1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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
    Last edited by a moderator: Apr 22, 2017
  4. Apr 22, 2006 #3

    Physics Monkey

    User Avatar
    Science Advisor
    Homework Helper

    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


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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

    User Avatar
    Science Advisor
    Homework Helper


    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


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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.


    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


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    You are most welcome.

    Yes, please pass over the favor to others!!
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Schrodinger equation in matrix form
  1. Schrodinger equation (Replies: 6)

  2. Schrodinger equation (Replies: 9)

  3. Schrodinger Equations (Replies: 1)

  4. Schrodinger Equation (Replies: 11)