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Homework Help: Quantum Mechanics - Spin.

  1. Mar 2, 2010 #1
    1. The problem statement, all variables and given/known data

    The spin of an electron is described by a vector: [tex]\psi = \left(\frac{\uparrow}{\downarrow}\right)[/tex] and the spin operator:

    [tex]\hat{S} = \hat{S_{x}}i + \hat{S_{y}}j + \hat{S_{z}}k[/tex]

    with components:

    [tex]\hat{S_{x}} = \frac{\hbar}{2} \left[ \begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array} \right][/tex]

    [tex]\hat{S_{y}} = \frac{\hbar}{2} \left[ \begin{array}{cc} 0 & -i \\ i & 0 \end{array} \right][/tex]

    [tex]\hat{S_{z}} = \frac{\hbar}{2} \left[ \begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array} \right][/tex]

    I.

    i. State the normalisation condition for: [tex]\psi[/tex].

    ii. Give the general expressions for the probabilities to find: [tex]\hat{S_{z}} = \pm \frac{\hbar}{2}[/tex] in a measurement of: [tex]\hat{S_{z}}[/tex]

    iii. Give the general expression for the expectation value of: [tex]\left<\hat{S_{z}}\right>[/tex]


    II.

    Calculate the commutators:

    [tex]\left[\hat{S_{y}},\hat{S_{z}}\right][/tex] and [tex]\left[\hat{S_{z}},\hat{S^{2}}}\right][/tex]

    .. Are [itex]\hat{S_{y}} [/itex] and [itex]\hat{S_{z}} [/itex] simultaneous observables? Are [itex]\hat{S_{z}}[/itex] and [itex]\hat{S^{2}}[/itex] simultaneous observables?

    III.

    i. Normalise the state: [tex]\frac{1}{1}[/tex]

    ii. Calculate the expectation values: [tex]\hat{S_{x}}[/tex], [tex]\hat{S_{y}}[/tex], and [tex]\hat{S_{z}}[/tex] for this normalised state.

    2. Relevant equations

    Within the question details and solution attempts.

    3. The attempt at a solution

    I:

    i. Normalisation condition: [tex]\langle\psi|\psi\rangle[/tex] .. yes?

    ii. I have calculated that:

    [tex]\left[\hat{S_{x}}, \hat{S_{y}}\right] = i\hbar S_{z}[/tex]

    and then know that eigenvalues of [itex]\hat{S_{z}}[/itex] are simply [itex]\frac{\hbar}{2}[/itex] times the eigenvalues of [itex]\sigma_{z}[/itex] (which is just the matrix part of [itex]\hat{S_{z}}[/itex] if that makes sense). Not sure what more to do now here though.

    iii. I don't know how to go about doing this part at the moment.

    II.

    I have calculated the commutators as:

    [tex]\left[\hat{S_{y}},\hat{S_{z}}\right] = \frac{\hbar^{2}}{2} \left[ \begin{array}{cc} 0 & i \\ i & 0 \end{array} \right][/tex]

    and

    [tex]\left[\hat{S_{z}},\hat{S^{2}}}\right] = 3 \hbar^{3} \left[ \begin{array}{cc} i & -1 \\ 1 & -i \end{array} \right][/tex]

    .. but I don't know in either case if they are simultaneous observables? Or indeed, how I would be able to determine if so for each case?

    III.

    I don't understand what I need to do, as far as normalising the state as required. Once I know how to do this, can then obviously have a go at calculating the expectation values.
     
  2. jcsd
  3. Mar 2, 2010 #2
    For part i, that function should be equal to 1: [itex]\langle\psi|\psi\rangle=1[/itex].

    For part, iii, note that

    [tex]\langle S_z\rangle=\langle\psi|S_z|\psi\rangle[/tex].

    That should help you a bit

    Not quite, but on the right track. You should be getting a number as a result, something like what you have as an answer for I. ii.

    This part requires you to actually calculate

    [tex]\langle\psi|\psi\rangle=1\longrightarrow|\psi\rangle_N=\frac{1}{\sqrt{\langle\psi|\psi\rangle}}|\psi\rangle[/tex]

    where [itex]|\psi\rangle_N[/itex] is the normalized state-ket.
     
