# I Expectation value with imaginary component?

1. Feb 18, 2017

### Kenneth Adam Miller

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 any body help point out where I have gone wrong?

Last edited: Feb 18, 2017
2. Feb 18, 2017

### Staff: Mentor

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. Feb 18, 2017

### PeroK

How do you intend to calculate $\langle S_x \rangle$?

4. Feb 18, 2017

### Kenneth Adam Miller

@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. Feb 18, 2017

### PeroK

Okay, but how are you going to calculate these probabilities?

6. Feb 18, 2017

### Kenneth Adam Miller

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: Feb 18, 2017
7. Feb 18, 2017

### PeroK

What does $\langle x+ | \psi \rangle$ give you?

8. Feb 18, 2017

### Staff: Mentor

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. Feb 18, 2017

### PeroK

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. Feb 18, 2017

### Kenneth Adam Miller

@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. Feb 18, 2017

### PeroK

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. Feb 18, 2017

### Kenneth Adam Miller

So, just going from $<x|phi>$ to $|<x|phi>|^2$ the imaginary component will be eliminated?

13. Feb 18, 2017

### PeroK

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. Feb 18, 2017

### Kenneth Adam Miller

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. Feb 18, 2017

### PeroK

Yes, that's it.

16. Feb 18, 2017

### Staff: Mentor

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. Feb 18, 2017

### Kenneth Adam Miller

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. Feb 18, 2017

### ftr

this looks incorrect

19. Feb 18, 2017

### Staff: Mentor

Why?

20. Feb 18, 2017

### ftr

(ib)^2= - b^2

21. Feb 18, 2017

### Staff: Mentor

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. Feb 18, 2017

### Kenneth Adam Miller

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. Feb 18, 2017

### ftr

The need is implicit, that is why I thought the OP will be confused.

24. Feb 18, 2017

### Staff: Mentor

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. Feb 18, 2017

### PeroK

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.

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