Simple QM Question: Eigenfunctions and Eigenvalues

[h0dgey84bc]

Hi,

Say if we have two energy eingenfunctions and correpsonding eigenvalues, as specified by:
H \psi_1 =E_1 \psi_1
H \psi_2 =E_2 \psi_2

I have written in my notes that since the Hamiltonian operator H is linear, then \phi=C_1 \psi_1 + C_2 \psi_2 must also be a solution. But

H \phi =H(C_1 \psi_1 + C_2 \psi_2) =C_1 E_1 \psi_1 +C_2 E_2 \psi_2 =E_1 C_1 \psi_1 + E_2 C_2\psi_2 != E \phi, where E is some constant. (!= is supposed to represent NOT equal to)

so how can this combination also be a solution (assuming the eigenfunctions are non degenerate)?

thanks

 

[George Jones]

I have written in my notes that since the Hamiltonian operator H is linear, then \phi=C_1 \psi_1 + C_2 \psi_2 must also be a solution.

This is true for solutions of the time-dependent Schrodinger equation, but it not, in general, true for solution of the time-independent Schrodinger equation. Consequently, the linear combination of two stationary states is (usually) not a stationary state.

 

[atyy]

H \psi_1 = E \psi_1

The above equation is linear - this is why degenerate eigenvalues can occur.

H \psi_1 =E_1 \psi_1
H \psi_2 =E_2 \psi_2

The above equations are two different linear equations, because E_1 and E_1 are different, so there is no reason for the sum of their solutions to remain solutions.

 

[Vanadium 50]

Note that C_1 \psi_1+ C_2 \psi_2 is a solution. However, it is (in general) not an energy eigenvalue. In fact, if one measured the energy, one would find a probability C_1^2/(C_1^2 + C_2^2) that it has energy E_1 and C_2^2/(C_1^2 + C_2^2) that it has energy E_2.

 

[h0dgey84bc]

ah thank you both! makes perfect sense now...

 

[Marty]

Note that C_1 \psi_1+ C_2 \psi_2 is a solution. However, it is (in general) not an energy eigenvalue. In fact, if one measured the energy, one would find a probability C_1^2/(C_1^2 + C_2^2) that it has energy E_1 and C_2^2/(C_1^2 + C_2^2) that it has energy E_2.

I agree with your answer as it pertains to the OP's question, but you touch on a sore point that went unresolved in a recent thread. Is this business of measuring the energy just a hypothetical argument, or is it actually possible to measure the energy of such a state? In the previous discussion we were considering the case of a hydrogen atom in a superposition of states.

 

[Vanadium 50]

Is this business of measuring the energy just a hypothetical argument, or is it actually possible to measure the energy of such a state?

It's possible. The difficulty is preparing a state like the OP described.

 

[Marty]

It's possible. The difficulty is preparing a state like the OP described.

I guess when I asked "is it possible to measure such a state" I wasn't looking for a yes or no answer; the implication was that I was looking for a description of how you would actually do it.

As for preparing such states, I would think they are the rule rather than the exception in nature. Any perturbation of a pure energy eigenstate will result in that state going into a superposition with other states mixed in.

 

[Vanadium 50]

This is getting pretty far afield of the OP's original question, and I have no idea what your objection is to the idea of measuring energy. I had thought it was a problem with conservation of energy with a mixed state, but I guess it's not. Since there's no short answer, I suggest we end this thread hijack.

 

[Marty]

This is getting pretty far afield of the OP's original question, and I have no idea what your objection is to the idea of measuring energy.

Then why don't you just tell me how it's done?

I had thought it was a problem with conservation of energy with a mixed state, but I guess it's not. Since there's no short answer, I suggest we end this thread hijack.

Why don't you just ban me from the group altogether?