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Adding a constant V to the wave equation

  1. Apr 12, 2005 #1
    The question says to show that the wave function picks up a time-dependent phase factor,
    [tex] e^\left(-i V_0 t / \hbar \right) [/tex],
    when you add a constant [itex] V_0 [/itex] to the potential energy. And then it asks what effect does this have on the expecation value of a dynamical variable?

    Assuming I only have been given the Schrodinger eqn thus far, and there has not been any discussion thus far about solutions for the equation, where do I start? I was thinking about replacing the V in the equation with [itex] V + V_0 [/itex], but I'm not sure how that would lead to getting the phase factor they talk about.

    [tex]
    i \hbar \left( \frac {\partial \psi}_{\partial t} \right) = - \left( \frac {\hbar^2}_{2m} \right) \left( \frac {\partial^2 \psi}_{\partial x^2} \right) + V \psi + V_0 \psi
    [/tex]

    Any ideas?
     
  2. jcsd
  3. Apr 12, 2005 #2

    StatusX

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    Just show that if [itex]\psi[/itex] satisfies the original wave equation, [itex]\psi e^{(-i V_0 t/\hbar)}[/itex] solves the modified potential equation.
     
  4. Apr 12, 2005 #3

    dextercioby

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    Not to break your confidence,but i'm sure this problem has been solved on these forums.But i'm not sure that somone indicated the real elegant solution.Assume a time-independent Hamiltonian (that way,the Dyson series can be restrained to the exponential).

    SE:[tex] \frac{d|\psi (t)\rangle}{dt}=\frac{1}{i\hbar} \hat{H}|\psi\rangle [/tex]

    had the simple solution (:wink:)

    [tex] |\psi (t)\rangle = \hat{U}(t,t_{0})|\psi (t_{0})\rangle [/tex]

    ,where (here comes the time-independence of the Hamiltonian)

    [tex] \hat{U}=\mbox{exp} \left(\frac{1}{i\hbar}\left(t-t_{0}\right)\hat{H}\right) [/tex]

    Okay?

    Good

    Now.Make the addition of a constant to the Hamiltonian.Then the new Hamiltonian is

    [tex] \hat{H}+V_{0}\hat{1} [/tex]

    It's not difficult to show that the new evolution operator [itex] \hat{\bar{U}} [/itex] is

    [tex] \hat{\bar{U}}=\exp\left(\frac{1}{i\hbar}\left(t-t_{0}\right)\left(\hat{H}+V_{0}\hat{1}\right)\right) [/tex]

    which can be written (using the property of the exponential of operators and the fact that the old hamiltonian commutes with the unit operator)

    [tex] \hat{\bar{U}}=\exp\left(\frac{1}{i\hbar}\left(t-t_{0}\right)\hat{H}\right)\exp\left(\frac{V_{0}}{i\hbar}\left(t-t_{0}\right) \hat{1}\right) [/tex]

    Okay?

    Good.

    Therefore the new state vector (solution of the new SE) is

    [tex] |\bar{\psi} (t)\rangle = \hat{\bar{U}}(t,t_{0})|\psi (t_{0})\rangle
    =\left[\exp\left(\frac{1}{i\hbar}\left(t-t_{0}\right)\hat{H}\right)\exp\left(\frac{V_{0}}{i\hbar}\left(t-t_{0}\right) \hat{1}\right)\right]|\psi (t_{0}\rangle [/tex]

    Now,use the fact that

    [tex] \exp \left(C\hat{1}\right) |\psi (t_{0}\rangle =\exp (C) |\psi (t_{0}\rangle [/tex]

    to write

    [tex] |\bar{\psi} (t)\rangle =\exp \left(\frac{V_{0}\left(t-t_{0}\right)}{i\hbar}\right) \exp\left(\frac{1}{i\hbar}\left(t-t_{0}\right)\hat{H}\right) |\psi (t_{0}\rangle =\exp \left(\frac{V_{0}\left(t-t_{0}\right)}{i\hbar}\right) |\psi (t)\rangle [/tex]

    which proves that the state vector picks up a time-dependent phase factor when a constant is added to the hamiltonian.

