- #1

says

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## 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.