Why is this NOT a solution to the time independent Schrodinger equation.

[pablo4429]

Homework Statement

Given a set of solutions |psi_n> to the TISE show that

a) |Capital-psi>=Sum(C_n|psi_n>) is not a solution to the TISE

Homework Equations

***H|psi_n>=E_n|psi_n>

The Attempt at a Solution

I don't really know where to start, I plugged |psi_n> into *** and I am not sure what is supposed to happen or if I am even supposed to do that.

Thanks all and sorry about the lack of tex formatting

 

[diazona]

Think about it this way: the equation involves an operator H. You already know how H acts on a state that can be labeled |\psi_n\rangle. But what other state do you have that you could apply H to?

 

[pablo4429]

Could it also operate on a linear combination of psi, i.e. c_n |psi_n>?

 

[ideasrule]

Yes. What do you get when you do that? Can you write H|Capital-psi> as some eigenvalue times |Capital-psi>?

 

[pablo4429]

I really don't know, this is a guess

|Capital_psi>=sum(|a><a|psi(t=0)>)

if that is right I don't know why...

 

[pablo4429]

or

|capital_psi>=sum(a_i|i>)

where a is the eigenvalue

 

[pablo4429]

does it need an exp(-iEt/h) tagged on the back of it for it to be a solution. Then the SE is separable and we can turn it into the equation I put in the "2" section of the first post?

 

[pablo4429]

So I think I get it, I hope

Without the time term, the *** equation above has t and x dependence all wrapped together which isn't explicitly solvable so the addition of that terms allows them all to cancel and separate the time stuff on one side which just turns into E and the spatial stuff on the other, the Hamiltonian acting on psi. If this is right them the damn formalism of the BraKet notation is what was confusing me.

 

[diazona]

No, I don't think that's it. The time independent Schrodinger equation (equation *** in your original post) doesn't involve time at all. That's why it's called time independent.

What do you get when you apply the operator H to |\Psi\rangle? Remember that
|\Psi\rangle = \sum_n c_n|\psi_n\rangle

 

[pablo4429]

No, I don't think that's it. The time independent Schrodinger equation (equation *** in your original post) doesn't involve time at all. That's why it's called time independent.

What do you get when you apply the operator H to |\Psi\rangle? Remember that
|\Psi\rangle = \sum_n c_n|\psi_n\rangle

I have no idea then! I don't know what you get. There is a d^2/dx^2(sum(c_n*psi_n)) and V(x)*sum(c_n*psi_n) but I am pretty sure that is not right and if it is I don't know what it means or where to go at all

 

[pablo4429]

I have to show why that (what you wrote) is not a solution and I simply don't see why or where I will show that

 

[diazona]

Try building up to it, then. Suppose that you have an operator which is defined to satisfy the relation
A|\psi_n\rangle = n(n + 1)|\psi_n\rangle
Consider the following questions in order:

What do you get when you apply the operator A to this state?
|\phi\rangle = c_i|\psi_i\rangle
Is the result a multiple of |\phi\rangle?

What do you get when you apply A to this state?
|\Phi\rangle = c_i|\psi_i\rangle + c_j|\psi_j\rangle
Is the result a multiple of |\Phi\rangle?

What do you get when you apply A to this state?
|\varphi\rangle = \sum_m c_m|\psi_m\rangle
Is the result a multiple of |\varphi\rangle?

Now what if you replace the defining relation for A with this?
A|\psi_n\rangle = E_n|\psi_n\rangle
What would A|\varphi\rangle be now?

Once you understand these, go back and do the same thing with H.