Quantum Mechanics - Spin.

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  • #1
Hart
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Homework Statement



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.

Homework Equations



Within the question details and solution attempts.

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.
 

Answers and Replies

  • #2
jdwood983
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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.

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

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?

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.

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.

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.
 
  • #3
Hart
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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?
 
  • #4
jdwood983
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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.

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]

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]

??

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.

Part III:

Could you clarify / explain a bit more that notation that you've used?

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]
 
  • #5
Hart
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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 canceled 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:
 
  • #6
jdwood983
383
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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:

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.

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?

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.

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?!

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.

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 canceled 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?!

Correct. You can also show that, in general

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

for i=x,y,z.

Part III:

Sorry, I still don't get what's going on in this question :confused:

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).
 
  • #7
Hart
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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.
 
  • #8
jdwood983
383
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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.

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]??

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]

Correct.

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.

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.
 
  • #9
jdwood983
383
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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].
 
  • #10
Hart
169
0
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]
 
  • #11
jdwood983
383
0
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]

Correct!

Then the expectation value.. using formula (?)

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


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]

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]
 
  • #12
Hart
169
0
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:
 
  • #13
Hart
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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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