Solution of time-dependent Schroedinger Equation

[Vagant]

Hello,

I am interested - is the solution of the time-dependent Schroedinger Equation on the picture attached correct? Are there any flaws in this solution?

Thank you. :)

 

[genneth]

You can't just divide by the wavefunction, so I'm afraid it's not quite correct.

 

[Vagant]

You can't just divide by the wavefunction, so I'm afraid it's not quite correct.

..but how it can be solved then?!

 

[genneth]

In general, this problem is very hard -- since the equation describes a huge amount of physics, a uniform solution to it would make life very much easier. Unfortunately, you have to solve it on a case-by-case basis, depending on what the hamiltonian H is. It is possible to write down formal solutions, but they are only a starting point, and not actually useful for getting the numbers out.

I know of two ways:

1) Find the eigenstates of H. These form a complete orthonormal basis, so it's possible to expand an arbitrary state as a linear combination of these. The equation is linear, and the time evolution of the eigenstates is easy to calculate (you can just use the method you used, since the operator H gets replaced by the eigenvalue). Then the state at some time t is just the same linear combination of the eigenstates at that time t.

2) Integrate both sides, and obtain an integral equation. Recursively substitute the left into the integrand, and obtain an infinitely nested set of integrals. Realise that this is a time-ordered path integral, and manipulate to obtain the propagator (some horribly messy looking thing).

As I said, neither actually solves the problem. You have to solve it for each H.

 

[pam]

It looks to me that your answer is correct if |psi> is an eigenstate of H.
It is also correct if you expand |psi> in eigenstates of H.
What other cases are there?

 

[Vagant]

It looks to me that your answer is correct if |psi> is an eigenstate of H.
It is also correct if you expand |psi> in eigenstates of H.
What other cases are there?

|psi> is a linear combination of eigenstates of H indeed. :)

 

[Vagant]

You can't just divide by the wavefunction, so I'm afraid it's not quite correct.

Even if ket |psi> is a linear combination of eigenkets of H?

 

[genneth]

|psi> is a linear combination of eigenstates of H. So, I can divide by |psi> in this case? Is such division - from (1) to (2), valid?

Like I said, for any Hamiltonian, if you know the eigenvectors and eigenvalues, you can express an arbitrary wavefunction as a linear combination:

Given the states $$\left|\psi_n\right>$$ where $$\hat{H}\left|\psi_n\right>=E_n \left|\psi_n\right>$$, you can write:

$$\left|\Psi\right> = \sum_n c_n \left|\psi_n\right>$$.

The time evolution of the eigenstates are easy:

$$i\hbar \frac{d\left|\psi_n(t)\right>}{dt} = \hat{H}\left|\psi_n(t)\right> = E_n \left|\psi_n(t)\right>$$

which has solutions: $$\left|\psi_n(t)\right> = e^{-iE_n t/\hbar} \left|\psi_n(0)\right>$$, because $$E_n$$ is just a number and not an operator, so you can use the usual algebraic manipulations.

Finally, since the Schroedinger equation is linear, you have that:

$$\left|\Psi(t)\right> = \sum_n c_n \left|\psi_n(t)\right> = \sum_n c_n e^{-iE_n t/\hbar} \left|\psi_n(0)\right>$$.

Note that crucially you need to be able to find the eigenstates and eigenvalues of the Hamiltonian, which is usually quite hard.

 

[Vagant]

Like I said, for any Hamiltonian, if you know the eigenvectors and eigenvalues, you can express an arbitrary wavefunction as a linear combination:

Given the states $$\left|\psi_n\right>$$ where $$\hat{H}\left|\psi_n\right>=E_n \left|\psi_n\right>$$, you can write:

$$\left|\Psi\right> = \sum_n c_n \left|\psi_n\right>$$.

The time evolution of the eigenstates are easy:

$$i\hbar \frac{d\left|\psi_n(t)\right>}{dt} = \hat{H}\left|\psi_n(t)\right> = E_n \left|\psi_n(t)\right>$$

which has solutions: $$\left|\psi_n(t)\right> = e^{-iE_n t/\hbar} \left|\psi_n(0)\right>$$, because $$E_n$$ is just a number and not an operator, so you can use the usual algebraic manipulations.

Finally, since the Schroedinger equation is linear, you have that:

$$\left|\Psi(t)\right> = \sum_n c_n \left|\psi_n(t)\right> = \sum_n c_n e^{-iE_n t/\hbar} \left|\psi_n(0)\right>$$.

Note that crucially you need to be able to find the eigenstates and eigenvalues of the Hamiltonian, which is usually quite hard.

Thank you! It's very clearly presented and helpful indeed. Luckily, I know eigenstates of my Hamiltonian, there are two eigen states only (on picture attached).

 

[Gigi]

It is solved in David Griffiths, Introduction to quantum mechanics. In the case of the time-dependent SE, psi is a function of x and t. Thus you use separation of variables ending with the time independent SE, whose solution is psi(x) as well as a time dependent ordinary differential equation that will give you the time factor. Try it.

 

[HOZEFATINWALA]

I am in need of solution of time dependent schrodinger equation in fortran 77 version, can u help?

 

[HOZEFATINWALA]

Can anyone help me in solving the time dependent schrodinger equation in fortran program?