Schrodinger equation in matrix form

[sensou]

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.

 

[Hyperreality]

I recommend you to use Latex, it is much more easier
https://www.physicsforums.com/showthread.php?t=8997".

I can't really decipher what you have wrote

 

[Physics Monkey]

Hi sensou,

I will give you a few hints on where to start. You obviously want to begin with the Schrodiner equation: i \hbar \frac{d}{dt} |\psi(t)\rangle = H | \psi(t) \rangle 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 |\psi(t) \rangle = \sum_n c_n(t) |\psi_n \rangle.

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

Now a question for you: Given a certain vector |\psi(t)\rangle, how do you find the coeffecients c_n(t)?

 

[sensou]

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
i \hbar \frac {d c_n(t)} {dt} = \sum_m H_n_m c_m(t)
H_n_m = \int_{-\infty}^{+\infty} \psi_n^* \hat{H} \psi_m \,dx
c_n(t) = \int \psi_n^* \Psi(x , t) \,dx
after a few steps i get:
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

there is a dot on the first c_m(t) but it is difficult to see

 

[nrqed]

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
c_n(t) = \int \psi_n^* \Psi(x , t) \,dx

That is right.

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

Now multiply everything (both right and left sides) of the equation by \psi_m(x) and integrate over x and you will get the equation you had to prove (notice that \sum_n c_n(t) \int \psi_m(x) \psi_n(x) = \sum_n \delta_{n,m} c_n(t) = c_n(t)

 

[sensou]

so, 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)} ?
if it is the i \hbar \dot{c_n(t)} 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 \delta_{n,m} that makes it ok. If m=n they are the same thing so \dot{c_n(t)} is the same as \dot{c_m(t)}
Thanks for all the help, especially if this is correct!

 

[Physics Monkey]

sensou,

That sum does indeed give i \hbar \dot{c}_n(t) because of the Kronecker delta which sets m = n, so you got it.

 

[nrqed]

so, 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)} ?
if it is the i \hbar \dot{c_n(t)} 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 \delta_{n,m} that makes it ok. If m=n they are the same thing so \dot{c_n(t)} is the same as \dot{c_m(t)}
Thanks for all the help, especially if this is correct!

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

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

but one should not sya that the result (the right hand side) is \dot{c_n} *OR* \dot{c_m}... 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 i \hbar \dot{c_{42}}. So for an arbitrary n, the result is i \hbar \dot{c_n}. 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

 

[sensou]

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 received was so good.

 

[nrqed]

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 received was so good.

You are most welcome.

Yes, please pass over the favor to others!