These solved problems practice the spin-1 / 2 1/2 1/2
Use Pauli Matrix Identity Index as the formula companion. For the conceptual pages, start with Spin-1/2 Hilbert Space , Pauli Matrices , Bloch Sphere , and Spin Rotations .
Use the S z S_z S z
∣ + z ⟩ = ( 1 0 ) , ∣ − z ⟩ = ( 0 1 ) . \lvert+z\rangle
=
\begin{pmatrix}
1\\
0
\end{pmatrix},
\qquad
\lvert-z\rangle
=
\begin{pmatrix}
0\\
1
\end{pmatrix}. ∣ + z ⟩ = ( 1 0 ) , ∣ − z ⟩ = ( 0 1 ) . The spin operators are
S i = ℏ 2 σ i . S_i=\frac{\hbar}{2}\sigma_i. S i = 2 ℏ σ i . For a unit vector n ^ \hat{\mathbf n} n ^
P ± ( n ^ ) = 1 2 ( I ± n ^ ⋅ σ ) . P_\pm(\hat{\mathbf n})
=
\frac12
\left(
I\pm\hat{\mathbf n}\cdot\boldsymbol\sigma
\right). P ± ( n ^ ) = 2 1 ( I ± n ^ ⋅ σ ) . Spinor rotations are
U ( n ^ , θ ) = exp ( − i θ 2 n ^ ⋅ σ ) . U(\hat{\mathbf n},\theta)
=
\exp
\left(
-\frac{i\theta}{2}\hat{\mathbf n}\cdot\boldsymbol\sigma
\right). U ( n ^ , θ ) = exp ( − 2 i θ n ^ ⋅ σ ) . An incoming beam is prepared in ∣ + z ⟩ \lvert+z\rangle ∣ + z ⟩
n ^ = ( sin θ , 0 , cos θ ) . \hat{\mathbf n}
=
(\sin\theta,0,\cos\theta). n ^ = ( sin θ , 0 , cos θ ) . Find the probabilities for the + n ^ +\hat{\mathbf n} + n ^ − n ^ -\hat{\mathbf n} − n ^
Solution The + n ^ +\hat{\mathbf n} + n ^
P + ( n ^ ) = 1 2 ( I + n ^ ⋅ σ ) . P_+(\hat{\mathbf n})
=
\frac12
\left(
I+\hat{\mathbf n}\cdot\boldsymbol\sigma
\right). P + ( n ^ ) = 2 1 ( I + n ^ ⋅ σ ) . The state ∣ + z ⟩ \lvert+z\rangle ∣ + z ⟩ z ^ \hat{\mathbf z} z ^
p + = 1 2 ( 1 + z ^ ⋅ n ^ ) = 1 2 ( 1 + cos θ ) = cos 2 θ 2 . p_+
=
\frac12
\left(
1+\hat{\mathbf z}\cdot\hat{\mathbf n}
\right)
=
\frac12(1+\cos\theta)
=
\cos^2\frac{\theta}{2}. p + = 2 1 ( 1 + z ^ ⋅ n ^ ) = 2 1 ( 1 + cos θ ) = cos 2 2 θ . Similarly,
p − = 1 2 ( 1 − cos θ ) = sin 2 θ 2 . p_-
=
\frac12(1-\cos\theta)
=
\sin^2\frac{\theta}{2}. p − = 2 1 ( 1 − cos θ ) = sin 2 2 θ . The half-angle appears because spinors double-cover ordinary directions.
