Spin-1 / 2 1/2 1/2 2 × 2 2\times2 2 × 2 S z S_z S z
S i = ℏ 2 σ i . S_i
=
\frac{\hbar}{2}\sigma_i. S i = 2 ℏ σ i . This card is organized around four recurring tasks:
construct the operator and eigenstates for a chosen measurement axis;
turn a spinor or density matrix into outcome probabilities;
rotate a spin state through a specified angle;
diagonalize and exponentiate a general Hermitian two-level Hamiltonian.
The conceptual explanation belongs at
Pauli Matrices . For an
exhaustive multiplication, trace, eigenvector, and Pauli-string lookup, use the
Pauli Matrices Table .
Use
{ ∣ + z ⟩ , ∣ − z ⟩ } , ∣ + z ⟩ = ( 1 0 ) , ∣ − z ⟩ = ( 0 1 ) . \lbrace
\lvert+z\rangle,
\lvert-z\rangle
\rbrace,
\qquad
\lvert+z\rangle
=
\begin{pmatrix}1\\0\end{pmatrix},
\qquad
\lvert-z\rangle
=
\begin{pmatrix}0\\1\end{pmatrix}. {∣ + z ⟩ , ∣ − z ⟩} , ∣ + z ⟩ = ( 1 0 ) , ∣ − z ⟩ = ( 0 1 ) . These satisfy
S z ∣ ± z ⟩ = ± ℏ 2 ∣ ± z ⟩ . S_z\lvert\pm z\rangle
=
\pm\frac{\hbar}{2}
\lvert\pm z\rangle. S z ∣ ± z ⟩ = ± 2 ℏ ∣ ± z ⟩ . In this basis,
σ x = ( 0 1 1 0 ) , σ y = ( 0 − i i 0 ) , σ z = ( 1 0 0 − 1 ) . \sigma_x
=
\begin{pmatrix}
0&1\\
1&0
\end{pmatrix},
\qquad
\sigma_y
=
\begin{pmatrix}
0&-i\\
i&0
\end{pmatrix},
\qquad
\sigma_z
=
\begin{pmatrix}
1&0\\
0&-1
\end{pmatrix}. σ x = ( 0 1 1 0 ) , σ y = ( 0 i − i 0 ) , σ z = ( 1 0 0 − 1 ) . Changing the ordered basis changes the displayed matrices. The abstract
operator and all physical probabilities remain unchanged if states and
operators are transformed consistently.
Task Formula Physical spin operator S i = ℏ σ i / 2 S_i=\hbar\sigma_i/2 S i = ℏ σ i /2 Pauli product σ i σ j = δ i j I + i ϵ i j k σ k \sigma_i\sigma_j=\delta_{ij}I+i\epsilon_{ijk}\sigma_k σ i σ j = δ ij I + i ϵ ij k σ k Directional observable S n = ℏ n ⋅ σ / 2 S_{\boldsymbol n}=\hbar\boldsymbol n\cdot\boldsymbol\sigma/2 S n = ℏ n ⋅ σ /2 Directional projectors P ± ( n ) = [ I ± n ⋅ σ ] / 2 P_\pm(\boldsymbol n)=[I\pm\boldsymbol n\cdot\boldsymbol\sigma]/2 P ± ( n ) = [ I ± n ⋅ σ ] /2 Outcome probability p ± = [ 1 ± r ⋅ n ] / 2 p_\pm=[1\pm\boldsymbol r\cdot\boldsymbol n]/2 p ± = [ 1 ± r ⋅ n ] /2 Active rotation U n ( θ ) = cos ( θ / 2 ) I − i sin ( θ / 2 ) n ⋅ σ U_{\boldsymbol n}(\theta)=\cos(\theta/2)I-i\sin(\theta/2)\boldsymbol n\cdot\boldsymbol\sigma U n ( θ ) = cos ( θ /2 ) I − i sin ( θ /2 ) n ⋅ σ General Hermitian matrix H = c 0 I + c ⋅ σ H=c_0I+\boldsymbol c\cdot\boldsymbol\sigma H = c 0 I + c ⋅ σ Two-level energies E ± = c 0 ± ∥ c ∥ E_\pm=c_0\pm\lVert\boldsymbol c\rVert E ± = c 0 ± ∥ c ∥
The compact multiplication rule is
