Expectation Value and Probabilities of Spin Operator Sy

  • #1
594
12

Homework Statement



(a) If a particle is in the spin state ## χ = 1/5 \begin{pmatrix}
i \\
3 \\
\end{pmatrix} ## , calculate the expectation value <Sy>


(b) If you measured the observable Sy on the particle in spin state given in (a), what values might you get and what is the probability of each?


Homework Equations


(a)
<Sy> = expectation value
<Sy> = < χ | Sy | χ >

## Sy = ħ/2 \begin{pmatrix}
0 & -i \\
i & 0 \\
\end{pmatrix} ##

|χ> = ##1/5 \begin{pmatrix}
i \\
3 \\
\end{pmatrix} ##

< χ | = ##1/5 \begin{pmatrix}
-i & 3\\
\end{pmatrix} ##

(b)
C = < φ | χ >
C*C (where * = complex conjugate) = probability

The Attempt at a Solution


(a) <Sy> =
## 1/5 \begin{pmatrix}
-i & 3\\
\end{pmatrix}
ħ/2 \begin{pmatrix}
0 & -i \\
i & 0 \\
\end{pmatrix} 1/5 \begin{pmatrix}
i \\
3 \\
\end{pmatrix} ##

## = ħ/50 \begin{pmatrix}
-i & 3\\
\end{pmatrix}
\begin{pmatrix}
-3i \\
-1 \\
\end{pmatrix} ##

<Sy> ## = -6ħ/50 = -3ħ/25 ##

If the particle is in the spin state χ (given above) then the expectation value is = -3ħ/25. I don't understand how the expectation value can be negative though?

(b)
The Sy spin operator has two eigenspinors, corrosponding to two eigenvalues of ±ħ/2

i) For the eigenspinor with the positive eigenvalue of ħ/2

## | φ > = 1/√2 \begin{pmatrix}
1 \\
i \\
\end{pmatrix} ##

## < φ | = 1/√2 \begin{pmatrix}
1 & -i \\
\end{pmatrix} ##

C = ## < φ | χ > = 1/√2 \begin{pmatrix}
1 & -i \\
\end{pmatrix} 1/5 \begin{pmatrix}
i\\
3\\
\end{pmatrix} ##

= ## < φ | χ > = 1/5√2 \begin{pmatrix}
1 & i \\
\end{pmatrix} \begin{pmatrix}
i\\
3\\
\end{pmatrix} ##

C = ## < φ | χ > = 1/5√2 \begin{pmatrix}
-2i \\
\end{pmatrix} ##

## C * C = 1/50 (-2i)(2i) = 4/50 = 2/25 ##

The probability of measuring χ for Sy with the eigenvalue ħ/2 = 2/25

ii) For the eigenspinor with the negative eigenvalue of -ħ/2
## | φ > = 1/√2 \begin{pmatrix}
1 \\
-i \\
\end{pmatrix} ##

## < φ | = 1/√2 \begin{pmatrix}
1 & i \\
\end{pmatrix} ##

C = ## < φ | χ > = 1/√2 \begin{pmatrix}
1 & i \\
\end{pmatrix} 1/5 \begin{pmatrix}
i\\
3\\
\end{pmatrix} ##

C = ## < φ | χ > = 1/5√2 \begin{pmatrix}
-4i \\
\end{pmatrix} ##

## C * C = 1/50 (-4i)(4i) = 16/50 = 8/25 ##

The probability of measuring χ for Sy with the eigenvalue -ħ/2 = 8/25

I think I've answered part (b) correctly, it's just the negative expectation value I'm a bit confused about.
 

Answers and Replies

  • #2
TSny
Homework Helper
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(a) If a particle is in the spin state ## χ = 1/5 \begin{pmatrix}
i \\
3 \\
\end{pmatrix} ## , calculate the expectation value <Sy>
Is this state vector normalized? Do your results for the probabilities in part (b) add to 1? Should they?

I don't understand how the expectation value can be negative though?
Why does this bother you?

Your work looks good, but you are not getting the correct numerical values due to using a state vector that is not normalized.
 