  4. Mar 3, 2010 #3
    Part I:

    i. : Of course, it should be 'Normalisation Condition: [itex]\langle\psi|\psi\rangle=1[/itex]'

    ii. & iii. : I still don't get how to do these parts of the question.

    Part II:

    I don't understand how to get the result of just a number. Can I do something like this (?) ..

    [tex]\left[\hat{S_{y}},\hat{S_{z}}\right] = \frac{\hbar^{2}}{2} \left[ \begin{array}{cc} 0 & i \\ i & 0 \end{array} \right] [/tex]

    [tex] ( 0 . 0 ) - ( i . i ) = - ( i^{2}) = 1
    [/tex]

    So:

    [tex]\left[\hat{S_{y}},\hat{S_{z}}\right] = \frac{\hbar^{2}}{2} \left[ \begin{array}{cc} 0 & i \\ i & 0 \end{array} \right] = \frac{\hbar^{2}}{2}[/tex]

    ??

    Part III:

    Could you clarify / explain a bit more that notation that you've used?
     
  5. Mar 3, 2010 #4
    An example how to get the probability (part ii) should be given in most textbooks and should be something like this:

    [tex]Probability=\frac{value\,you\,are\,looking\,for}{sum\,of\,the\,total\,possible\,values}[/tex]

    The expectation value will be

    [tex]\langle\hat{S}_z\rangle=\psi^*\hat{S}_z\psi=\left(1,\,1\right)\cdot\frac{\hbar}{2}\left(\begin{array}{cc}1&0\\0&-1\end{array}\right)\cdot\left(\begin{array}{c}1\\1\end{array}\right)[/tex]

    Whoops, I sorta misread and then mis-spoke here. What you should find here is the same sort of thing as what you have as an answer in Part I ii:

    [tex][\hat{S}_x,\,\hat{S}_y]=i\hbar\hat{S}_z[/tex]

    You will likely want to double check on your second commutator. In both cases, if the commutation relation gives zero, they are simultaneous observables; if you get a non-zero answer they are not simultaneous observables.

    The notation is

    [tex]|\psi\rangle=\left(\begin{array}{c}1 \\ 1\end{array}\right)[/tex]

    so then you have the normalized wave function is

    [tex]\psi_N=\frac{1}{\sqrt{\psi^*\psi}}\psi \equiv |\psi\rangle_N=\frac{1}{\sqrt{\langle\psi|\psi\rangle}}|\psi\rangle= \psi_N=\frac{1}{\sqrt{\left(1,\,1\right)\cdot\left(\begin{array}{c}1\\1\end{array}\right)}}\left(\begin{array}{c}1\\1\end{array}\right)[/tex]
     
  6. Mar 3, 2010 #5
    Part I:

    ii. I've looked in the two quantum mechanics textbooks I have to hand (Rae & Griffiths) but I can't seem to find a bit about these spin probabilities :frown:

    iii : OK, so:

    [tex]\langle\hat{S}_z\rangle=\psi^*\hat{S}_z\psi=\left( 1,\,1\right)\cdot\frac{\hbar}{2}\left(\begin{array}{cc}1&0\\0&-1\end{array}\right)\cdot\left(\begin{array}{c}1\\1 \end{array}\right) = \left( \begin{array}{c}\left(\frac{\hbar}{2}\right)\\ \left( \frac{\hbar}{2}\right) \end{array} \right) \left( 1 , 1\right)[/tex]

    .. correct? Then I assume can combine those just like this:

    [tex]\langle\hat{S}_z\rangle=\psi^*\hat{S}_z\psi= \left( \begin{array}{c}\left(\frac{\hbar}{2}\right)\\ \left( \frac{\hbar}{2}\right) \end{array} \right) \left( 1 , 1\right) = \left[ \begin{array}{cc} \left(\frac{\hbar}{2}\right)& \left(\frac{\hbar}{2}\right)\\ \left(\frac{\hbar}{2}\right)& \left(\frac{\hbar}{2}\right) \end{array} \right] = \left(\frac{\hbar}{2}\right)\left[ \begin{array}{cc} 1 & 1 \\ 1& 1 \end{array} \right] = 0[/tex]

    .. yes?


