Given a general spin-1/2 state ∣ Ψ ⟩ = ( α β ) |\Psi\rangle = \begin{pmatrix}\alpha \\ \beta\end{pmatrix} ∣Ψ ⟩ = ( α β ) ∣ α ∣ 2 + ∣ β ∣ 2 = 1 |\alpha|^2 + |\beta|^2 = 1 ∣ α ∣ 2 + ∣ β ∣ 2 = 1 y y y z z z
Setup: eigenstates of σ y \sigma_y σ y # The Pauli matrix σ y = ( 0 − i i 0 ) \sigma_y = \begin{pmatrix}0 & -i \\ i & 0\end{pmatrix} σ y = ( 0 i − i 0 ) ± 1 \pm 1 ± 1 ± ℏ / 2 \pm\hbar/2 ± ℏ/2
∣ ↑ y ⟩ = 1 2 ( 1 i ) , ∣ ↓ y ⟩ = 1 2 ( 1 − i ) |\uparrow_y\rangle = \frac{1}{\sqrt{2}}\begin{pmatrix}1\\i\end{pmatrix}, \qquad |\downarrow_y\rangle = \frac{1}{\sqrt{2}}\begin{pmatrix}1\\-i\end{pmatrix} ∣ ↑ y ⟩ = 2 1 ( 1 i ) , ∣ ↓ y ⟩ = 2 1 ( 1 − i )
Our state ∣ Ψ ⟩ = ( α β ) |\Psi\rangle = \begin{pmatrix}\alpha\\\beta\end{pmatrix} ∣Ψ ⟩ = ( α β ) S z S_z S z ∣ Ψ ⟩ = α ∣ ↑ z ⟩ + β ∣ ↓ z ⟩ |\Psi\rangle = \alpha|\uparrow_z\rangle + \beta|\downarrow_z\rangle ∣Ψ ⟩ = α ∣ ↑ z ⟩ + β ∣ ↓ z ⟩ y y y σ y \sigma_y σ y
Probability of s y = + ℏ / 2 s_y = +\hbar/2 s y = + ℏ/2 # By the Born rule, P ( s y = + ℏ 2 ) = ∣ ⟨ ↑ y ∣ S y ∣ Ψ ⟩ ∣ 2 P(s_y = +\tfrac{\hbar}{2}) = |\langle\uparrow_y | S_y |\Psi\rangle|^2 P ( s y = + 2 ℏ ) = ∣ ⟨ ↑ y ∣ S y ∣Ψ ⟩ ∣ 2
P = 1 2 ∣ ( 1 − i ) ( 0 − i i 0 ) ( α β ) ∣ 2 P = \frac{1}{2}\left|\begin{pmatrix}1 & -i\end{pmatrix}\begin{pmatrix}0 & -i\\i & 0\end{pmatrix}\begin{pmatrix}\alpha\\\beta\end{pmatrix}\right|^2 P = 2 1 ( 1 − i ) ( 0 i − i 0 ) ( α β ) 2
First apply σ y \sigma_y σ y
σ y ∣ Ψ ⟩ = ( 0 − i i 0 ) ( α β ) = ( − i β i α ) \sigma_y|\Psi\rangle = \begin{pmatrix}0 & -i\\i & 0\end{pmatrix}\begin{pmatrix}\alpha\\\beta\end{pmatrix} = \begin{pmatrix}-i\beta\\i\alpha\end{pmatrix} σ y ∣Ψ ⟩ = ( 0 i − i 0 ) ( α β ) = ( − i β i α )
Then project:
P ( s y = + ) = 1 2 ∣ ( 1 − i ) ( − i β i α ) ∣ 2 = 1 2 ∣ − i β − i 2 α ∣ 2 = 1 2 ∣ α − i β ∣ 2 P(s_y = +) = \frac{1}{2}\left|\begin{pmatrix}1 & -i\end{pmatrix}\begin{pmatrix}-i\beta\\i\alpha\end{pmatrix}\right|^2 = \frac{1}{2}|-i\beta - i^2\alpha|^2 = \frac{1}{2}|\alpha - i\beta|^2 P ( s y = + ) = 2 1 ( 1 − i ) ( − i β i α ) 2 = 2 1 ∣ − i β − i 2 α ∣ 2 = 2 1 ∣ α − i β ∣ 2
Probability of s y = − ℏ / 2 s_y = -\hbar/2 s y = − ℏ/2 # Similarly:
P ( s y = − ) = 1 2 ∣ ( 1 i ) ( − i β i α ) ∣ 2 = 1 2 ∣ α + i β ∣ 2 P(s_y = -) = \frac{1}{2}\left|\begin{pmatrix}1 & i\end{pmatrix}\begin{pmatrix}-i\beta\\i\alpha\end{pmatrix}\right|^2 = \frac{1}{2}|\alpha + i\beta|^2 P ( s y = − ) = 2 1 ( 1 i ) ( − i β i α ) 2 = 2 1 ∣ α + i β ∣ 2
Sanity check: P ( + ) + P ( − ) = 1 2 ( ∣ α − i β ∣ 2 + ∣ α + i β ∣ 2 ) = 1 2 ( 2 ∣ α ∣ 2 + 2 ∣ β ∣ 2 ) = 1 P(+) + P(-) = \frac{1}{2}(|\alpha - i\beta|^2 + |\alpha + i\beta|^2) = \frac{1}{2}(2|\alpha|^2 + 2|\beta|^2) = 1 P ( + ) + P ( − ) = 2 1 ( ∣ α − i β ∣ 2 + ∣ α + i β ∣ 2 ) = 2 1 ( 2∣ α ∣ 2 + 2∣ β ∣ 2 ) = 1