σ i σ j = δ i j I + i ϵ i j k σ k . \sigma_i\sigma_j
=
\delta_{ij}I
+
i\epsilon_{ijk}\sigma_k. σ i σ j = δ ij I + i ϵ ij k σ k . Its symmetric and antisymmetric parts are
{ σ i , σ j } = 2 δ i j I , [ σ i , σ j ] = 2 i ϵ i j k σ k . \lbrace\sigma_i,\sigma_j\rbrace
=
2\delta_{ij}I,
\qquad
[\sigma_i,\sigma_j]
=
2i\epsilon_{ijk}\sigma_k. { σ i , σ j } = 2 δ ij I , [ σ i , σ j ] = 2 i ϵ ij k σ k . Consequently,
[ S i , S j ] = i ℏ ϵ i j k S k [S_i,S_j]
=
i\hbar\epsilon_{ijk}S_k [ S i , S j ] = i ℏ ϵ ij k S k and
S 2 = S x 2 + S y 2 + S z 2 = 3 4 ℏ 2 I . S^2
=
S_x^2+S_y^2+S_z^2
=
\frac{3}{4}\hbar^2I. S 2 = S x 2 + S y 2 + S z 2 = 4 3 ℏ 2 I . For three-component vectors a \boldsymbol a a b \boldsymbol b b
( a ⋅ σ ) ( b ⋅ σ ) = ( a ⋅ b ) I + i ( a × b ) ⋅ σ . (\boldsymbol a\cdot\boldsymbol\sigma)
(\boldsymbol b\cdot\boldsymbol\sigma)
=
(\boldsymbol a\cdot\boldsymbol b)I
+
i(\boldsymbol a\times\boldsymbol b)
\cdot\boldsymbol\sigma. ( a ⋅ σ ) ( b ⋅ σ ) = ( a ⋅ b ) I + i ( a × b ) ⋅ σ . This identity is usually faster and less error-prone than multiplying entries.
For a real unit vector n \boldsymbol n n
( n ⋅ σ ) 2 = I . (\boldsymbol n\cdot\boldsymbol\sigma)^2=I. ( n ⋅ σ ) 2 = I . That single result controls directional eigenvalues, spectral projectors, and
matrix exponentials.
Parameterize a unit vector by
n = ( sin θ cos ϕ , sin θ sin ϕ , cos θ ) . \boldsymbol n
=
(\sin\theta\cos\phi,
\sin\theta\sin\phi,
\cos\theta). n = ( sin θ cos ϕ , sin θ sin ϕ , cos θ ) . The dimensionless directional operator is
σ n = n ⋅ σ = ( cos θ e − i ϕ sin θ e i ϕ sin θ − cos θ ) . \sigma_{\boldsymbol n}
=
\boldsymbol n\cdot\boldsymbol\sigma
=
\begin{pmatrix}
\cos\theta & e^{-i\phi}\sin\theta\\
e^{i\phi}\sin\theta & -\cos\theta
\end{pmatrix}. σ n = n ⋅ σ = ( cos θ e i ϕ sin θ e − i ϕ sin θ − cos θ ) . It has eigenvalues + 1 +1 + 1 − 1 -1 − 1
∣ n , + ⟩ = ( cos ( θ / 2 ) e i ϕ sin ( θ / 2 ) ) , \lvert\boldsymbol n,+\rangle
=
\begin{pmatrix}
\cos(\theta/2)\\
e^{i\phi}\sin(\theta/2)
\end{pmatrix}, ∣ n , + ⟩ = ( cos ( θ /2 ) e i ϕ sin ( θ /2 ) ) , ∣ n , − ⟩ = ( − e − i ϕ sin ( θ / 2 ) cos ( θ / 2 ) ) . \lvert\boldsymbol n,-\rangle
=
\begin{pmatrix}
-e^{-i\phi}\sin(\theta/2)\\
\cos(\theta/2)
\end{pmatrix}. ∣ n , − ⟩ = ( − e − i ϕ sin ( θ /2 ) cos ( θ /2 ) ) . They satisfy
σ n ∣ n , ± ⟩ = ± ∣ n , ± ⟩ . \sigma_{\boldsymbol n}
\lvert\boldsymbol n,\pm\rangle
=
\pm
\lvert\boldsymbol n,\pm\rangle. σ n ∣ n , ± ⟩ = ± ∣ n , ± ⟩ . Their phases are conventional. Multiplying either eigenvector by its own
constant phase changes no projector or probability.
