What is the Inner Product of <+z|+n>?

[danJm]

Homework Statement

given |+n> = cos (θ/2)|+z> + e^(i*phi)sin(θ/2)|-z>

i'm asked to find out what <+z|+n>=

Homework Equations

<+z|+z> = 1
<+z|-z> = 0

The Attempt at a Solution

I am just unsure what <+z| is..

computing the inner product of the 2 quantum states to get the probability amplitude is not the issue. I know i need to use the <+z|+z> = 1, <+z|-z> = 0 to find it, but I'm just not sure how to approach it.

i feel like I'm just missing something really simple, please any advice would be greatly appreciated.

 

[capandbells]

computing the inner product of the 2 quantum states to get the probability amplitude is not the issue.

It seems to me that it's ALL of the issue. The inner product <z+|n+> is all you're asked to compute.

Spoiler

(By the way, you DO know what <z+| is. Since you know what it does to the basis {<z+|,<z-|}, you know what it does to all vectors in the space.)

 

[danJm]

ok.. well the solution i come up with is <+z|+n> = cos(theta/2)

i just have no idea if that's right

 

[capandbells]

Yeah, that's all you need.

 

[I like Serena]

Welcome to PF, danJm!

ok.. well the solution i come up with is <+z|+n> = cos(theta/2)

i just have no idea if that's right

As to why it is right...

You are applying <+z| to |+n>.
You can replace |+n> by the expression you have for it.

What you need to use is that the operation of <a| on |b> is a so called inner product.
The axioms for an inner product state that it is linear for addition and scalar multiplication in the first argument.
See for instance: wiki.

Note that you still need to be careful with a scalar in the second argument, which you have, since that requires a conjugate (see wiki page).
Do you see why that is not a problem here?

 

[danJm]

i assumed the second term would go away because of <+z|-z> = 0

 

[I like Serena]

Yes it does.

The proper way to do it, is to apply the axioms/propositions of the inner product.
First you can split it in a summation of 2 inner products.
Then you can get the scalars out.
Since they are in the second argument, you need their conjugates.

That leaves you with:

(cos(θ/2))* <+z|+z> + (e^(i*phi)sin(θ/2))* <+z|-z>​

where * denotes the conjugate.

In this expression you can substitute your values for <+z|+z> and <+z|-z>.

 

[danJm]

further, the expectation value for this problem would just be
<+z|+z>|^2 = cos^2(θ/2)h(bar)/2
yea?

 

[I like Serena]

You seem to have left out some information.
Can you supply us with the full question?

In particular |<+z|+z>|^2 = 1^2 = 1.

 

[danJm]

Suppose that a measurement of S_z is carried out on a particle in the state |+n> what is the probability that the measurement yields (i)h(bar)/2 and (ii)-h(bar)/2?

i wrote the inner product incorrectly, i assume that S_z = |<+z|+n>|²

so that would follow cos²(θ/2)(h(bar)/2)

 

[I like Serena]

Hmm, I can't quite say.
I do not know what was intended with S_z.

From your context I tentatively assume S_z can either be $\frac \hbar 2$ or $- \frac \hbar 2$.
Do those outcomes perhaps correspond to the states |+z> respectively |-z>?

If that is the case, the observable S_z might be $\frac \hbar 2$ if the state of the particle is |+z>.Btw, your question does not seem to include an "expectation value"...?

Either way, in the probability for such a measurement, $\hbar$ would not play any role.
It seems you are mixing up probabilities and expectation values...?

 

[danJm]

ah, you i did, wow, thanks for the help.
i believe Sz is the spin in the z direction.

so the probability to find the particle in the +ℏ/2 is cos^2(θ/2)
and for the -ℏ/2 = e^(2*i*phi)sin^2(θ/2)

 

[I like Serena]

Ah, we're starting to get there... :)

The probability for -ℏ/2 would be |<-z|+n>|².

What is <-z|+n>?
And what is |<-z|+n>|?

Btw, did you know that probabilities are supposed to be real?
And that they are supposed to add up to 1?

 

[danJm]

ah, this is when you square it, you do the complex conjugate. leaving me with sin^2(θ/2) for the probability of -ℏ/2

ya i knew that, i just failed at thinking... sigh..

 

[I like Serena]

Congratulations!