# Momentum operator eigenfunction

Homework Helper
This might be trivial for some people but this has been bothering lately.
If P is momentum operator and p its eigenvalue then the eigenfunction is up(x) = exp(ipx/h). where h is the reduced Planck constant (sorry cant find a way to make the proper notation).
While it can also be proved that <x|p> = exp(ipx/h) (omitting constant pre factor), where |p> is also momentum operator eigenfunction in different notation, in this sense one would say that up(x) = |p>. But why is <x|p> = up(x) instead?
I will appreciate anyone who is trying to enlighten me in this problem. Thanks in advance.

Last edited:

atyy
This might be trivial for some people but this has been bothering lately.
If P is momentum operator and p its eigenvalue then up(x) = exp(ipx/h). where h is the reduced planck constant (sorry cant find a way to make the proper notation).
While it can also be proved that <x|p> = exp(ipx/h) (omitting constant pre factor), where |p> is also momentum operator eigenfunction, in this sense one would say that up(x) = |p>. But why is <x|p> = up(x) instead?
I will appreciate anyone who is trying to enlighten me in this problem. Thanks in advance.

|p> is the momentum eigenfunction.

<x|p> is the representation of the momentum eigenfunction in the position basis.

Homework Helper
I have also thought the same way, but then it suggests that P = h/i ∂/∂x is also representation of momentum operator in space coordinate, because it yields solution of eigenfunction up(x) = exp(ipx/h) = <x|p>. Does this means that there is other forms of P depending on which basis its eigenfunction is evaluated?

atyy
I have also thought the same way, but then it suggests that P = h/i ∂/∂x is also representation of momentum operator in space coordinate, because it yields solution of eigenfunction up(x) = exp(ipx/h) = <x|p>. Does this means that there is other forms of P depending on which basis its eigenfunction is evaluated?

Yes, that's right.

Homework Helper
Ah yeah I think that makes sense, when I want to evaluate <P> for wavefunction already represented momentum space, I don't need to replace P with h/i ∂/∂x, instead I shoud substitute P = p.

vanhees71
Gold Member
2021 Award
It is very important to distinguish the different descriptions. Dirac notation is the representation independent description of quantum theory in terms of an abstract (rigged) Hilbert space. The vectors ##|\psi \rangle## are abstract objects which come to life just from the definition of their properties through the axioms declaring what a Hilbert space is. The same holds for linear self-adjoint operators like ##\hat{p}## and ##\hat{x}##, which are defined in an abstract way by their commutator relations (Lie algebra of the translation group in 1 dimension, the socalled Heisenberg algebra,
$$[x,p]=\mathrm{i} \hbar \hat{1}.$$
It turns out from this algebra alone that both ##\hat{x}## and ##\hat{p}## have entire ##\mathbb{R}## as their spectrum, i.e., only continuous "generalized eigenvalues".

Now you can work in a specific representation by choosing an arbitrary complete set of (generalized) eigenvectors of an observable like the position. It's not a true Hilbert-space vector but a distribution, living in the dual space of the dense subspace of the Hilbert space, where the position and momentum operators are defined. It's "normalized to a ##\delta## distribution":
$$\langle x'|x \rangle=\delta(x-x').$$
The true normalizable Hilbertspace vectors are mapped to square-Lebesgue-integrable functions (vulgo "the wave function"),
$$\psi(x)=\langle x|\psi \rangle.$$
The Heisenberg algebra in position representation is given by
$$\tilde{x} \psi(x):=\langle x|\hat{x} \psi \rangle=\langle \hat{x} x|\psi \rangle=x \langle x|\psi=x \psi(x)$$
and
$$\tilde{p} \psi(x)=-\mathrm{i} \hbar \partial_x \psi(x).$$
Now you can look for the generalized momentum eigenfunctions in position representation, leading to
$$\tilde{p} u_p(x)=p u_p(x) \; \Rightarrow \; u_p(x)=N_p \exp(\mathrm{i} p x/\hbar).$$
Again, that's a distribution, normalizable to a ##\delta## distribution. Using the completeness relation for the generalized position eigenstates,
$$\int_{\mathbb{R}} \mathrm{d} x |x \rangle \langle x|=\hat{1},$$
you get
$$\langle p|p' \rangle=\int_{\mathbb{R}} \mathrm{d} x \langle p| x\rangle \langle x|p' \rangle=\int_{\mathbb{R}} \mathrm{d} x u_p^*(x) u_{p'}(x) = \int_{\mathbb{R}} \mathrm{d} x N_{p}^* N_{p'} \exp[\mathrm{i} x(p'-p)/\hbar]=|N_p|^2 2 \pi \hbar \delta(p-p'),$$
as is well-known from the maths of the Fourier transformation. Thus up to an irrelevant phase factor we have found the properly normalized generalized momentum eigenvector in position representation to be
$$u_p(x)=\frac{1}{\sqrt{2 \pi \hbar}} \exp(\mathrm{i} p x/\hbar).$$
Now you can also work in momentum representation. The corresponding wave function is
$$\tilde{\psi}(p)=\langle p|\psi \rangle=\int_{\mathbb{R}} \mathrm{d} x \langle p | x \rangle \langle x | \psi \rangle=\int_{\mathbb{R}} \mathrm{d} x u_{p}^*(x) \psi(x)= \int_{\mathbb{R}} \mathrm{d} x \frac{1}{\sqrt{2 \pi \hbar}} \exp(-i p x/\hbar) \psi(x).$$
In the same way you find the inverse relation,
$$\psi(x)=\langle x|\psi \rangle=\int_{\mathbb{R}} \mathrm{d} p \langle x|p \rangle \langle p|\psi \rangle=\int_{\mathbb{R}} \mathrm{d} p u_p(x) \tilde{\psi}(p)= \int_{\mathbb{R}} \mathrm{d} p \frac{1}{\sqrt{2 \pi \hbar}} \exp(\mathrm{i} p x/\hbar) \tilde{\psi}(p),$$
as also expected from the maths of Fourier transformations.

Homework Helper
Thanks vanhees71, your post has deepened m understanding.
I have gotten part of those stuffs in my undergrad except for Hilbert space thing. And looks like I need to self-study the next level of quantum mechanics since I don't get quantum mech for graduate in my current program but I will have to be dealing with it again for my master thesis. So, any suggestion for not-so-hard-to-understand books on graduate quantum mechanics?

vanhees71
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