Let
∣ ψ ⟩ = cos θ 2 ∣ + z ⟩ + e i ϕ sin θ 2 ∣ − z ⟩ . \lvert\psi\rangle
=
\cos\frac{\theta}{2}\lvert+z\rangle
+
e^{i\phi}\sin\frac{\theta}{2}\lvert-z\rangle. ∣ ψ ⟩ = cos 2 θ ∣ + z ⟩ + e i ϕ sin 2 θ ∣ − z ⟩ . Compute ⟨ σ x ⟩ \langle\sigma_x\rangle ⟨ σ x ⟩ ⟨ σ y ⟩ \langle\sigma_y\rangle ⟨ σ y ⟩ ⟨ σ z ⟩ \langle\sigma_z\rangle ⟨ σ z ⟩
Solution Write
a = cos θ 2 , b = e i ϕ sin θ 2 . a=\cos\frac{\theta}{2},
\qquad
b=e^{i\phi}\sin\frac{\theta}{2}. a = cos 2 θ , b = e i ϕ sin 2 θ . Then
⟨ σ x ⟩ = a ∗ b + b ∗ a = 2 Re ( a ∗ b ) = sin θ cos ϕ . \langle\sigma_x\rangle
=
a^*b+b^*a
=
2\operatorname{Re}(a^*b)
=
\sin\theta\cos\phi. ⟨ σ x ⟩ = a ∗ b + b ∗ a = 2 Re ( a ∗ b ) = sin θ cos ϕ . Similarly,
⟨ σ y ⟩ = − i a ∗ b + i b ∗ a = 2 Im ( a ∗ b ) = sin θ sin ϕ . \langle\sigma_y\rangle
=
-ia^*b+ib^*a
=
2\operatorname{Im}(a^*b)
=
\sin\theta\sin\phi. ⟨ σ y ⟩ = − i a ∗ b + i b ∗ a = 2 Im ( a ∗ b ) = sin θ sin ϕ . Finally,
⟨ σ z ⟩ = ∣ a ∣ 2 − ∣ b ∣ 2 = cos 2 θ 2 − sin 2 θ 2 = cos θ . \langle\sigma_z\rangle
=
\lvert a\rvert^2-\lvert b\rvert^2
=
\cos^2\frac{\theta}{2}
-
\sin^2\frac{\theta}{2}
=
\cos\theta. ⟨ σ z ⟩ = ∣ a ∣ 2 − ∣ b ∣ 2 = cos 2 2 θ − sin 2 2 θ = cos θ . Thus the Bloch vector is
n ^ = ( sin θ cos ϕ , sin θ sin ϕ , cos θ ) . \hat{\mathbf n}
=
(\sin\theta\cos\phi,\sin\theta\sin\phi,\cos\theta). n ^ = ( sin θ cos ϕ , sin θ sin ϕ , cos θ ) .
A beam is prepared in ∣ + z ⟩ \lvert+z\rangle ∣ + z ⟩ x x x + x +x + x z z z + z +z + z − z -z − z
Solution After the + x +x + x
∣ + x ⟩ = 1 2 ( ∣ + z ⟩ + ∣ − z ⟩ ) . \lvert+x\rangle
=
\frac{1}{\sqrt2}
\left(
\lvert+z\rangle+\lvert-z\rangle
\right). ∣ + x ⟩ = 2 1 ( ∣ + z ⟩ + ∣ − z ⟩ ) . The final z z z ∣ + z ⟩ , ∣ − z ⟩ \lvert+z\rangle,\lvert-z\rangle ∣ + z ⟩ , ∣ − z ⟩
p ( + z ) = ∣ ⟨ + z ∣ + x ⟩ ∣ 2 = 1 2 , p(+z)
=
\left|
\langle+z\vert+x\rangle
\right|^2
=
\frac12, p ( + z ) = ∣ ⟨ + z ∣ + x ⟩ ∣ 2 = 2 1 , and
