Quantum Operators (or just operators in general)

In summary: 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?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.
  • #1
Plutoniummatt
46
0

Homework Statement



[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]

Homework Equations



Non?

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 can't be satisfied simultaneously? Where am i going wrong? thanks
 
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  • #2
Plutoniummatt said:
... 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.
 
  • #3
You could try using Gram-Schmidt to solve the problem.
 
  • #4
kuruman said:
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[tex]\phi_1[/tex] + b[tex]\phi_2[/tex] | a[tex]\phi_1[/tex] + b[tex]\phi_2[/tex]> = 1


Sorry I haven't really gotten grips with this operator stuff.
 
  • #5
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.
 
  • #6
kuruman said:
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]



How is this proved?
 
  • #7
kuruman said:
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.
That's only if they're orthogonal, right?
 
  • #8
Correct. That's only if they are orthogonal. My mistake.

*** Correction ***

Should read orthonormal.
 
Last edited:
  • #9
vela said:
That's only if they're orthogonal, right?

Orthonormal?
 
  • #10
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
kuruman said:
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?
 
  • #12
Plutoniummatt said:
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.
 
  • #13
kuruman said:
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?
 
  • #14
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
vela said:
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?
 
  • #16
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
vela said:
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?
 
  • #18
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
Thanks guys! have been very helpful!
 
  • #20
Plutoniummatt said:
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

[tex]ae^{i\phi_a}; \:be^{i\phi_b}[/tex]

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.
 
  • #21
Finally, vela and I converged. :smile:
 
  • #22
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
 
  • #23
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.
 

1. What is a quantum operator?

A quantum operator is a mathematical representation of an observable quantity in quantum mechanics. It is represented by a linear transformation of the state vector of a quantum system, and it operates on the wavefunction of the system to determine the possible outcomes of a measurement.

2. How are quantum operators used in quantum mechanics?

Quantum operators are used to describe the behavior of quantum systems and predict the outcomes of measurements. They are also used in calculations to determine the probabilities of different states of a system and the evolution of a system over time.

3. What are some examples of quantum operators?

Some common examples of quantum operators include the position operator, momentum operator, and energy operator. These correspond to the observable quantities of position, momentum, and energy, respectively, in a quantum system.

4. How are quantum operators different from classical operators?

Quantum operators differ from classical operators in that they operate on wavefunctions rather than on classical variables. They also follow different rules and properties, such as being non-commutative, and can have complex eigenvalues.

5. Can quantum operators be measured or observed directly?

No, quantum operators cannot be directly measured or observed. They are mathematical representations of observable quantities and are only used to predict the outcomes of measurements on a quantum system.

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