Quantum Operators (or just operators in general)

[Plutoniummatt]

Homework Statement

$$\phi_1$$ 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) $$\phi_1$$+$$\phi_2$$

Homework Equations

Non?

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 can't be satisfied simultaneously? Where am i going wrong? thanks

 

[kuruman]

... if I let b = 1, ...

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]

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

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 haven't 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

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

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

 

[Plutoniummatt]

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

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

How is this proved?

 

[vela]

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.

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

 

[kuruman]

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

*** Correction ***

Should read orthonormal.

 

[Plutoniummatt]

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

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]

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.

a² + b² = 1

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

 

[kuruman]

a² + b² = 1

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

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]

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.

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]

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.

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]

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.

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?

 

[vela]

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.

 

[Plutoniummatt]

Thanks guys! have been very helpful!

 

[kuruman]

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?

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 φ_a-φ_b. Still two equations and two unknowns.

 

[kuruman]

Finally, vela and I converged.

 

[Plutoniummatt]

ok guys, so:

a + bc = 0


aa* + ab*c + ba*c + bb* = 1
2aa* + bb* - ccbb* = 1

are my equations, where do I go from here? If a and b are in general complex, how do I solve for them? does the phase not matter like whatsoever? so I can just write aa* as a^2

 

[vela]

Use the first equation to eliminate a from the second equation. You should write bb* = |b|². The best you can do is get the relative phase between a and b, so you can assume one is real.