Expectation value with imaginary component?

In summary: This is two.But the need for complex conjugation is not the source of your confusion, is it? You're just not sure what |a+ib|^2 means.In summary, the conversation discusses a problem in quantum mechanics where the state and observation probability both have imaginary components. The solution is to take the squared modulus of the complex number, which eliminates the imaginary part and results in a real probability value. The need for complex conjugation is implicit, but the main issue is understanding what |a+ib|^2 represents.
  • #1
Kenneth Adam Miller
20
0
Hello, I'm a beginner at quantum mechanics. I'm working through problems of the textbook A Modern Approach to Quantum Mechanics without a professor since I am not going to college right now, so I need a brief bit of help on problem 1.10. Everything else I have gotten right so far, but I am having trouble understanding how to apply the examples provided in order to attain an observation probability that does not have an imaginary component.

To explain, the state:

##|phi> = 1/2 * | +z > + i*sqrt(3)/2 * | -z >##

And we wish to know ##<S_x>##, where ##|+x > = 1/sqrt(2)*|+z> + 1/sqrt(2)*|-z>##

Since there is not an imaginary component for |+x>, I don't see how I can calculate ##<+x | phi >##; there is not an i to negate for ##|+x>## to acquire the complex conjugate <+x|. Perhaps I don't understand the complex conjugation part well enough - equal magnitude and opposite sign, yes?

So, then when I'm calculating ##<+x | phi >## according to the example, I wind up with an expectation value with an imaginary component. Which I don't think is correct.

Can anybody help point out where I have gone wrong?
 
Last edited:
Physics news on Phys.org
  • #2
Please use the PF LaTeX feature. You can find help on it in the Info section. It makes posts involving math much easier to read.
 
  • #3
How do you intend to calculate ##\langle S_x \rangle ##?
 
  • #4
@PeroK Well, I was going to calculate the probability of encountering the state X given that it is in state PHI as P, and from that calculate 1 - P to get the probability of encountering state -X given state Phi. From that, I was going to multiply each by h/2 and -h/2 respectively.

@PeterDonis sorry, I'm new to this site, I'll try and edit my post here shortly.
 
  • #5
Kenneth Adam Miller said:
@PeroK Well, I was going to calculate the probability of encountering the state X given that it is in state PHI as P, and from that calculate 1 - P to get the probability of encountering state -X given state Phi. From that, I was going to multiply each by h/2 and -h/2 respectively.

Okay, but how are you going to calculate these probabilities?
 
  • #6
Here's what I have: ##<x | \phi > = (1/\sqrt(2)*<z| + 1/\sqrt(2)<-z|)(1/2 * |z> + i*\sqrt(3)/2*|-z>)##

[Moderator's note: edited for LaTeX formatting.]
 
Last edited by a moderator:
  • #7
Kenneth Adam Miller said:
Here's what I have: ##<x | phi > = (1/sqrt(2)*<z| + 1/sqrt(2)<-z|)(1/2 * |z> + i*sqrt(3)/2*|-z>)##

What does ##\langle x+ | \psi \rangle## give you?
 
  • #8
Kenneth Adam Miller said:
Perhaps I don't understand the complex conjugation part well enough - equal magnitude and opposite sign, yes?

Equal magnitude and opposite sign of the imaginary part. No change to the real part. So if a number is real (zero imaginary part), what would that mean?
 
  • #9
What does ##\langle x \!+ | \psi \rangle## give you?

Hint: it doesn't give you the probablity of getting ##|x \!+ \rangle##. So, if it's not the probability itself, what does it give you?
 
  • #10
@PeterDonis I asked, but if I am right, then there is no change since there is no imaginary part to negate. In which case, how does the imaginary component cancel out of the amplitude.

@PeroK I know what ##<x+|phi>## is; that's the amplitude, and you take ##|A|^2## to get a probability value for that amplitude. Knowing what to calculate is not the problem. When I carry out the calculations, I get a imaginary part remaining, and even if I take ##|A|^2##, I still don't get rid of that part.

The amplitude I get is: ##(1/(sqrt(2)*2) + i*sqrt(3)/(sqrt(2)*2)##
 
  • #11
Kenneth Adam Miller said:
@PeroK I know what ##<x+|phi>## is; that's the amplitude, and you take ##|A|^2## to get a probability value for that amplitude. Knowing what to calculate is not the problem. When I carry out the calculations, I get a imaginary part remaining, and even if I take ##|A|^2##, I still don't get rid of that part.

The amplitude I get is: ##(1/(sqrt(2)*2) + i*sqrt(3)/(sqrt(2)*2)##

Okay. The magnitiude of a complex number must be real. In fact:

##|a + ib|^2 = a^2 + b^2##

You don't need complex conjugation, per se.
 
  • #12
So, just going from ##<x|phi>## to ##|<x|phi>|^2## the imaginary component will be eliminated?
 
  • #13
Kenneth Adam Miller said:
So, just going from ##<x|phi>## to ##|<x|phi>|^2## the imaginary component will be eliminated?

You may need to revise some complex numbers. The magnitude of a complex number should be clear. I posted above what the magintude of a complex number is. I wouldn't describe that as eliminating the imaginary part, any more than eliminating the real part.
 
