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Suppose you add a constantV0 to the potential energy (by 鈥渃onstant鈥 I mean independent ofxas well as t). In classical mechanics this doesn鈥檛 change anything, but what about quantum mechanics? Show that the wave function picks up a time-dependent phase factor:exp(-iV0t/h). What effect does this have on the expectation value of a dynamical variable?

Short Answer

Expert verified

This constant will give the wave function a time-dependent phase factor. However, it will have no effect on the dynamical variables' expected values.

Step by step solution

01

The Schrödinger equation

The Schr枚dinger equation is given by,

iht=-h2m2X2+痴蠄 鈥︹ (1)

This equation is without a constant V0.

02

Schrödinger equation after adding constant V0

If a constant is added to the potential energy, the Schr枚dingerequation will be,

iht=-h2m2X2+(V+V0) 鈥︹ (2)

A time-dependent phase factor will be picked up by the wave function that is e-N0t/h. For doing the calculation a new wave function will be taken that is equal to the old wavefunction which is multiplied with the phase factor.

=蠄别-iv0t/h

03

Replacement of first equation with wave function

Now, equation (1) will be replaced with this new wave function (the originalSchr枚dingerequation before adding the constant to the potential energy). So, the equation will be:

iht=-h2m2X2+Vihte-iv0t/h=-h2m2X2e-iv0t/h+Ve-iv0t/hihte-iv0t/h+V0e-iv0t/h=-h2m2X2e-iv0t/h+Ve-iv0t/hiht=-h2m2X2+V-V0iht=-h2m2X2+(V+V0)

04

Calculating the expectation value of a particle using the new potential energy

The negative sign is absorbed into the constant in this case. The extra phase factor will also have no effect on the dynamical variables' expected values.

Q=-*Qx,-ihXdxQ=-e-iv0t/h*Qx,-ihXe-iv0t/hdxQ=-e-iv0t/hQx,-ihXe-iv0t/hdxQ=-*Qx,-ihXdx

Because the dynamical variables do not depend on time (just on position), the extra phase component will have no effect on the dynamical variables' expected value.

Hence, it鈥檚 the same as the old potential energy.

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