p ( − z ) = ∣ ⟨ − z ∣ + x ⟩ ∣ 2 = 1 2 . p(-z)
=
\left|
\langle-z\vert+x\rangle
\right|^2
=
\frac12. p ( − z ) = ∣ ⟨ − z ∣ + x ⟩ ∣ 2 = 2 1 . The intermediate x x x S z S_z S z
Apply a spinor rotation about the z z z α \alpha α ∣ + x ⟩ \lvert+x\rangle ∣ + x ⟩
Solution The rotation operator is
U ( z ^ , α ) = ( e − i α / 2 0 0 e i α / 2 ) . U(\hat z,\alpha)
=
\begin{pmatrix}
e^{-i\alpha/2}&0\\
0&e^{i\alpha/2}
\end{pmatrix}. U ( z ^ , α ) = ( e − i α /2 0 0 e i α /2 ) . Since
∣ + x ⟩ = 1 2 ( ∣ + z ⟩ + ∣ − z ⟩ ) , \lvert+x\rangle
=
\frac{1}{\sqrt2}
\left(
\lvert+z\rangle+\lvert-z\rangle
\right), ∣ + x ⟩ = 2 1 ( ∣ + z ⟩ + ∣ − z ⟩ ) , we get
U ( z ^ , α ) ∣ + x ⟩ = 1 2 ( e − i α / 2 ∣ + z ⟩ + e i α / 2 ∣ − z ⟩ ) . U(\hat z,\alpha)\lvert+x\rangle
=
\frac{1}{\sqrt2}
\left(
e^{-i\alpha/2}\lvert+z\rangle
+
e^{i\alpha/2}\lvert-z\rangle
\right). U ( z ^ , α ) ∣ + x ⟩ = 2 1 ( e − i α /2 ∣ + z ⟩ + e i α /2 ∣ − z ⟩ ) . Removing the global phase e − i α / 2 e^{-i\alpha/2} e − i α /2
1 2 ( ∣ + z ⟩ + e i α ∣ − z ⟩ ) . \frac{1}{\sqrt2}
\left(
\lvert+z\rangle
+
e^{i\alpha}\lvert-z\rangle
\right). 2 1 ( ∣ + z ⟩ + e i α ∣ − z ⟩ ) . This is the Bloch-sphere direction
( cos α , sin α , 0 ) . (\cos\alpha,\sin\alpha,0). ( cos α , sin α , 0 ) . The spinor uses half-angles, but the Bloch vector rotates by the physical angle α \alpha α
Let
H = − γ ℏ B 0 2 σ z , H
=
-\frac{\gamma\hbar B_0}{2}\sigma_z, H = − 2 γ ℏ B 0 σ z , and define ω = γ B 0 \omega=\gamma B_0 ω = γ B 0 ∣ + x ⟩ \lvert+x\rangle ∣ + x ⟩ ⟨ σ x ⟩ ( t ) \langle\sigma_x\rangle(t) ⟨ σ x ⟩ ( t ) ⟨ σ y ⟩ ( t ) \langle\sigma_y\rangle(t) ⟨ σ y ⟩ ( t )
Solution The time-evolution operator is
U ( t ) = exp ( − i H t ℏ ) = exp ( i ω t 2 σ z ) . U(t)
=
\exp\left(-\frac{iHt}{\hbar}\right)
=
\exp\left(\frac{i\omega t}{2}\sigma_z\right). U ( t ) = exp ( − ℏ i H t ) = exp ( 2 iω t σ z ) . Acting on ∣ + x ⟩ \lvert+x\rangle ∣ + x ⟩
