B Build a "full" wave function without data in simple problems

1. Dec 15, 2016

DoobleD

Is it possible to build the full wave function for a simple problem in QM, such as an infinite well, without any experimental data ?

I'm learning about QM, and I saw how to compute energy states (the wave function for each allowed energy level) in some usual QM basic problems. But then, I was always given the probabilities for energy state, so that I could build the "full" wave function (sum of all possible energy state, each multiplied by its probability); OR, I was given a full wave function so that I could compute the probability for each state by taking the inner product of the full wave function with each energy state.

Is it possible to build a full wave function, which is the sum of all possible energy state each with its probability coefficient, from scratch ?

In the case of the infinite well, or an other simple problem, what would be the procedure once you computed the possible energy state ?

Here is the question reformulated below with equations which I think is much clearer :

Last edited: Dec 15, 2016
2. Dec 15, 2016

Staff: Mentor

That doesn't make sense. The eigenstates of the Hamiltonian forms a basis which you can use to describe the state of a given physical system. But that state can be anything, depending on the actual physical situation at hand.

To make a comparison with classical physics: Could you find the position and velocity of a particle from scratch? (What would that even mean?)

3. Dec 15, 2016

DoobleD

Right, I should have said, is it possible without knowing in advance the full wave function or the probability of each eigenstate ?

Of course we always need data like well length etc. I meant, no "solution" data.

4. Dec 15, 2016

Staff: Mentor

If there is an uncertainty in the state, then you have to use the density matrix formalism, but I get the feeling this is not what you are thinking about.

The wave function is a description of the (quantum mechanical) state of a system. The wave function of a particle will depend on what state the particle is in, in what state it was prepared. Again, the analogy with classical physics, where the state of a point particle is given by its position and velocity at time t, which depends on what happened before that time t.

5. Dec 15, 2016

Staff: Mentor

In classical mechanics, given a description of the forces acting on a particle, the position and velocity of the particle at time t depend on its position and velocity at time t=0.

In quantum mechanics, give a description of the "forces acting on a particle" (more precisely its potential energy as a function of position), the particle's wave function at time t, $\Psi(x,t)$, depends on its wave function at time t=0, $\Psi(x,0)$.

In both cases, the state of the particle at time t depends on its state at time t=0 (the initial conditions).

6. Dec 15, 2016

DoobleD

That makes sense, thank you guys. But then I wonder : how do you know the initial conditions in QM ? In classical mechanics I can measure a particule velocity at t = 0, but in QM if I try to measure it I'll have an eigenstate only, and I'll make the wave function collapse.

How does a physicist get the phi(x, 0) in practice ? DrClaude gave an element if answer I think by "the particle is prepared". Can we prepare a particule to be in a state phi(x) without making it an eigenstate ?

7. Dec 15, 2016

Staff: Mentor

No (setting aside the possibility of preparing it in a mixed state, which isn't what you're asking about).

But that's exactly the initial condition you're looking for: the complete wave function at $t=0$. It's an eigenstate of whatever observable you measured to prepare the particle, but it can still be (and in general is) an superposition of the energy eigenstates that you found by solving the Schrodinger equation. So the general recipe is to take the initial state determined by the preparation procedure, write it as a superposition of energy eigenstates each with their own weight, and use that to calculate $\psi(x,t)$.

8. Dec 15, 2016

DoobleD

Awesome. Thank you !

(And long live Physics Forum)