposition are the $\langle\psi \vert p\rangle$ for each $p$ (a complex number).

To simplify things, $\int \mathrm{d}^{3}p\lvert p\rangle \langle p\rvert$ is an operator. Let $\int \mathrm{d}^{3}p\lvert p\rangle\langle p\rvert = A$, then $A\lvert\psi\rangle = \lvert\psi\rangle$ for an arbitrary $\lvert\psi\rangle$ then $A$ must be the identity operator.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.