    Your business is to show what happens in this case with the physical quantities (quantum observables).

    Daniel.
     
    Last edited: Apr 12, 2005
  5. Apr 12, 2005 #4
    dextercioby:

    The Hamiltonian hasn't even been formally introduced yet for us, and neither has the bra-ket notation. This leaves me with just calculus to try to solve this with (any way to translate your approach to this constraint?).

    If this has been solved without that notation somewhere else, could someone point me in that direction? The search function seems to glitch out if I try searching for more than 3 terms.

    -

    StatusX:

    To clearify your suggestion, your saying let:

    [tex]
    \psi_2 = \psi_1 e^{(-i V_0 t/\hbar)}
    [/tex]

    in the equation

    [tex]
    i \hbar \left( \frac {\partial \psi_2}_{\partial t} \right) = - \left( \frac {\hbar^2}_{2m} \right) \left( \frac {\partial^2 \psi_2}_{\partial x^2} \right) + V \psi_2 + V_0 \psi_2
    [/tex]

    and prove that each side of the equation is equivalent, right?
     
    Last edited: Apr 12, 2005
  6. Apr 12, 2005 #5

    StatusX

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    That equation doesn't make sense as it stands. Make sure to distinguish between the new and the old wave functions. But yes, I think you know what to do. This works if all you have to do is verify this new function solves the equation. To actually derive it, you'll need to do something similar to what dextercioby suggested.
     
  7. Apr 12, 2005 #6
    Made a quick edit to those equations.... Does that look correct now?
     
  8. Apr 12, 2005 #7

    dextercioby

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    https://www.physicsforums.com/showthread.php?t=61864&highlight=constant+Schrodinger

    That was the thread that came into my mind,but it didn't have the solution.

    SOLUTION #2 (sucks compared to the first :yuck: ).

    Assume time independent hamiltonian,then the wave-function

    [tex] \psi_{1}(t)=e^{\frac{1}{i\hbar} E_{1} t} \psi_{1} [/tex] (1)

    is the solution of the time-dep.SE and [itex] \psi_{1} [/itex] is the solution of the time-indep Hamiltonian's spectral eq. (forget about degeneration,this is bastardizing QM)

    [tex] H_{1}\psi_{1}=E_{1}\psi_{1} [/tex] (2)

    Now,the new Hamiltonian

    [tex] H_{2}=:H_{1}+V_{0} [/tex] (3)

    commutes with the old one (u can check) and is time-independent as well.Therefore,it has a commune set of eigenfunctions [itex] \psi_{2} [/itex] with the old one

    [tex] \psi_{2}=\psi_{1} [/tex] (4)

    What about the spectrum...???????Well,the new eigenvalues [itex] E_{2} [/itex] will be

    [tex] H_{2}\psi_{2}=E_{2}\psi_{2}\Rightarrow (H_{1}+V_{0})\psi_{2}=E_{2}\psi_{2} [/tex] (5)

    Now use (4) and (2) to write

    [tex] (E_{1}+V_{0})\psi_{1}=E_{2}\psi_{1} [/tex] (6)

    Therefore

    [tex] E_{2}=E_{1}+V_{0} [/tex] (7)

    The solution to the new time-dependent SE will be (cf.(1))

    [tex] \psi_{2}(t)=e^{\frac{1}{i\hbar} E_{2}t} \psi_{2} [/tex] (8)

    which can be written,according to (7) and (1)

    [tex] \psi_{2}(t)=e^{\frac{V_{0}t}{i\hbar}} \psi_{1} [/tex] (9)

    Q.e.d.

    :yuck: I'd vote for the first solution with my eyes closed... :wink:


    Daniel.
     
    Last edited: Apr 12, 2005
  9. Apr 12, 2005 #8
    StatusX:

    I tried your approach, but I couldn't get the sides to become equivalent. I assume I have my equation mis-labeled in some manner.

