Determine state of particle: Quantum Mechanics (Phase)

[grandpa2390]

Homework Statement

Homework Equations

The Attempt at a Solution

This is the Solution. I am having trouble understanding parts of it.

The first part I don't get is why the e^i... goes with the -z. Did my professor just choose one at random, or is there a specific reason?

The second part I am not understanding is how he got that |<+x|>|^2 = 1/2...
I tried squaring the expression for <+x| > but I am not coming up with that value.

 

[TSny]

The first part I don't get is why the e^i... goes with the -z. Did my professor just choose one at random, or is there a specific reason?

It is an arbitrary choice. A state is determined only up to an arbitrary overall phase factor. So, the most you can expect to do is determine the relative phase between the up and down z states in $|\psi \rangle$. You could just as well put the relative phase factor $e^{i \delta}$ with the $|+z \rangle$ ket.

The second part I am not understanding is how he got that |<+x|>|^2 = 1/2

It has to do with the information given about the probability for finding the particle in the $|+x \rangle$ state.

 

[grandpa2390]

It has to do with the information given about the probability for finding the particle in the $|+x \rangle$ state.

Could you elaborate?

How did he get to

 

[TSny]

If z is a complex number, then |z|² = z*z.

 

[grandpa2390]

If z is a complex number, then |z|² = z*z.

you are not making any sense. As I said, I did square the expression in the previous step. but I did not get that result.

 

[grandpa2390]

If z is a complex number, then |z|² = z*z.

are you say to multiply it by the complex conjugate?

 

[DrClaude]

are you say to multiply it by the complex conjugate?

Yes. Maybe it will be clearer using LaTeX:
$$
|z|^2 = z^* z
$$

(Edit: That's not necessarily clearer. I understand better why many mathematicians prefer the notation $\bar{z}$ to $z^*$.)

 

[grandpa2390]

Yes. Maybe it will be clearer using LaTeX:
$$
|z|^2 = z^* z
$$

(Edit: That's not necessarily clearer. I understand better why many mathematicians prefer the notation $\bar{z}$ to $z^*$.)

Thanks.
Lol, Written by hand, it makes sense. But on the computer, I am so used to seeing that symbol used for regular multiplication.

Thank you both.

 

[TSny]

OK, good. Sorry for the confusion regarding the complex conjugate notation.
Moreover, when I first read the following

The second part I am not understanding is how he got that |<+x|>|^2 = 1/2...
I tried squaring the expression for <+x| > but I am not coming up with that value.

I didn't notice the three dots ... after the 1/2. So, I thought you were asking why $|\langle +x|\psi \rangle|^2= 1/2$ . Thus, my response in post #2.
Oh well, glad it all got sorted out.

 

[grandpa2390]

OK, good. Sorry for the confusion regarding the complex conjugate notation.
Moreover, when I first read the following

I didn't notice the three dots ... after the 1/2. So, I thought you were asking why $|\langle +x|\psi \rangle|^2= 1/2$ . Thus, my response in post #2.
Oh well, glad it all got sorted out.

I suspected that. so I posted the whole thing the second time (as I should have done the first time ;)

thanks