  • #3
594
12
The results for the probabilities don't add to 1, which I thought they should. I'm guessing they will add to one once I normalise χ.

I'm not entirely sure how to normalise χ though?
 
  • #4
TSny
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What does it mean for a state vector to be normalized?
 
  • #5
594
12
## [ 1 / |A| ] * A ##

|A| = √(i2+32) = √ 8
|A| = √(-1+9) = √8

1/(5√8) ## \begin{pmatrix}
i \\
3 \\
\end{pmatrix} ##

Sorry, I posted that before I'd finished writing. I believe that is the normalised state vector though.
 
  • #6
TSny
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You need to replace the factor of 1/5 in the state |χ> by some other number, A, that normalizes the state. So, write the state as
|χ> = A ## \begin{pmatrix}
i \\
3 \\
\end{pmatrix} ##
where you need to determine a value of A that normalizes the state (<χ|χ> = 1). You can take A to be a real number.

Be careful in forming <χ|. What happens to the entry ##i## in going from |χ> to <χ|?
 
  • #7
594
12
< χ | = A ## \begin{pmatrix}
- i & 3 \\
\end{pmatrix} ##

| χ > = A ## \begin{pmatrix}
i\\
3 \\
\end{pmatrix} ##

< χ | χ > = 1 ⇒ A=1/10
 
  • #8
TSny
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< χ | = A ## \begin{pmatrix}
- i & 3 \\
\end{pmatrix} ##
Yes, taking A to be real.

< χ | χ > = 1 ⇒ A=1/10
Almost. Note that A occurs in both < χ | and | χ >. Check your work.
 
  • Like
Likes says
  • #9
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12
opps!

A= √(1/10)
 
  • #10
TSny
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Right.
 
  • #11
594
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C = < φ | ψ>
C*C

Ok, I re-calculated with the normalised state vector and got:

4/20 for the eigenspinor with the positive eigenvalue of ħ/2
16/20 for the eigenspinor with the negative eigenvalue of -ħ/2

*I haven't put in my working here just to save on formatting time

4/20 + 16/20 = 1

Question though relating to part (a)
1) why don't I have to use the normalised state vector to get the expectation value?
2) And how can the expectation value be negative? I thought we would always at the very least expect the value to be a positive integer
 
  • #12
TSny
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4/20 for the eigenspinor with the positive eigenvalue of ħ/2
16/20 for the eigenspinor with the negative eigenvalue of -ħ/2
Looks right. Of course you can reduce these fractions.

Question though relating to part (a)
1) why don't I have to use the normalised state vector to get the expectation value?
You do need to used the normalized state. Or, you could use the more general formula for non-normalized states: ##\langle S_y \rangle = \frac {\langle \chi |S_y| \chi \rangle}{\langle \chi | \chi \rangle}##

2) And how can the expectation value be negative? I thought we would always at the very least expect the value to be a positive integer
Note that you have found that the probability of measuring ##-\hbar/2## is four times greater than the probability of getting ##+\hbar/2##. What does "expectation value" mean in terms of the probabilites?
 
  • #13
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12
Ok, so I've re-calculated the expectation value with the normalised state vector.

## 1/√10 \begin{pmatrix}
-i & 3 \\
\end{pmatrix} *
ħ/2 \begin{pmatrix}
0 & -i \\
i & 0 \\
\end{pmatrix} *
1/√10 \begin{pmatrix}
i \\
3 \\
\end{pmatrix}
##

## = -6ħ/20 = -3ħ/10 ##

So that is the expectation value. Expectation value is the mean / average of all the results we would get. The probability in part (b) is four times greater for -ħ/2 and thus the expectation value should be more skewed towards being a negative number?
 
  • #15
TSny
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The wavevector in Griffiths example is already normalized with the factor of 1/√6. It is a little odd that the vector that you are given in your problem has a factor of 1/5 (unless it is meant to test you on the idea of normalization :oldsmile:).
 
  • Like
Likes says
  • #16
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ok. Thanks for your help @TSny
 

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