    Moreover, it is my assumption that the book (the same one mentioned in the other post linked by dextercioby) is looking for a derivation. Is there any other approaches? If not, at this point, I would suffice with a clearified way of proving that is a solution.

    -

    dextercioby:

    Okay, I'm going to try to recap your approach without the use of the Hamiltonian - please correct or clearify what I get wrong:

    Assume that

    [tex] \psi_{1}(t)=e^{\frac{1}{i\hbar} E_{1} t} \psi_{1} [/tex] (1)

    is a solution to the wave function. Moreover, assume

    [tex] \psi_{1} [/tex]
    and
    [tex] \psi_{2} [/tex]

    are both solutions to the Schrodinger eqn. Moreover, assume that the potential energy for those solutions are

    [tex] E_{1} = V [/tex]
    and
    [tex] E_{2} = V + V_{0} [/tex]

    respectively. Therefore,

    [tex] E_{2} = E_{1} + V_{0} [/tex]

    The solution for the 2nd wave function is therefore:

    [tex] \psi_{2}(t)=e^{\frac{1}{i\hbar} E_{2}t} \psi_{2} [/tex]

    Which is simplified to simply:

    [tex] \psi_{2}(t)=e^{\frac{V_{0}t}{i\hbar}} \psi_{1} [/tex]

    -----

    My only question for that approach then is how can I assume (1) if that has not been suggested thus far in the text? I mean, the time-independent schrodinger eqn hasn't even been introduced, and yet it is assumed that this problem is answerable with the material presented thus far (which is simply the schrodinger eqn, how to normalize it, the expectation values of x and p, and the rate of change of the expecation values over time)?
     
  10. Apr 12, 2005 #9

    dextercioby

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    If u don't use that essential commutation relation between the 2 hamiltonians,u can't make the connection between the 2 wavefunctions...That's because the 2 spectral equations are different,because the operators are different.

    U assumed that [itex]\psi_{1}[/itex] and [itex] \psi_{2} [/itex] are eigenfunction of the same operator,which is incorrect.

    I find the two solutions provided enough to get full mark.Especially the first,which is elegant.

    Daniel.
     
  11. Apr 12, 2005 #10

    dextercioby

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    You don't know that the new spectral equation has the same eigenstates [itex] \psi_{n} [/itex] as the the old one,unless proving those 2 hamiltonians commute...Which is basically my second solution.

    Daniel.

    EDIT:Status X,why did u delete your post?? :surprised
     
  12. Apr 12, 2005 #11

    StatusX

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    I think this is the best you can do given you haven't gotten into operators or commutators or anything. I assume you have covered seperation of variables and the time-independent schrodinger equation:

    [tex]\frac{-\hbar^2}{2m}\frac{\partial^2}{\partial x^2} \psi_n(x) +V(x) \psi_n(x) = E_n \psi_n(x) [/tex]*

    Now, assume every possible solution can be expressed as a linear combination of these solutions, with the appropriate time factor tagged on:

    [tex] \Psi_1(x,t) = \sum_{n=1}^{\infty} c_n \psi_n(x) e^{-i E_n t/\hbar} [/tex]

    Now look at the new equation:

    [tex]\frac{-\hbar^2}{2m}\frac{\partial^2}{\partial x^2} \psi_n(x) +(V(x)+V_0) \psi_n(x) = (E_n)' \psi_n(x) [/tex]

    which can be rewritten as:

    [tex]\frac{-\hbar^2}{2m}\frac{\partial^2}{\partial x^2} \psi_n(x) +V(x)\psi_n(x) = ((E_n)' -V_0)\psi_n(x)[/tex]*

    Now, since the nth eigenstate for the first potential had an eigenvalue of En, the En' here will be related to the old eigenvalues by:

    [tex](E_n)' = E_n +V_0[/tex]

    So:

    [tex] \Psi_2(x,t) = \sum_{n=1}^{\infty} c_n \psi_n(x) e^{-i((E_n)') t/\hbar} = \sum_{n=1}^{\infty} c_n \psi_n(x) e^{-i(E_n+V_0) t/\hbar} =e^{(-i V_0 t/\hbar)} \sum_{n=1}^{\infty} c_n \psi_n(x) e^{-iE_n t/\hbar}[/tex]

    [tex] \Psi_2(x,t) = e^{(-i V_0 t/\hbar)} \Psi_1(x,t) [/tex]

    EDIT: Sorry, I realized I made a mistake and I wasn't sure how long it would take to fix it. It should be right now. As for your comment, the operator on the left sides of the starred equations above are identical, so they have identical eignfunctions and eigenvalues.
     
