Momentum operator eigenfunction

  • #1
blue_leaf77
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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.
 
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  • #2
atyy
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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.
 
  • #3
blue_leaf77
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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?
 
  • #4
atyy
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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.
 
  • #5
blue_leaf77
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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.
Ok thanks for your guide.
 
  • #6
vanhees71
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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.
 
  • #7
blue_leaf77
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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?
 
  • #8
vanhees71
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Sakurai, Modern Quantum Mechanics
Ballentine, Quantum Mechanics
Weinberg, Lectures on Quantum Mechanics
 

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