- #1

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\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?

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- Thread starter intervoxel
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- #1

- 195

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\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?

- #2

Matterwave

Science Advisor

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- #3

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Ok, thank you.

- #4

SpectraCat

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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 i

- #5

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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.

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

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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 i^{2}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

- 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.

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