    Part II:

    I have checked through all my calculations and recalculated the commutators as:

    Firstly:

    [tex]\left[\hat{S_{y}},\hat{S_{z}}\right] = \frac{i\hbar^{2}}{2} \left[ \begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array}\right] = \frac{i\hbar^{2}}{2} \left( \hat{S_{x}}\right) [/tex]

    So [itex]\hat{S_{y}}[/itex] and [itex]\hat{S_{z}}[/itex] are not simultaneous observables as the result [itex]\neq 0[/itex].. correct?!

    Secondly:

    [tex]\hat{S_{z}^{2}} = \left[ \begin{array}{cc} \left(\frac{9\hbar^{2}}{16}\right) & 0 \\ 0 & \left(\frac{9\hbar^{2}}{16}\right) \end{array} \right][/tex]

    (just included for completion)

    .. which seemed to have horrible factors, but they cancelled with those in the other commutator [itex]\hat{S_{z}}[/itex] so that was good, giving the result..

    [tex]\left[\hat{S_{z}},\hat{S^{2}}}\right] = \left[ \begin{array}{cc} 0 & 0 \\ 0 & 0 \end{array} \right][/tex]

    .. then since [itex](0-0)-(0-0) = 0 [/itex] ..

    Hence [itex]\hat{S_{z}}[/itex] and [itex]\hat{S^{2}}[/itex] are simultaneous observables as the result [itex]= 0[/itex].. correct?!

    Part III:

    Sorry, I still don't get what's going on in this question :confused:
     
  7. Mar 3, 2010 #6
    See Griffiths Section 4.4, page 154. Everything below appears there. I don't know of a QM book that doesn't cover the spin-half system.

    You may have multiplied your matrices incorrectly:

    [tex]\langle\hat{S}_z\rangle=\psi^*\hat{S}_z\psi=\left( 1,\,1\right)\cdot\frac{\hbar}{2}\left(\begin{array}{cc}1&0\\0&-1\end{array}\right)\cdot\left(\begin{array}{c}1\\1 \end{array}\right) = \left( 1 , 1\right)\cdot\left( \begin{array}{c}\frac{\hbar}{2}\\ - \frac{\hbar}{2}\end{array} \right)=\frac{\hbar}{2}-\frac{\hbar}{2}=0[/tex]

    I am also unsure how you turned the 4x4 matrix of all 1's into zero.

    Not quite, recall that

    [tex]\hat{S}_x=\frac{\hbar}{2}\sigma=\frac{\hbar}{2}\left(\begin{array}{cc}0&1\\1&0\end{array}\right)[/tex]

    So you actually have the expected result:

    [tex][\hat{S}_y,\,\hat{S}_z]=i\hbar\hat{S}_x[/tex]

    However, you do still end up with a non-zero value, so you are correct in saying that [tex]\hat{S}_y[/tex] and [tex]\hat{S}_z[/tex] are not simultaneous observables.

    Correct. You can also show that, in general

    [tex][\hat{S}_i,\,\hat{S}^2]=0[/tex]

    for i=x,y,z.

    You need to find [itex]\psi_N[/itex], the normalized wave function, by using the expression I gave in my previous post. After you do that, you need to calculate the expectation value of the 3 components of spin, using the equation I gave you for Part I, iii (keeping in mind to correctly multiply your matrices).
     
  8. Mar 4, 2010 #7
    I. ii.

    From Griffiths' QM:

    " If measure [itex]S_{z}[/itex] on a particle in the general state [itex]\chi = \left(\frac{a}{b}\right)[/itex] then can get [itex]+ \frac{\hbar}{2}[/itex] with probability [itex]|a|^{2}[/itex] or [itex]-\frac{\hbar}{2}[/itex] with probability [itex]|b|^{2}[/itex]. Since these are the only possibilities: [itex]|a|^{2} + |b|^{2} = 1[/itex] "

    .. I understand that, but not quite sure how need to adapt it further.

