Expectation value with imaginary component?

[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 anybody help point out where I have gone wrong?

 

[PeterDonis]

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.

 

[PeroK]

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

 

[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.

 

[PeroK]

@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?

 

[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.]

 

[PeroK]

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?

 

[PeterDonis]

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?

 

[PeroK]

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

Hint: it doesn't give you the probability of getting $|x \!+ \rangle$. So, if it's not the probability itself, what does it give you?

 

[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: $(1/(sqrt(2)*2) + i*sqrt(3)/(sqrt(2)*2)$

 

[PeroK]

@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.

 

[Kenneth Adam Miller]

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

 

[PeroK]

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.

 

[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
$|<x | phi>|^2 = (1/8 + 3/8) = 1/2$

 

[PeroK]

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.

 

[PeterDonis]

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?

 

[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.

 

[ftr]

|a+ib|2=a2+b2|a+ib|2=a2+b2|a + ib|^2 = a^2 + b^2

this looks incorrect

 

[PeterDonis]

this looks incorrect

Why?

 

[ftr]

Why?

(ib)^2= - b^2

 

[PeterDonis]

(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.

 

[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.

 

[ftr]

You don't need complex conjugation

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

 

[PeterDonis]

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 $r^2$.

 

[PeroK]

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.

 

[Kenneth Adam Miller]

AH!

$|z|^2 = zz*$

Didn't know this!