Quantum Operators (or just operators in general)

Plutoniummatt

1. The problem statement, all variables and given/known data

[tex]\phi_1[/tex] and [tex]\phi_2[/tex] are normalized eigenfunctions of observable A which are degenerate, and hence not necessarily orthogonal, if <[tex]\phi_1[/tex] | [tex]\phi_2[/tex]> = c and c is real, find linear combos of [tex]\phi_1[/tex] and [tex]\phi_2[/tex] which are normalized and orthogonal to: a) [tex]\phi_1[/tex]; b) [tex]\phi_1[/tex]+[tex]\phi_2[/tex]

2. Relevant equations

Non?

3. The attempt at a solution

attempt at a)
from the fact that <[tex]\phi_1[/tex] | a[tex]\phi_1[/tex] + b[tex]\phi_2[/tex]> = 0

i got a + bc = 0

if I let b = 1, then a = -c,

then if i use the fact that it must be normalised:
<a[tex]\phi_1[/tex] + b[tex]\phi_2[/tex] | a[tex]\phi_1[/tex] + b[tex]\phi_2[/tex]> = 1,

both equations cant be satisfied simultaneously? Where am i going wrong? thanks
1. The problem statement, all variables and given/known data



2. Relevant equations



3. The attempt at a solution

kuruman

You are forgetting that the wavefunction a|φ₁>+b|φ₂> must be normalized. If you let b = 1, then necessarily a = 0.

vela

You could try using Gram-Schmidt to solve the problem.

Plutoniummatt

Isn't the normalization taken into account when I did:
<a[tex]\phi_1[/tex] + b[tex]\phi_2[/tex] | a[tex]\phi_1[/tex] + b[tex]\phi_2[/tex]> = 1


Sorry I havent really gotten grips with this operator stuff.

kuruman

It has nothing to do with operators; it has to do with normalization. For a two-component linear combination, a normalized wavefunction (with real coefficients) is always

[tex]a|\phi_1>+\sqrt{1-a^2}| \phi_2>[/tex]

Now pick a = 1 and see what happens to this linear combination.

Plutoniummatt

How is this proved?

vela

That's only if they're orthogonal, right?

kuruman

Correct. That's only if they are orthogonal. My mistake.

*** Correction ***

Should read orthonormal.

Plutoniummatt

Orthonormal?

kuruman

However, you have

a + bc = 0 and the normalization condition

a² + b² = 1

Two equations and two unknowns. It is picking b =1 that you cannot do.

Plutoniummatt

a² + b² = 1

Where did you get that from? the cross terms in normalization don't vanish?

kuruman

I seem to have a one-track mind here. Yes, throw in the cross term 2ab (for real coefficients). You still have two equations and two unknowns.

Plutoniummatt

How do we know for sure the coefficients are real?

vela

You have two choices: If you set b=1, then a=-c, and the linear combination you have is orthogonal, but not normalized. So you have to normalize -c|1>+|2>. If you don't set b=1, you can eliminate a using a+bc=0 and then solve for b using the normalization condition. What you can't do is require b=1 in the normalized solution. That's where you're running into problems.

Plutoniummatt

Yeah, I realize now that I cannot do that, is it possible to do it without resorting to the Gram-Schmit process? Kuruman was suggesting using simultaneous equations, but are the coefficients a and b necessarily real?

vela

You don't need to use Gram-Schmidt. It just seemed like an obvious method to me when I first read the problem.

Just pick an approach and normalize the resulting state the usual way.

Plutoniummatt

I end up with conjugates of the coefficients, so I don't have to worry about them? meaning they're real then? If so, how come?