# I About general solutions to Schrodinger equation

1. Mar 1, 2017

### betelgeuse91

Hi,

I am wondering why every general solution to Schrodinger equation can be built from separable solutions. In other words, I don't follow that why every solution to Schrodinger equation can be written as
$$\Psi(x,t) = \sum c_n\Psi_n(x,t)=\sum c_n\psi_n(x)\phi_n(t)$$
I know that the right hand side is a solution to Schrodinger equation but this does not mean that every solution should be of this type. I also know that separable solutions form eigenbasis of time-independent Schrodinger equation but the above fact still does not follow, as they only span the solution space of the time-independent Schrodinger equation, not the general time-dependent Schrodinger equation.

What am I missing here? Thank you for your help.

2. Mar 1, 2017

### blue_leaf77

To the best of my knowledge, the above equation only applies for time-independent SE. From where did you read that it also applies for the time-dependent one? Did you not miss that the context the source is made on is implicitly the time-independent case?

3. Mar 1, 2017

### PeroK

It's usually taken as an assumption that the eigenfunctions of the given Hamiltonian, ($\psi_n$) form a complete set (*), in the sense that any function can be expressed as a linear combination of them. So, whatever the wavefunction is initially it can be expressed as a linear combination of these functions.

Each of these components evolves over time with the appropriate time-component ($\phi_n(t)$), so the function evolves as the linear combination you have given.

That process covers all the possibilities.

You could, however, potentially solve the Schodinger equation another way and get a different looking solution. But, whetever solution you got would be equivalent to one of the general form given.

(*) PS somewhere in the theory of PDE's there will be a proof of this.

4. Mar 1, 2017

### dextercioby

Well, the time evolution is a tricky problem, because it assumes a parametrization of a family of self-adjoint operators, thus a parametrization of the solutions of spectral equations. Normally, time-dependent Hamiltonians don't commute at different times, thus, for each value of "t", you need to solve the spectral equation of the Hamiltonian evaluated at "t". That's why one adopts the approximate method called Dyson series to evaluate how a solution of the spectral equation of $H(t_0)$ evolves into a solution of the spectral equation of $H(t_1)$.

The Ansatz $\Psi (x,t) = \sum \int c_{n,\alpha} \psi_{n,\alpha} (x,t) = \sum \int c_{n,\alpha} \psi_{n,\alpha} (x)\phi (t)$ holds only when the Hamiltonian is time independent.

Last edited: Mar 1, 2017
5. Mar 1, 2017

### PeroK

Perhaps the OP can correct me if I'm wrong, but I think he is using the time-dependent Schrodinger equation to mean:

$i\hbar \frac{\partial \Psi}{\partial t} = H \Psi(x, t)$

And, the time-independent equation to be:

$H \psi(x) = E \psi(x)$

He is not talking about a time-dependent Hamiltonian. The above is the terminology in Griffiths, for example.

6. Mar 1, 2017

### betelgeuse91

I found this on Griffiths page 25, "Moreover (as is typically the case with separation of variables) we will be able at the end to patch together the separable solutions in such a way as to construct the most general solution."

7. Mar 1, 2017

See post #3.

8. Mar 1, 2017

### betelgeuse91

I see that the eigenfunctions of the Hamiltonian indeed form a complete set such that any function that satisfies time-independent Schrodinger equation $H\psi=E\psi$ can be written as linear combination of them. I also see that the time evolution of that function corresponds to the time evolution of each of its component. But this is a story of time-independent Schrodinger equation only. How does this make connections to general solutions of time-dependent Schrodinger equation $H\Psi=i\hbar\frac{\partial\Psi}{\partial t}$?

9. Mar 1, 2017

### betelgeuse91

I apologize that I forgot to mention that the Hamiltonian concerned here is time-independent, as PeroK pointed out.

10. Mar 1, 2017

### PeroK

The product $\Psi_n(x, t) = \psi_n(x)\phi_n(t)$ is a solution to the time-dependent equation.

