Expectation Value and Probabilities of Spin Operator Sy

[says]

Homework Statement

(a) If a particle is in the spin state $χ = 1/5 \begin{pmatrix} i \\ 3 \\ \end{pmatrix}$ , calculate the expectation value <S_y>(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)
<S_y> = expectation value
<S_y> = < χ | S_y | 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) <S_y> =
$1/5 \begin{pmatrix} -i & 3\\ \end{pmatrix} ħ/2 \begin{pmatrix} 0 & -i \\ i & 0 \\ \end{pmatrix} 1/5 \begin{pmatrix} i \\ 3 \\ \end{pmatrix}$

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

<S_y> $= -6ħ/50 = -3ħ/25$

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 S_y 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 S_y 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.

 

[TSny]

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

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.

 

[says]

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?

 

[TSny]

What does it mean for a state vector to be normalized?

 

[says]

$[ 1 / |A| ] * A$

|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.

 

[TSny]

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|?

 

[says]

< χ | = A $\begin{pmatrix} - i & 3 \\ \end{pmatrix}$

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

< χ | χ > = 1 ⇒ A=1/10

 

[TSny]

< 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.

 

[says]

opps!

A= √(1/10)

 

[TSny]

Right.

 

[says]

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

 

[TSny]

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?

 

[says]

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?

 

[says]

I was following this Griffiths example to do this question. He doesn't normalise the state vector, which confuses me a bit in relation to my problem

 

[TSny]

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 ).

 

[says]

ok. Thanks for your help @TSny