Expectation Value and Probabilities of Spin Operator Sy

In summary, the expectation value for the spin operator Sy on a particle in spin state χ = 1/√10(i,3), is -3ħ/10. The probabilities of obtaining the eigenvalues ±ħ/2 are 4/20 and 16/20 respectively. This means that the probability of obtaining the eigenvalue -ħ/2 is four times greater than the probability of obtaining the eigenvalue ħ/2. The expectation value being negative is a result of this unequal probability distribution. The state vector given in the problem is not normalized, but can be normalized by multiplying by the factor 1/√10.
  • #1
says
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.
 
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  • #2
says said:
(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
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
What does it mean for a state vector to be normalized?
 
  • #5
## [ 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
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
< χ | = A ## \begin{pmatrix}
- i & 3 \\
\end{pmatrix} ##

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

< χ | χ > = 1 ⇒ A=1/10
 
  • #8
says said:
< 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
opps!

A= √(1/10)
 
  • #10
Right.
 
  • #11
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
says said:
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
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?
 
  • #14
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
 

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  • #15
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
ok. Thanks for your help @TSny
 

Related to Expectation Value and Probabilities of Spin Operator Sy

1. What is the expectation value of the spin operator?

The expectation value of the spin operator is a measure of the average value that would be obtained if the spin of a particle was measured multiple times. It is calculated by taking the sum of the possible spin values multiplied by their corresponding probabilities.

2. How is the spin operator related to probability?

The spin operator is related to probability in that it represents the possible spin states of a particle and the probabilities of measuring each state. The square of the spin operator gives the probability of measuring a specific spin value.

3. What is the difference between expectation value and probability in spin operators?

The expectation value of the spin operator is a measure of the average value that would be obtained from multiple measurements, while probability in spin operators represents the likelihood of measuring a specific spin value. The expectation value takes into account all possible spin states and their probabilities, while probability focuses on a single spin state.

4. How is the expectation value of the spin operator calculated?

The expectation value of the spin operator is calculated by taking the sum of the possible spin values multiplied by their corresponding probabilities. This can be represented mathematically as ⟨S⟩ = Σ SiPi, where Si represents the possible spin values and Pi represents their corresponding probabilities.

5. What is the significance of the expectation value in spin operators?

The expectation value in spin operators is significant because it allows us to make predictions about the behavior of particles in quantum systems. By calculating the expectation value, we can determine the most likely spin state that will be measured and make further calculations based on this information.

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