Probabilities along z z z # For σ z = ( 1 0 0 − 1 ) \sigma_z = \begin{pmatrix}1&0\\0&-1\end{pmatrix} σ z = ( 1 0 0 − 1 ) ∣ ↑ z ⟩ = ( 1 0 ) |\uparrow_z\rangle = \begin{pmatrix}1\\0\end{pmatrix} ∣ ↑ z ⟩ = ( 1 0 ) ∣ ↓ z ⟩ = ( 0 1 ) |\downarrow_z\rangle = \begin{pmatrix}0\\1\end{pmatrix} ∣ ↓ z ⟩ = ( 0 1 )
P ( s z = + ) = ∣ ( 1 0 ) ( 1 0 0 − 1 ) ( α β ) ∣ 2 = ∣ ( 1 0 ) ( α − β ) ∣ 2 = ∣ α ∣ 2 P(s_z = +) = \left|\begin{pmatrix}1&0\end{pmatrix}\begin{pmatrix}1&0\\0&-1\end{pmatrix}\begin{pmatrix}\alpha\\\beta\end{pmatrix}\right|^2 = \left|\begin{pmatrix}1&0\end{pmatrix}\begin{pmatrix}\alpha\\-\beta\end{pmatrix}\right|^2 = |\alpha|^2 P ( s z = + ) = ( 1 0 ) ( 1 0 0 − 1 ) ( α β ) 2 = ( 1 0 ) ( α − β ) 2 = ∣ α ∣ 2
P ( s z = − ) = ∣ ( 0 1 ) ( α − β ) ∣ 2 = ∣ − β ∣ 2 = ∣ β ∣ 2 P(s_z = -) = \left|\begin{pmatrix}0&1\end{pmatrix}\begin{pmatrix}\alpha\\-\beta\end{pmatrix}\right|^2 = |-\beta|^2 = |\beta|^2 P ( s z = − ) = ( 0 1 ) ( α − β ) 2 = ∣ − β ∣ 2 = ∣ β ∣ 2
Which is just the Born rule applied directly — reassuringly, α \alpha α β \beta β s z = ± ℏ / 2 s_z = \pm\hbar/2 s z = ± ℏ/2
Verification with known states# Case 1: ∣ Ψ ⟩ = ∣ ↑ y ⟩ |\Psi\rangle = |\uparrow_y\rangle ∣Ψ ⟩ = ∣ ↑ y ⟩ α = 1 2 \alpha = \frac{1}{\sqrt{2}} α = 2 1 β = i 2 \beta = \frac{i}{\sqrt{2}} β = 2 i # P ( s y = + ) = 1 2 ∣ 1 2 − i ⋅ i 2 ∣ 2 = 1 2 ∣ 1 2 + 1 2 ∣ 2 = 1 2 ⋅ 2 = 1 ✓ P(s_y = +) = \frac{1}{2}\left|\frac{1}{\sqrt{2}} - i\cdot\frac{i}{\sqrt{2}}\right|^2 = \frac{1}{2}\left|\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}\right|^2 = \frac{1}{2}\cdot 2 = 1 \checkmark P ( s y = + ) = 2 1 2 1 − i ⋅ 2 i 2 = 2 1 2 1 + 2 1 2 = 2 1 ⋅ 2 = 1 ✓
P ( s y = − ) = 1 2 ∣ 1 2 + i ⋅ i 2 ∣ 2 = 1 2 ⋅ 0 = 0 ✓ P(s_y = -) = \frac{1}{2}\left|\frac{1}{\sqrt{2}} + i\cdot\frac{i}{\sqrt{2}}\right|^2 = \frac{1}{2}\cdot 0 = 0 \checkmark P ( s y = − ) = 2 1 2 1 + i ⋅ 2 i 2 = 2 1 ⋅ 0 = 0 ✓
Case 2: ∣ Ψ ⟩ = ∣ ↑ z ⟩ |\Psi\rangle = |\uparrow_z\rangle ∣Ψ ⟩ = ∣ ↑ z ⟩ α = 1 \alpha = 1 α = 1 β = 0 \beta = 0 β = 0 # P ( s y = + ) = 1 2 ∣ 1 − 0 ∣ 2 = 1 2 , P ( s y = − ) = 1 2 ∣ 1 + 0 ∣ 2 = 1 2 P(s_y = +) = \frac{1}{2}|1 - 0|^2 = \frac{1}{2}, \qquad P(s_y = -) = \frac{1}{2}|1 + 0|^2 = \frac{1}{2} P ( s y = + ) = 2 1 ∣1 − 0 ∣ 2 = 2 1 , P ( s y = − ) = 2 1 ∣1 + 0 ∣ 2 = 2 1
A spin-up z z z y y y S y S_y S y S z S_z S z [ S y , S z ] = i ℏ S x ≠ 0 [S_y, S_z] = i\hbar S_x \neq 0 [ S y , S z ] = i ℏ S x = 0 z z z y y y
Case 3: ∣ Ψ ⟩ = ∣ ↓ z ⟩ |\Psi\rangle = |\downarrow_z\rangle ∣Ψ ⟩ = ∣ ↓ z ⟩ α = 0 \alpha = 0 α = 0 β = 1 \beta = 1 β = 1 # P ( s y = + ) = 1 2 ∣ − i ∣ 2 = 1 2 , P ( s y = − ) = 1 2 ∣ i ∣ 2 = 1 2 P(s_y = +) = \frac{1}{2}|{-i}|^2 = \frac{1}{2}, \qquad P(s_y = -) = \frac{1}{2}|i|^2 = \frac{1}{2} P ( s y = + ) = 2 1 ∣ − i ∣ 2 = 2 1 , P ( s y = − ) = 2 1 ∣ i ∣ 2 = 2 1
Same result — again expected by symmetry.