    I. iii.

    so..

    [tex]\langle\hat{S}_z\rangle=\left<\psi^*\hat{S}_z\psi\right>=\left( 1 , 1\right)\cdot\left( \begin{array}{c}\frac{\hbar}{2}\\ - \frac{\hbar}{2}\end{array} \right)=\frac{\hbar}{2}-\frac{\hbar}{2}=0[/tex]

    III:

    I'm still not doing very well with this part of the question, really don't think I get it, but this is what I have so far:

    [tex]\psi_N=\frac{1}{\sqrt{\left(1,\,1\right)\cdot\left (\begin{array}{c}1\\1\end{array}\right)}}\left(\begin{array}{c}1\\1\end{array}\right)[/tex]

    so..

    [tex]\psi_N=\frac{1}{\sqrt{\left(1,\,1\right)\cdot\left (\begin{array}{c}1\\1\end{array}\right)}}\left(\begin{array}{c}1\\1\end{array}\right) = 1[/tex] ??!?

    .. which is hopefully somewhere along the right lines, though I don't think I should be getting that result, but I can't figure out how to do the calculations to get any other result.
     
  9. Mar 4, 2010 #8
    Do you understand eigenvalue problems?

    [tex]\hat{S}_z\chi=\lambda\chi\rightarrow\det\left[\hat{S}_z-\lambda\mathbb{I}\right]=0,[/tex]

    then solving for [itex]\lambda[/itex]??

    Correct.

    From what you wrote, you haven't done anything different than what I wrote in post 4. Do you remember how to multiply matrices? For a 1x2 and 2x1 matrix, you multiply them as:

    [tex](a,\,b)\cdot\left(\begin{array}{c}c\\d\end{array}\right)=ac+bd[/tex]

    And I'm not sure how/why you keep writing a matrix as a single number, it is not.
     
  10. Mar 4, 2010 #9
    After getting your new wavefunction, [tex]\psi_N[/tex], you need to use the same calculation as before to find the expectation values of [tex]\hat{S}_z[/tex]:

    [tex]\langle\hat{S}_z\rangle=\psi^*_N\hat{S}_z\psi_N[/tex]

    And similarly for [tex]\hat{S}_y[/tex] and [tex]\hat{S}_x[/tex].
     
  11. Mar 4, 2010 #10
    I.

    [tex]\hat{S_{z}} = \frac{\hbar}{2} \left[ \begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array}\right] = \left[ \begin{array}{cc} \frac{\hbar}{2} & 0 \\ 0 & -\frac{\hbar}{2} \end{array}\right][/tex]

    [tex]\implies \hat{S_{z}} = \left[ \begin{array}{cc} \frac{\hbar}{2}-\lambda & 0 \\ 0 & -\frac{\hbar}{2}-\lambda \end{array}\right] \implies \lambda^{2} - \frac{\hbar^{2}}{4} = 0[/tex]

    [tex] \implies \lambda = \pm \frac{\hbar}{2}[/tex] .. which are the Eigenvalues of [itex]\hat{S_{z}}[/itex]

    Then the expectation value.. using formula (?)


    III.

    [tex]\psi_N= \frac{1}{\sqrt{2}}\left(\begin{array}{c}1\\1\end{array}\right) = \left(\begin{array}{c}\left(\frac{1}{\sqrt{2}}\right)\\\left(\frac{1}{\sqrt{2}}\right)\end{array}\right)[/tex]

    [tex]\langle\hat{S}_z\rangle=\psi_{N}^*\hat{S}_z\psi_{N}=\left(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right)\cdot\frac{\hbar}{2}\left(\begin{array}{cc}1&0\\0&-1\end{array}\right)\cdot\left(\begin{array}{c}\left(\frac{1}{\sqrt{2}}\right)\\\left(\frac{1}{\sqrt{2}}\right)\end{array}\right) = \left(\frac{1}{\sqrt{2}} , \frac{1}{\sqrt{2}}\right)\cdot\left( \begin{array}{c}\left(\frac{\hbar}{2\sqrt{2}}\right)\\ \left(-\frac{\hbar}{2\sqrt{2}}\right)\end{array} \right)= \left(\frac{\hbar}{2}\right) - \left(\frac{\hbar}{2}\right) = 0[/tex]

    ..yes?