To show you more how this works, let's assume first that the initial state is an eigenfunction:

$\Psi(x, 0) = \psi_n(x)$

Now, that initial solution must evolve according to:

$\Psi(x, t) = \psi_n(x)\phi_n(t)$

Likewise, if the initial state is a sum of two eigenfunctions:

$\Psi(x, 0) = \psi_n(x) + \psi_m(x)$

Then, that function (if it obeys the Schrodinger equation) must evolve according to:

$\Psi(x, t) = \psi_n(x)\phi_n(t) + \psi_m(x)\phi_m(t)$

And, similarly for any sum (including an infinite sum).

So, any solution must be of the general form:

1) take the initial state at time 0, and express this as a linear combination of the eigenfunctions.

2) Apply the appropriate time evolution to each eigenfunction, to give the time-dependent solution.

This relies, of course, on the theory of PDE's about uniqueness of solutions.

11. Mar 1, 2017

### betelgeuse91

It looks like that what you showed here is that the time evolution of linear combination of separable solutions is $\Psi(x,t) = \sum c_n\psi_n(x)\phi_n(t)$. But this only states about evolution of separable solutions, not evolution of general solutions. Can you state the PDE theorem here please? Maybe that's what I am missing.

12. Mar 1, 2017

### PeroK

It's nothing to do with separable solutions, it's to do with uniqueness. Take, for example, the quadratic equation and you get two solutions using the quadratic formula. Now, you're effectively saying: that's solutions from the quadratic formula, but what about other solutions? But, there can be no other solutions, because a quadratic equation has only two. It doesn't matter how you obtain them, once you've got them, they are the only ones.

It's the same with PDE's. If there is a unique solution, then it doesn't matter how you find it (in this case building it up from separable components). Once you've found it, it must be the only solution.

Note, also that the general solution:
$$\Psi(x,t) = \sum c_n\Psi_n(x,t)=\sum c_n\psi_n(x)\phi_n(t)$$
Is not itself a separable function. It is a sum of separable components, but cannot itself be written in the form $\Psi(x,t) = f(x)g(t)$

In other words, it is the most general solution and is not separable.

13. Mar 1, 2017

### betelgeuse91

! I see. So the set of solutions we found as above is indeed the set of all the possible solutions by uniqueness. Thank you very much, now I understand what was going on!

14. Mar 1, 2017

### dextercioby

Well, the story is not so simple. First of all, the Schroedinger equation, when the Hamiltonian is known and expressed in terms of other dynamical observables (let's choose momenta and coordinates, as is the case for the homogenous D-dim. harmonic oscillator or 3-D Hydrogen atom) is a linear partial differential equation of second degree. For example:

$$\frac{\partial\Psi (t,x,y,z)}{\partial t} = \frac{1}{i\hbar} \left(-\frac{\hbar^2}{2m}\Delta \Psi(t,x,y,z) + K (x^2 + y^2 + z^2) \Psi(t,x,y,z) \right)$$

But in mathematics, a PDE (linear or not) is well-formulated only in the presence of initial conditions or boundary conditions. These initial conditions apply to physical equations in which time is a variable, so one needs to have given (i.e. known) $\Psi (t_0,x,y,z) = f(x,y,z)$ on one hand, then Cauchy (i.e. both Dirichlet and Neumann) boundary conditions such as $\Psi (t,\infty,y,z) = 0$ and $\left[ \partial_{x} \Psi (t,x,y,z)\right]_{x=\infty} = 0$. ONLY UNDER THESE conditions, as per the Cauchy-Kovalevskaya theorem, the PDE has a unique solution.

This means that no matter what resolution technique I use (separation of variables, change of variables, method of characteristics, reduction to an ODE through Laplace or Fourier transformation), once I found the solution which solves the equation & the boundary conditions I have, then this is it. Physicists are very fond of separation of variables, but, according to the symmetries (and here Sophus Lie was the pioneer) or other criteria, it's not the only way to solve a PDE.

Last edited: Mar 1, 2017
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