∣ ψ ( t ) ⟩ = 1 2 ( e i ω t / 2 ∣ + z ⟩ + e − i ω t / 2 ∣ − z ⟩ ) . \lvert\psi(t)\rangle
=
\frac{1}{\sqrt2}
\left(
e^{i\omega t/2}\lvert+z\rangle
+
e^{-i\omega t/2}\lvert-z\rangle
\right). ∣ ψ ( t )⟩ = 2 1 ( e iω t /2 ∣ + z ⟩ + e − iω t /2 ∣ − z ⟩ ) . Up to a global phase, this is
1 2 ( ∣ + z ⟩ + e − i ω t ∣ − z ⟩ ) . \frac{1}{\sqrt2}
\left(
\lvert+z\rangle
+
e^{-i\omega t}\lvert-z\rangle
\right). 2 1 ( ∣ + z ⟩ + e − iω t ∣ − z ⟩ ) . Therefore the Bloch azimuthal angle is − ω t -\omega t − ω t
⟨ σ x ⟩ ( t ) = cos ω t , ⟨ σ y ⟩ ( t ) = − sin ω t . \langle\sigma_x\rangle(t)=\cos\omega t,
\qquad
\langle\sigma_y\rangle(t)=-\sin\omega t. ⟨ σ x ⟩ ( t ) = cos ω t , ⟨ σ y ⟩ ( t ) = − sin ω t . If γ \gamma γ ω \omega ω
Show that a 2 π 2\pi 2 π 1 / 2 1/2 1/2 − 1 -1 − 1
Solution For any axis n ^ \hat{\mathbf n} n ^
U ( n ^ , 2 π ) = cos π I − i sin π n ^ ⋅ σ = − I . U(\hat{\mathbf n},2\pi)
=
\cos\pi\,I
-i\sin\pi\,\hat{\mathbf n}\cdot\boldsymbol\sigma
=
-I. U ( n ^ , 2 π ) = cos π I − i sin π n ^ ⋅ σ = − I . Thus
∣ ψ ⟩ ↦ − ∣ ψ ⟩ . \lvert\psi\rangle
\mapsto
-\lvert\psi\rangle. ∣ ψ ⟩ ↦ − ∣ ψ ⟩ . For a projector P P P
⟨ − ψ ∣ P ∣ − ψ ⟩ = ⟨ ψ ∣ P ∣ ψ ⟩ . \langle-\psi\vert P\vert-\psi\rangle
=
\langle\psi\vert P\vert\psi\rangle. ⟨ − ψ ∣ P ∣ − ψ ⟩ = ⟨ ψ ∣ P ∣ ψ ⟩ . The ray is unchanged. The sign can matter only when it becomes a relative phase compared with another branch or reference path.
Let ∣ + a ^ ⟩ \lvert+\hat{\mathbf a}\rangle ∣ + a ^ ⟩ ∣ + b ^ ⟩ \lvert+\hat{\mathbf b}\rangle ∣ + b ^ ⟩ a ^ \hat{\mathbf a} a ^ b ^ \hat{\mathbf b} b ^
∣ ⟨ + b ^ ∣ + a ^ ⟩ ∣ 2 = 1 2 ( 1 + a ^ ⋅ b ^ ) . \left|
\langle+\hat{\mathbf b}\vert+\hat{\mathbf a}\rangle
\right|^2
=
\frac12
\left(
1+\hat{\mathbf a}\cdot\hat{\mathbf b}
\right). ⟨ + b ^ ∣ + a ^ ⟩ 2 = 2 1 ( 1 + a ^ ⋅ b ^ ) .
