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Schrodinger equation in matrix form

  1. Apr 22, 2006 #1
    I have been asked to show that the Schroding equation is equivalent to:
    i(hbar)d/dt(cn(t))=sum over m (Hnm*cm(t))
    where Hnm=integral over all space of (complex conjugate of psin)*Hamiltonian operating on psim

    psi=sum over n (cn(t)*psin)
    But i don't know how to even start this question.
     
  2. jcsd
  3. Apr 22, 2006 #2
  4. Apr 22, 2006 #3

    Physics Monkey

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    Hi sensou,

    I will give you a few hints on where to start. You obviously want to begin with the Schrodiner equation: [tex] i \hbar \frac{d}{dt} |\psi(t)\rangle = H | \psi(t) \rangle [/tex] in bra-ket notation (if you are unfamiliar with this notation, let me know). You have also expanded your wavefunction in terms of a basis set [tex] |\psi(t) \rangle = \sum_n c_n(t) |\psi_n \rangle [/tex].

    Here is the hint: The coeffecients [tex] c_n(t) [/tex] are time dependent, while the basis vectors [tex] |\psi_n \rangle [/tex] are independent of time.

    Now a question for you: Given a certain vector [tex] |\psi(t)\rangle [/tex], how do you find the coeffecients [tex] c_n(t) [/tex]?
     
  5. Apr 22, 2006 #4
    i haven't used bra-ket notation before but i can understand what you wrote
    i will try to use latex code now hopefully it will turn out correctly
    what i wrote in the initial question was
    [tex] i \hbar \frac {d c_n(t)} {dt} = \sum_m H_n_m c_m(t) [/tex]
    [tex] H_n_m = \int_{-\infty}^{+\infty} \psi_n^* \hat{H} \psi_m \,dx [/tex]
    [tex] c_n(t) = \int \psi_n^* \Psi(x , t) \,dx [/tex]
    after a few steps i get:
    [tex] i \hbar \int_{-\infty}^{+\infty} \sum_m \dot{c_m} (t) \psi_n^* \psi_m \,dx = \sum_m c_m(t) \int_{-\infty}^{+\infty} \psi_n^* \hat{H} \psi_m \,dx[/tex]

    there is a dot on the first [tex] c_m(t) [/tex] but it is difficult to see
     
    Last edited: Apr 23, 2006
  6. Apr 22, 2006 #5

    nrqed

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    That is right.

    So the way to do it is this:
    Start with Schrodinger's equation applied to Psi(x,t). Write
    [itex] \Psi(x,t) = \sum_n c_n(t) \psi_n(x) [/itex].

    Now multiply everything (both right and left sides) of the equation by [itex] \psi_m(x)[/itex] and integrate over x and you will get the equation you had to prove (notice that [itex] \sum_n c_n(t) \int \psi_m(x) \psi_n(x) = \sum_n \delta_{n,m} c_n(t) = c_n(t) [/itex]
     
  7. Apr 22, 2006 #6
    so, [tex] i \hbar \int \sum_m \dot{c_m} (t) \psi_n^* \psi_m \,dx = i \hbar \sum_m \delta_{n,m} \dot{c_m(t)} = i \hbar \dot{c_m(t)} or \dot{c_n(t)} ?[/tex]
    if it is the [tex] i \hbar \dot{c_n(t)} [/tex] then my answer works out
    but i don't think the fact that it is c_n rather than c_m makes sense to me
    or is it because of the [tex] \delta_{n,m} [/tex] that makes it ok. If m=n they are the same thing so [tex] \dot{c_n(t)} [/tex] is the same as [tex] \dot{c_m(t)} [/tex]
    Thanks for all the help, especially if this is correct!
     
    Last edited: Apr 22, 2006
  8. Apr 23, 2006 #7

    Physics Monkey

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    sensou,

    That sum does indeed give [tex] i \hbar \dot{c}_n(t) [/tex] because of the Kronecker delta which sets [tex] m = n [/tex], so you got it.
     
  9. Apr 23, 2006 #8

    nrqed

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    Yes, this is correct. But let me explain a little.

    Indeed, [itex] \sum_m \delta_{n,m} \dot{c_m(t)} = \dot{c_n(t)} [/itex]

    but one should not sya that the result (the right hand side) is [itex]\dot{c_n} [/itex] *OR* [itex]\dot{c_m} [/itex]... It can only be with the expression with the index *n*. Notice that m is summed over (it is then called a dummy index). After the summation is done, no m should appear in the final expression (it's a bit like a definite integral over x...the final answer should not contain x). What the delta does is that it ''kills'' (sets to zero) al the terms where m is not equal to n, leaving only the term in the sum where m is equal to n.

    The way to think about the equation then is :for a given n (let`s say 42), what is the result of doing this integral. The result is [itex]i \hbar \dot{c_{42}} [/itex]. So for an arbitrary n, the result is [itex]i \hbar \dot{c_n} [/itex]. You might say ''well, the delta function sets m=n anyway so hwy can't I write the answer with ''m''. This is a common source of confusion. The problem is that if I give you a value fo n an dyou do the calculation and send me back the answer with an index m, I will say ''what do you mean by m? I gave you a value of n, I have no idea what m stands for!'' And you will say ''well, m is equal to n''. And I will say ''then why not writing n!?!?''

    This may sound as illy point but to be more serious, when you do long calculations with a lt of indices and summation and so on, there is only one way to correctly do it when there is a Kronecker delta summed over: you *must* set the dummy index (the one summed over) equal to the value imposed by the delta. The resulting expression (after the sum has been carried out) muts *NOT* contain the dummy variable anymore.


    Regards

    Patrick
     
    Last edited: Apr 23, 2006
  10. Apr 23, 2006 #9
    ahh i think i understand now. Thanks for the deeper insight into the question and helping me understand it better. I am new to this forum and i plan to stay and help others if i can since the help i recieved was so good.
     
  11. Apr 23, 2006 #10

    nrqed

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    You are most welcome.

    Yes, please pass over the favor to others!!
     
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