    Last edited: Apr 12, 2005
  13. Apr 13, 2005 #12
    StatusX:

    Unfortunately, both seperation of variables and the time-independent Schrodinger equation have both not been formally covered...

    Sheesh; I mean, if there are all these methods of solving this problem using those concepts, then why was the question asked before they are even presented? Is there any way to solve it without them, or is this author making seriously skewed assumptions about what can be answered with the presented concepts thus far? Am I just misreading this or something? Here is the word-for-word question, just like in the link:

    "Suppose you add a constant [itex] V_0 [/itex] to the potential energy (by "constant" I mean independent of x as well as t). In classical mechanics this doesn't change anything, but what about quantum mechanics? Show that the wave function picks up a time-dependent phase factor: [itex] exp(-i V_{0} t / \hbar ) [/itex]. What effect does this have on the expectation value of a dynamical variable?"

    dextercioby:

    I appreciate your efforts - it's clear that by Ch 2, I could use the approach presented by your second solution. Until then, I'm not sure what this author is thinking.

    --------

    In general, if there is no way to solve this without using some of the techniques you two have presented, I'll just go ahead and use one of them, explaining that there was no other way to approach the problem.
     
  14. Apr 13, 2005 #13

    StatusX

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    If you don't know any ways to solve the equation, I can't think of a way to derive the result. You can check that it works, though, by plugging the new solution into the new equation. Just be careful to label the new and old wavefunctions, and use the fact that the left and right sides of the first equation are equal when you plug in the old solution.
     
  15. Apr 13, 2005 #14

    dextercioby

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    Nope,Status X.They're different operators acting on different (a priori) wavefunctions...I can say that your third equation is incorrect,as it is not justified...


    Daniel.
     
  16. Apr 13, 2005 #15

    dextercioby

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    I have the book and looked through it.Ha hasn't covered the material to account for a proof.One bad thing for the book... :yuck:

    Daniel.
     
  17. Apr 13, 2005 #16
    Alright, thanks for your help. I may try mimicking the seperation of variable technique in the next chapter to reach a point where I can use this approach anyway. If not, I'll just plug in like you suggested.

    well, after spending w-a-y too much time on this problem, I'm off to bed... thanks to both of you for all your help
     
  18. Apr 13, 2005 #17
    Yeah, really. This book is terrible about correlating the material with the questions. Even Prob 1.3 took me forever (my professor, a M.A. in theoretical physics couldn't even initially figure it out). Hopefully it gets better as the chapters progress, but so far I'm not digging the approach. I have had other overviews before that were better organized than this.
     
  19. Apr 13, 2005 #18

    dextercioby

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    What can i say,"good luck" with & post all the problems u encounter.I'm sure someone around here would know a bit about QM & would be willing to help.


    Daniel.
     
  20. Apr 13, 2005 #19

    StatusX

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    My third equation (by which I assume you mean the third line of latex) is just the time independent schroedinger's equation for the new potential. The En' are just the energies for this new hamiltonian, and are not initially connected to any of the previous equations. Moving V0 to the other side, the left side equation obviously has the same eigenfunctions and eigenvalues as the left side of the first equation, since they are one in the same. I really don't see any problems here.
     
  21. Apr 13, 2005 #20

    dextercioby

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    Nope,that equation should HAVE OTHER WAVEFUNCTIONS,becasue the operator is different...A priori.Then,u can equate those sets of eigenfunctions,by proving the 2 hamiltnians commute,ergo form a csco...

    Daniel.
     
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