Common Cartesian eigenstates in the stated basis are
State Column vector ∣ + x ⟩ \lvert+x\rangle ∣ + x ⟩ 2 − 1 / 2 ( 1 , 1 ) T 2^{-1/2}(1,1)^{\mathsf T} 2 − 1/2 ( 1 , 1 ) T ∣ − x ⟩ \lvert-x\rangle ∣ − x ⟩ 2 − 1 / 2 ( 1 , − 1 ) T 2^{-1/2}(1,-1)^{\mathsf T} 2 − 1/2 ( 1 , − 1 ) T ∣ + y ⟩ \lvert+y\rangle ∣ + y ⟩ 2 − 1 / 2 ( 1 , i ) T 2^{-1/2}(1,i)^{\mathsf T} 2 − 1/2 ( 1 , i ) T ∣ − y ⟩ \lvert-y\rangle ∣ − y ⟩ 2 − 1 / 2 ( 1 , − i ) T 2^{-1/2}(1,-i)^{\mathsf T} 2 − 1/2 ( 1 , − i ) T ∣ + z ⟩ \lvert+z\rangle ∣ + z ⟩ ( 1 , 0 ) T (1,0)^{\mathsf T} ( 1 , 0 ) T ∣ − z ⟩ \lvert-z\rangle ∣ − z ⟩ ( 0 , 1 ) T (0,1)^{\mathsf T} ( 0 , 1 ) T
The spectral projectors of σ n \sigma_{\boldsymbol n} σ n
P ± ( n ) = 1 2 ( I ± n ⋅ σ ) . P_\pm(\boldsymbol n)
=
\frac12
\left(
I\pm\boldsymbol n\cdot\boldsymbol\sigma
\right). P ± ( n ) = 2 1 ( I ± n ⋅ σ ) . They obey
P ± 2 = P ± , P + P − = 0 , P + + P − = I . P_\pm^2=P_\pm,
\qquad
P_+P_-=0,
\qquad
P_++P_-=I. P ± 2 = P ± , P + P − = 0 , P + + P − = I . For a state ρ \rho ρ S n = ± ℏ / 2 S_{\boldsymbol n}=\pm\hbar/2 S n = ± ℏ/2
p ± = tr [ ρ P ± ( n ) ] . p_\pm
=
\operatorname{tr}
\left[
\rho P_\pm(\boldsymbol n)
\right]. p ± = tr [ ρ P ± ( n ) ] . If
ρ = 1 2 ( I + r ⋅ σ ) , ∥ r ∥ ≤ 1 , \rho
=
\frac12
\left(
I+\boldsymbol r\cdot\boldsymbol\sigma
\right),
\qquad
\lVert\boldsymbol r\rVert\le1, ρ = 2 1 ( I + r ⋅ σ ) , ∥ r ∥ ≤ 1 , then
p ± = 1 2 ( 1 ± r ⋅ n ) . p_\pm
=
\frac12
\left(
1\pm\boldsymbol r\cdot\boldsymbol n
\right). p ± = 2 1 ( 1 ± r ⋅ n ) . The directional mean and variance are
⟨ S n ⟩ = ℏ 2 r ⋅ n , \langle S_{\boldsymbol n}\rangle
=
\frac{\hbar}{2}
\boldsymbol r\cdot\boldsymbol n, ⟨ S n ⟩ = 2 ℏ r ⋅ n , ( Δ S n ) 2 = ℏ 2 4 [ 1 − ( r ⋅ n ) 2 ] . (\Delta S_{\boldsymbol n})^2
=
\frac{\hbar^2}{4}
\left[
1-(\boldsymbol r\cdot\boldsymbol n)^2
\right]. ( Δ S n ) 2 = 4 ℏ 2 [ 1 − ( r ⋅ n ) 2 ] . For a pure state pointing along unit vector m \boldsymbol m m r = m \boldsymbol r=\boldsymbol m r = m
Pr ( + n ∣ + m ) = 1 + m ⋅ n 2 = cos 2 γ 2 , \Pr(+\boldsymbol n\mid+\boldsymbol m)
=
\frac{1+\boldsymbol m\cdot\boldsymbol n}{2}
=
\cos^2\frac{\gamma}{2}, Pr ( + n ∣ + m ) = 2 1 + m ⋅ n = cos 2 2 γ , where γ \gamma γ cos 2 γ \cos^2\gamma cos 2 γ
Let
∣ ψ ⟩ = α ∣ + z ⟩ + β ∣ − z ⟩ , ∣ α ∣ 2 + ∣ β ∣ 2 = 1. \lvert\psi\rangle
=
\alpha\lvert+z\rangle
+
\beta\lvert-z\rangle,
\qquad
\lvert\alpha\rvert^2+\lvert\beta\rvert^2=1. ∣ ψ ⟩ = α ∣ + z ⟩ + β ∣ − z ⟩ , ∣ α ∣ 2 + ∣ β ∣ 2 = 1. The pure-state Bloch vector is
r = ( 2 Re ( α ∗ β ) 2 Im ( α ∗ β ) ∣ α ∣ 2 − ∣ β ∣ 2 ) . \boldsymbol r
=
\begin{pmatrix}
2\operatorname{Re}(\alpha^*\beta)\\
2\operatorname{Im}(\alpha^*\beta)\\
\lvert\alpha\rvert^2-\lvert\beta\rvert^2
\end{pmatrix}. r = 2 Re ( α ∗ β ) 2 Im ( α ∗ β ) ∣ α ∣ 2 − ∣ β ∣ 2 . Conversely, up to global phase, a pure state with polar angles
( θ , ϕ ) (\theta,\phi) ( θ , ϕ )
∣ 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 ⟩ . The relative phase controls the transverse components. A global phase
e i χ e^{i\chi} e i χ r \boldsymbol r r Bloch Sphere
canonical page.
