# Homework Help: Quantum Operators (or just operators in general)

1. Feb 15, 2010

### Plutoniummatt

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

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

2. Relevant equations

Non?

3. The attempt at a solution

attempt at a)
from the fact that <$$\phi_1$$ | a$$\phi_1$$ + b$$\phi_2$$> = 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$$\phi_1$$ + b$$\phi_2$$ | a$$\phi_1$$ + b$$\phi_2$$> = 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

2. Feb 15, 2010

### kuruman

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

3. Feb 15, 2010

### vela

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

4. Feb 15, 2010

### Plutoniummatt

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

Sorry I havent really gotten grips with this operator stuff.

5. Feb 15, 2010

### 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

$$a|\phi_1>+\sqrt{1-a^2}| \phi_2>$$

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

6. Feb 15, 2010

### Plutoniummatt

How is this proved?

7. Feb 15, 2010

### vela

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

8. Feb 15, 2010

### kuruman

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

*** Correction ***

Last edited: Feb 15, 2010
9. Feb 15, 2010

### Plutoniummatt

Orthonormal?

10. Feb 15, 2010

### 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.

11. Feb 15, 2010

### Plutoniummatt

a2 + b2 = 1

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

12. Feb 15, 2010

### 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.

13. Feb 15, 2010

### Plutoniummatt

How do we know for sure the coefficients are real?

14. Feb 15, 2010

### vela

Staff Emeritus
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.

15. Feb 15, 2010

### 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?

16. Feb 15, 2010

### vela

Staff Emeritus
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.

17. Feb 15, 2010

### 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?

18. Feb 15, 2010

### vela

Staff Emeritus
The normalization condition always turns out to be real because the squared terms always involve the multiplication of a complex number by its conjugate and the cross terms are complex conjugates of each other so when they combine, the imaginary part cancels out.

In this problem, because c is assumed to be real, a and b can be real simultaneously. You could always multiply the state by an arbitrary phase and get complex coefficients if you want, but the relative phase of a and b will remain unchanged.

19. Feb 15, 2010

### Plutoniummatt

Thanks guys! have been very helpful!

20. Feb 15, 2010

### kuruman

They don't have to be. You can use coefficients

$$ae^{i\phi_a}; \:be^{i\phi_b}$$

where a and b are real. Then the normalization condition will be in terms of a, b and the (arbitrary) phase difference φab. Still two equations and two unknowns.