Solution The projector onto spin up along b ^ \hat{\mathbf b} b ^
P + ( b ^ ) = 1 2 ( I + b ^ ⋅ σ ) . P_+(\hat{\mathbf b})
=
\frac12
\left(
I+\hat{\mathbf b}\cdot\boldsymbol\sigma
\right). P + ( b ^ ) = 2 1 ( I + b ^ ⋅ σ ) . The state ∣ + a ^ ⟩ \lvert+\hat{\mathbf a}\rangle ∣ + a ^ ⟩ a ^ \hat{\mathbf a} a ^
⟨ + a ^ ∣ σ ∣ + a ^ ⟩ = a ^ . \langle+\hat{\mathbf a}\vert
\boldsymbol\sigma
\vert+\hat{\mathbf a}\rangle
=
\hat{\mathbf a}. ⟨ + a ^ ∣ σ ∣ + a ^ ⟩ = a ^ . Therefore
∣ ⟨ + b ^ ∣ + a ^ ⟩ ∣ 2 = ⟨ + a ^ ∣ P + ( b ^ ) ∣ + a ^ ⟩ = 1 2 ( 1 + a ^ ⋅ b ^ ) . \left|
\langle+\hat{\mathbf b}\vert+\hat{\mathbf a}\rangle
\right|^2
=
\langle+\hat{\mathbf a}\vert
P_+(\hat{\mathbf b})
\vert+\hat{\mathbf a}\rangle
=
\frac12
\left(
1+\hat{\mathbf a}\cdot\hat{\mathbf b}
\right). ⟨ + b ^ ∣ + a ^ ⟩ 2 = ⟨ + a ^ ∣ P + ( b ^ ) ∣ + a ^ ⟩ = 2 1 ( 1 + a ^ ⋅ b ^ ) . If the angle between the directions is γ \gamma γ cos 2 ( γ / 2 ) \cos^2(\gamma/2) cos 2 ( γ /2 )
Use
Θ = − i σ y K , \Theta=-i\sigma_yK, Θ = − i σ y K , where K K K S z S_z S z Θ ∣ + z ⟩ \Theta\lvert+z\rangle Θ ∣ + z ⟩ Θ ∣ − z ⟩ \Theta\lvert-z\rangle Θ ∣ − z ⟩ Θ 2 ∣ + z ⟩ \Theta^2\lvert+z\rangle Θ 2 ∣ + z ⟩
Solution In the S z S_z S z
∣ + z ⟩ = ( 1 0 ) , ∣ − z ⟩ = ( 0 1 ) . \lvert+z\rangle
=
\begin{pmatrix}
1\\
0
\end{pmatrix},
\qquad
\lvert-z\rangle
=
\begin{pmatrix}
0\\
1
\end{pmatrix}. ∣ + z ⟩ = ( 1 0 ) , ∣ − z ⟩ = ( 0 1 ) . Since these basis vectors are real, K K K
− i σ y = ( 0 − 1 1 0 ) . -i\sigma_y
=
\begin{pmatrix}
0&-1\\
1&0
\end{pmatrix}. − i σ y = ( 0 1 − 1 0 ) . Thus
Θ ∣ + z ⟩ = ∣ − z ⟩ , \Theta\lvert+z\rangle
=
\lvert-z\rangle, Θ ∣ + z ⟩ = ∣ − z ⟩ , and
Θ ∣ − z ⟩ = − ∣ + z ⟩ . \Theta\lvert-z\rangle
=
-\lvert+z\rangle. Θ ∣ − z ⟩ = − ∣ + z ⟩ . Applying Θ \Theta Θ
Θ 2 ∣ + z ⟩ = Θ ∣ − z ⟩ = − ∣ + z ⟩ . \Theta^2\lvert+z\rangle
=
\Theta\lvert-z\rangle
=
-\lvert+z\rangle. Θ 2 ∣ + z ⟩ = Θ ∣ − z ⟩ = − ∣ + z ⟩ . The minus sign is the spin-1 / 2 1/2 1/2 Θ 2 = − I \Theta^2=-I Θ 2 = − I
Confusing the analyzer direction with a property called simply “spin.”
Treating σ i \sigma_i σ i S i = ℏ σ i / 2 S_i=\hbar\sigma_i/2 S i = ℏ σ i /2
Forgetting the half-angle in spinor rotations.
Treating a 2 π 2\pi 2 π
Losing the sign of γ \gamma γ
Using Θ = − i σ y \Theta=-i\sigma_y Θ = − i σ y K K K
J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics , 3rd ed., Cambridge University Press, 2020.
R. Shankar, Principles of Quantum Mechanics , 2nd ed., Springer, 1994.
L. E. Ballentine, Quantum Mechanics: A Modern Development , 2nd ed., World Scientific, 2014.
C. Cohen-Tannoudji, B. Diu, and F. Laloë, Quantum Mechanics , Wiley, 1977.