  • #14
Ok, so if I did the calculations right using what you've said all the way through, I should get:

##
<x | phi> = (1/sqrt(2)*<z+| + 1/sqrt(2)*<-z|)(1/2*|z+> + i*sqrt(3)/2*|-z>) = (1/(2*sqrt(2)) + i*sqrt(3))##

And so
##
|<x | phi>|^2 = (1/8 + 3/8) = 1/2
##
 
  • #15
Kenneth Adam Miller said:
Ok, so if I did the calculations right using what you've said all the way through, I should get:

##
<x | phi> = (1/sqrt(2)*<z+| + 1/sqrt(2)*<-z|)(1/2*|z+> + i*sqrt(3)/2*|-z>) = (1/(2*sqrt(2)) + i*sqrt(3))##

And so
##
|<x | phi>|^2 = (1/8 + 3/8) = 1/2
##

Yes, that's it.
 
  • #16
Kenneth Adam Miller said:
I am right, then there is no change since there is no imaginary part to negate

Yes, because the amplitude you are dealing with is real. And when you take the complex conjugate, what happens to the real part? And what does that imply about the complex conjugate of a real number?
 
  • #17
The real part is unaffected. And that the complex conjugate of a real number is itself unaffected.

Yeah, so this is the first time that I have started exercising this knowledge, so although I've read it several times, I haven't had to apply it.
 
  • #18
PeroK said:
|a+ib|2=a2+b2|a+ib|2=a2+b2|a + ib|^2 = a^2 + b^2

this looks incorrect
 
  • #20
PeterDonis said:
Why?

(ib)^2= - b^2
 
  • #21
ftr said:
(ib)^2= - b^2

That's true, but irrelevant to what was being said. ##|a + ib|^2## is the squared modulus of the complex number ##a + ib##, not its algebraic square.
 
  • #22
Yeah, I struggled with this problem because of the issue with understanding ##|a + ib| ^2## has only two terms. I thought there were three.
 
  • #23
PeroK said:
You don't need complex conjugation

The need is implicit, that is why I thought the OP will be confused.
 
  • #24
Kenneth Adam Miller said:
I struggled with this problem because of the issue with understanding ##|a + ib| ^2## has only two terms.

This is one situation in which it helps to know about the alternate representation of complex numbers in terms of modulus and phase: ##a + ib = r e^{i \theta}##. This works just like polar coordinates vs. Cartesian coordinates in a plane: ##r = \sqrt{a^2 + b^2}##, ##\theta = \tan^{-1} ( b / a )##. Then the squared modulus is just ##r^2##.
 
  • #25
Kenneth Adam Miller said:
Yeah, I struggled with this problem because of the issue with understanding ##|a + ib| ^2## has only two terms. I thought there were three.

There's a big difference between:

##(a+ib)^2 = a^2 - b^2 + 2iab##

And:

##|a +ib|^2 = a^2 + b^2 \ \ ## (which is by definition of the complex modulus - where ##a, b## are real)

Note that it is very useful to know and remember that also:

##|z|^2 = zz^*##

But that's not generally the definition of the complex modulus.
 
  • #26
AH!

##|z|^2 = zz*##

Didn't know this!
 

1. What is the expectation value with an imaginary component?

The expectation value with an imaginary component is a mathematical concept used in quantum mechanics to calculate the average value of a quantum mechanical operator. It takes into account both the real and imaginary components of the operator.

2. How is the expectation value with an imaginary component calculated?

The expectation value with an imaginary component is calculated by taking the integral of the operator multiplied by the complex conjugate wavefunction, and then dividing by the integral of the wavefunction squared. This is represented by the formula: ⟨A⟩ = ∫Ψ* A Ψ dτ / ∫Ψ* Ψ dτ, where A is the operator and Ψ is the wavefunction.

3. What is the significance of the expectation value with an imaginary component?

The expectation value with an imaginary component is significant because it represents the most probable outcome of a measurement when the system is in a given quantum state. It is also used to calculate the uncertainty in the measurement of a physical quantity.

4. How does the expectation value with an imaginary component differ from the classical expectation value?

The expectation value with an imaginary component differs from the classical expectation value because it takes into account the probabilistic nature of quantum mechanics. In classical mechanics, the expectation value is equivalent to the average value, but in quantum mechanics, it represents the most probable outcome of a measurement.

5. Can the expectation value with an imaginary component be a complex number?

Yes, the expectation value with an imaginary component can be a complex number. This is because it takes into account the complex nature of quantum mechanics and the possibility of obtaining a complex number as the outcome of a measurement.

Similar threads

  • Quantum Physics
Replies
14
Views
2K
Replies
2
Views
574
Replies
3
Views
280
  • Quantum Physics
Replies
9
Views
1K
  • Quantum Physics
Replies
13
Views
754
  • Quantum Physics
Replies
20
Views
3K
  • Quantum Physics
Replies
1
Views
226
Replies
3
Views
1K
Replies
2
Views
1K
  • Quantum Physics
Replies
4
Views
1K
Back
Top