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Quantum bra/ket problem

  1. Oct 7, 2012 #1
    1. The problem statement, all variables and given/known data
    given |+n> = cos (θ/2)|+z> + e^(i*phi)sin(θ/2)|-z>

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

    2. Relevant equations

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

    3. 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.
     
  2. jcsd
  3. Oct 7, 2012 #2
    It seems to me that it's ALL of the issue. The inner product <z+|n+> is all you're asked to compute.
    (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.)
     
  4. Oct 7, 2012 #3
    ok.. well the solution i come up with is <+z|+n> = cos(theta/2)

    i just have no idea if that's right
     
  5. Oct 7, 2012 #4
    Yeah, that's all you need.
     
  6. Oct 7, 2012 #5

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    Welcome to PF, danJm! :smile:

    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?
     
  7. Oct 7, 2012 #6
    i assumed the second term would go away because of <+z|-z> = 0
     
  8. Oct 7, 2012 #7

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    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>.
     
  9. Oct 7, 2012 #8
    further, the expectation value for this problem would just be
    <+z|+z>|^2 = cos^2(θ/2)h(bar)/2
    yea?
     
  10. Oct 7, 2012 #9

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

    In particular |<+z|+z>|^2 = 1^2 = 1.
     
  11. Oct 7, 2012 #10
    Suppose that a measurement of Sz 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 Sz = |<+z|+n>|2

    so that would follow cos2(θ/2)(h(bar)/2)
     
  12. Oct 7, 2012 #11

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

    From your context I tentatively assume Sz 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 Sz 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...?
     
  13. Oct 7, 2012 #12
    ah, ya 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)
     
  14. Oct 7, 2012 #13

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    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?
     
  15. Oct 7, 2012 #14
    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..
     
  16. Oct 7, 2012 #15

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    Congratulations! :smile:
     
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