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Quantum mechanics, adding a phase factor

  1. Apr 8, 2013 #1
    Here's the question:

    1. In classical mechanics, we know that the evolution of a system, that is the trajectories of the particles and objects, does not depend on where we chose the xero-point of the potential energy. Here we analyse what happens in quantum mechanics. Assume that you add a constant C0 to the potential energy of an arbitrary potential V(x)

    Show that the wave function now differs from the one you would get without the C0 added by having a phase factor [tex]e^{-i \frac{C_0 t}{\hbar}}[/tex] multiplied to it.
    How will this phase-factor affect the expectation value of a dynamical variable?



    So I was thinking of using: [tex]\Psi (x,t) = \psi (x) \varphi (x)[/tex]
    to solve it.

    Can someone guide me through this (like with a hint or 2)? I'm quite confused about quantum mechanics.
     
  2. jcsd
  3. Apr 8, 2013 #2

    DrClaude

    User Avatar

    Staff: Mentor

    Start with an eigenstate ##\psi## of a Hamiltonian ##\hat{H}##, with eigenenergy ##E##. What is the time evolution of ##\psi##? (This should be in your textbook.)

    Then, arbitrarily shift the zero of energy by ##C_0##. How is the time evolution of ##\psi## affected?

    Then, calculate the expectation value of an operator ##\hat{A}##,
    $$
    \langle \hat{A} \rangle = \int_{-\infty}^{\infty} \psi(x,t)^* \hat{A} \psi(x,t) dx
    $$
    for both wave functions (energy shifted and not shifted), and compare.
     
  4. Apr 10, 2013 #3
    Sorry, I'm not getting what you mean. Maybe I should have posted this in the Introductory Physics section of this forum.

    In my textbook, it says I can use a separation of variables to solve the Schrodinger equation.
    [tex]\Psi (x,t) = \psi (x) + \varphi (t) [/tex]


    [tex]i \hbar \dfrac{\partial \Psi (x,t)}{\partial t} = -\dfrac{\hbar ^2}{2m} \dfrac{\partial ^2 \Psi (x,t)}{\partial x^2} + V \Psi (x,t)[/tex]

    [tex]i \hbar \psi (x) \dfrac{\partial \varphi (t)}{\partial t} = -\dfrac{\hbar ^2}{2m} \dfrac{\partial ^2 \psi (x)}{\partial x^2} \varphi (t) + V \psi (x) \varphi (t)[/tex]

    And then dividing through by ##\psi (x) + \varphi (t) ## I will get:

    [tex]i \hbar \dfrac{1}{\varphi (t)} \dfrac{\partial \varphi (t)}{\partial t} = -\dfrac{\hbar ^2}{2m} \dfrac{1}{\psi (x)} \dfrac{\partial ^2 \psi (x)}{\partial x^2} + V[/tex]

    It then uses a separation constant E to separate out the left and right hand sides (since one depends on time and the other depends on position).

    [tex]E = -\dfrac{\hbar ^2}{2m} \dfrac{1}{\psi (x)} \dfrac{\partial ^2 \psi (x)}{\partial x^2} + V[/tex]


    So I guess what you're saying is to multiply everything by ## \psi (x) ## whch will give me:

    [tex]E \psi (x) = -\dfrac{\hbar ^2}{2m} \dfrac{\partial ^2 \psi (x)}{\partial x^2} + V \psi (x)[/tex]

    And the right hand side can be defined as the Hamiltonian operator ## \hat{H} ## with an eigenstate ## \psi (x) ##

    [tex]E \psi (x) = \hat{H} \psi (x)[/tex]

    And the time evolution of ## \psi (x) ## is constant because it is only dependent on position and not time?

    "Then, arbitrarily shift the zero of energy by C0. How is the time evolution of ψ affected?"
    What do you mean? I'm confused as to where to go from here.
    Do I add C0 to V which will give me:
    [tex](E + C_0) \psi (x) = -\dfrac{\hbar ^2}{2m} \dfrac{\partial ^2 \psi (x)}{\partial x^2} + (V+C_0) \psi (x)[/tex]
    and then solve this second order differential equation?
     
  5. Apr 10, 2013 #4

    DrClaude

    User Avatar

    Staff: Mentor

    You're getting along fine. You just need to go back to the time-dependent Schrödinger equation (TDSE) to get the full wave function:

    Indeed, for ##\psi (x)## an eigenfunction of ##\hat{H}##, the separation constant ##E## is the energy. Use this to solve for ##\varphi(t)## in the equation above.

    Yes, it corresponds to that equation, but you don't need to actually solve it. Again go back to solving for ##\varphi(t)##, and see how the full, time-dependent wave function chnages with this ##C_0## added to the energy.
     
  6. Apr 10, 2013 #5
    Right so with the time dependent part,

    [tex]E = i \hbar \dfrac{1}{\varphi (t)} \dfrac{\partial \varphi (t)}{\partial t}[/tex]

    If I re-arrange:

    [tex]\dfrac{E}{i \hbar} = \dfrac{1}{\varphi (t)} \dfrac{\partial \varphi (t)}{\partial t}[/tex]

    And the integrate with respect to (t) I'll get:

    [tex]\dfrac{E}{i \hbar} t = \ln(\varphi (t))[/tex]

    And then re-arranging I get:

    [tex]\varphi (t) = e^{\dfrac{-i E t}{ \hbar}}[/tex]


    So if the separation constant E is the energy, I'll get my other differential equation with C0 added to it:

    [tex]E + C_0 = i \hbar \dfrac{1}{\varphi (t)} \dfrac{\partial \varphi (t)}{\partial t}[/tex]

    [tex]\dfrac{E + C_0}{i \hbar} = \dfrac{1}{\varphi (t)} \dfrac{\partial \varphi (t)}{\partial t}[/tex]

    [tex]\dfrac{E + C_0}{i \hbar} t = \ln (\varphi (t))[/tex]

    And then re-arranging:

    [tex]e^\dfrac{-iEt -iC_0 t}{\hbar} = \varphi (t)[/tex]

    Which becomes ##e^\dfrac{-iEt}{\hbar} e^\dfrac{-iC_0 t}{\hbar} = \varphi (t)##

    This is just like the original one, but with the phase factor of ##e^\dfrac{-iC_0 t}{\hbar}## added to it. So I guess that's the answer?

    With the second part of the question: "How will this phase-factor affect the expectation value of a dynamical variable?"

    Does it mean to calculate ##<x> = \int \varphi (t)^* (x) \varphi (t) dx ## and compare the one with and without the C0?
     
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