    Then obviously need to repeat the method for [itex]\hat{S_{x}}[/itex] and [itex]\hat{S_{y}}[/itex]
     
  12. Mar 4, 2010 #11
    Correct!

    You have already calculated the expectation value, not sure what you're asking here.


    Mostly correct, there is a small multiplication error:

    [tex]\frac{1}{\sqrt{2}}\cdot\frac{\hbar}{2\sqrt{2}}=\frac{\hbar}{4}\,\,\mathrm{not}\,\,\frac{\hbar}{2}[/tex]
     
  13. Mar 5, 2010 #12
    Thank's for pointing out that small maths mistake, should've noticed that! :redface:

    I think that it should actually be:

    [tex]\langle\hat{S}_z\rangle=\psi_{N}^*\hat{S}_z\psi_{N}= \left(\frac{1}{\sqrt{2}} , \frac{1}{\sqrt{2}}\right)\cdot\left( \begin{array}{c}\left(\frac{\hbar}{2\sqrt{2}}\right)\\ \left(\frac{\hbar}{2\sqrt{2}}\right)\end{array} \right)= \left(\frac{\hbar}{4}\right) - \left(\frac{\hbar}{4}\right) = 0[/tex]

    i.e. I put in minus sign by mistake towards the end, but still end up with: result = 0.

    I've calculated the remaining two expectation values as:

    [tex]\langle\hat{S}_x\rangle=\psi_{N}^*\hat{S}_x\psi_{N }=\left(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right)\cdot\frac{\hbar}{2}\left(\begin{array}{cc}0&1\\ 1&0\end{array}\right)\cdot\left(\begin{array}{c}\left(\frac{1}{\sqrt{2}}\right)\\\left(\frac{1}{\sqrt{ 2}}\right)\end{array}\right) = \left(\frac{1}{\sqrt{2}} , \frac{1}{\sqrt{2}}\right)\cdot\left( \begin{array}{c}\left(\frac{\hbar}{2\sqrt{2}}\right)\\ \left(\frac{\hbar}{2\sqrt{2}}\right)\end{array} \right)= \left(\frac{\hbar}{4}\right) - \left(\frac{\hbar}{4}\right) = 0[/tex]

    [tex]\langle\hat{S}_y\rangle=\psi_{N}^*\hat{S}_y\psi_{N }=\left(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right)\cdot\frac{\hbar}{2}\left(\begin{array}{cc}0&i\\ -i&0\end{array}\right)\cdot\left(\begin{array}{c}\left(\frac{1}{\sqrt{2}}\right)\\\left(\frac{1}{\sqrt{2}}\right)\end{array}\right) = \left(\frac{1}{\sqrt{2}} , \frac{1}{\sqrt{2}}\right)\cdot\left( \begin{array}{c}\left(\frac{i\hbar}{2\sqrt{2}}\right)\\ \left(\frac{-i\hbar}{2\sqrt{2}}\right)\end{array} \right)= \left(\frac{i\hbar}{4}\right) - \left(\frac{-i\hbar}{4}\right) = \frac{i\hbar}{2}[/tex]

    .. these correct?

    Then going back to part i, my last post wasn't very clear. I was confused because I thought I'd just answered the part which was:

    ii. Give the general expressions for the probabilities to find: [itex]\hat{S_{z}} = \pm \frac{\hbar}{2}[/itex] in a measurement of: [itex]\hat{S_{z}}[/itex]

    .. then didn't think I'd done the next part too, which was:

    iii. Give the general expression for the expectation value of: [itex]\left<\hat{S_{z}}\right>[/itex]

    ?! :confused:
     
  14. Mar 6, 2010 #13
    Just looked back through the posts, ignore my last comment about part I, I see that I have actually completed all parts.

    Unsure though if the calculations for part III above (i.e. expectation values) are correct?
     
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