Expected value of momentum

  • Thread starter intervoxel
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  • #1
intervoxel
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[itex]
\langle p\rangle=-i\hbar\int \Psi^*\frac{\partial}{\partial x}\Psi dx
[/itex]

Apparently, in general, we get a complex number for momentum. What does it mean, since p is an observable?
 

Answers and Replies

  • #2
Matterwave
Science Advisor
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The values are real, and can only be real, the imaginary numbers will cancel out. The momentum operator, p, being hermitian, guarantees this.
 
  • #3
intervoxel
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Ok, thank you.
 
  • #4
SpectraCat
Science Advisor
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In a little more detail, if the wavefunction is purely real, then you will always find that <p>=0 (try this out if you like, for example on the Harmonic oscillator wavefunctions).

If the wavefunction is complex, then you can separate the real and imaginary parts ... the imaginary part will of course be multiplied by i, and so you will end up with i2 in your final answer, which will be completely real. (The real part will yield a result of zero, as covered above).
 
  • #5
sweet springs
1,225
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Hi.

Apparently, in general, we get a complex number for momentum. What does it mean, since p is an observable?

By partial integration, as ψ vanishes at infinity x,
∫ψ*∂x ψ dx = [ψ*ψ] - ∫(∂x ψ*)ψ dx = - ∫ψ∂xψ* dx.
By adding the both and dividing by two,
-ih'∫ψ*∂x ψ dx = -ih'{ ∫ψ*∂x ψ dx - ∫ψ∂xψ* dx}/2
= -ih'/2 ∫ψ*∂x ψ dx + ih'/2 ∫ψ∂xψ* dx
= -ih'/2 ∫ψ*∂x ψ dx + its conjugate complex
= real number

Regards.
 
Last edited:
  • #6
intervoxel
195
1
In a little more detail, if the wavefunction is purely real, then you will always find that <p>=0 (try this out if you like, for example on the Harmonic oscillator wavefunctions).

If the wavefunction is complex, then you can separate the real and imaginary parts ... the imaginary part will of course be multiplied by i, and so you will end up with i2 in your final answer, which will be completely real. (The real part will yield a result of zero, as covered above).

Another way is to take into consideration that the derivative of a real function has opposite parity. So the integrand is odd, resulting in a null integral in the symmetric limits. I think.
 
  • #7
intervoxel
195
1
Hi.

By partial integration, as ψ vanishes at infinity x,
∫ψ*∂x ψ dx = [ψ*ψ] - ∫(∂x ψ*)ψ dx = - ∫ψ∂xψ* dx.
By adding the both and dividing by two,
-ih'∫ψ*∂x ψ dx = -ih'{ ∫ψ*∂x ψ dx - ∫ψ∂xψ* dx}/2
= -ih'/2 ∫ψ*∂x ψ dx + ih'/2 ∫ψ∂xψ* dx
= -ih'/2 ∫ψ*∂x ψ dx + its conjugate complex
= real number

Regards.

Interesting proof, thank you.
 
  • #8
varga
64
0
Since everyone seem to agree this momentum is "real", can someone please explain what kind of momentum is that - spatial, angular, spin? Does it apply to any particle or object or just photons or just bound electrons, or what? How do we measure it experimentally?
 

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