An active right-handed rotation of a spin-1 / 2 1/2 1/2 ϑ \vartheta ϑ n \boldsymbol n n
U n ( ϑ ) = exp ( − i ϑ ℏ n ⋅ S ) . U_{\boldsymbol n}(\vartheta)
=
\exp\!\left(
-\frac{i\vartheta}{\hbar}
\boldsymbol n\cdot\mathbf S
\right). U n ( ϑ ) = exp ( − ℏ i ϑ n ⋅ S ) . Since S = ℏ σ / 2 \mathbf S=\hbar\boldsymbol\sigma/2 S = ℏ σ /2
U n ( ϑ ) = cos ϑ 2 , I − i sin ϑ 2 n ⋅ σ . U_{\boldsymbol n}(\vartheta)
=
\cos\frac{\vartheta}{2},I
-
i\sin\frac{\vartheta}{2}
\boldsymbol n\cdot\boldsymbol\sigma. U n ( ϑ ) = cos 2 ϑ , I − i sin 2 ϑ n ⋅ σ . Special cases are
U z ( ϑ ) = ( e − i ϑ / 2 0 0 e i ϑ / 2 ) U_z(\vartheta)
=
\begin{pmatrix}
e^{-i\vartheta/2}&0\\
0&e^{i\vartheta/2}
\end{pmatrix} U z ( ϑ ) = ( e − i ϑ /2 0 0 e i ϑ /2 ) and
U x ( ϑ ) = ( cos ( ϑ / 2 ) − i sin ( ϑ / 2 ) − i sin ( ϑ / 2 ) cos ( ϑ / 2 ) ) . U_x(\vartheta)
=
\begin{pmatrix}
\cos(\vartheta/2)&-i\sin(\vartheta/2)\\
-i\sin(\vartheta/2)&\cos(\vartheta/2)
\end{pmatrix}. U x ( ϑ ) = ( cos ( ϑ /2 ) − i sin ( ϑ /2 ) − i sin ( ϑ /2 ) cos ( ϑ /2 ) ) . The spinor obeys
U n ( 2 π ) = − I , U n ( 4 π ) = I . U_{\boldsymbol n}(2\pi)=-I,
\qquad
U_{\boldsymbol n}(4\pi)=I. U n ( 2 π ) = − I , U n ( 4 π ) = I . The minus sign after 2 π 2\pi 2 π 2 π 2\pi 2 π
With the active convention above,
U n ( ϑ ) ( a ⋅ σ ) U n † ( ϑ ) = [ R n ( ϑ ) a ] ⋅ σ , U_{\boldsymbol n}(\vartheta)
(\boldsymbol a\cdot\boldsymbol\sigma)
U_{\boldsymbol n}^\dagger(\vartheta)
=
\left[
\mathcal R_{\boldsymbol n}(\vartheta)
\boldsymbol a
\right]
\cdot\boldsymbol\sigma, U n ( ϑ ) ( a ⋅ σ ) U n † ( ϑ ) = [ R n ( ϑ ) a ] ⋅ σ , where R n \mathcal R_{\boldsymbol n} R n
Every Hermitian 2 × 2 2\times2 2 × 2
H = c 0 I + c ⋅ σ , c 0 ∈ R , c ∈ R 3 . H
=
c_0I+\boldsymbol c\cdot\boldsymbol\sigma,
\qquad
c_0\in\mathbb R,
\qquad
\boldsymbol c\in\mathbb R^3. H = c 0 I + c ⋅ σ , c 0 ∈ R , c ∈ R 3 . For c ≠ 0 \boldsymbol c\ne\boldsymbol0 c = 0
c = ∥ c ∥ , c ^ = c c . c=\lVert\boldsymbol c\rVert,
\qquad
\widehat{\boldsymbol c}
=
\frac{\boldsymbol c}{c}. c = ∥ c ∥ , c = c c . The energies and projectors are
E ± = c 0 ± c , P ± = 1 2 ( I ± c ^ ⋅ σ ) . E_\pm
=
c_0\pm c,
\qquad
P_\pm
=
\frac12
\left(
I\pm
\widehat{\boldsymbol c}\cdot\boldsymbol\sigma
\right). E ± = c 0 ± c , P ± = 2 1 ( I ± c ⋅ σ ) . The exact propagator is
U ( t ) = e − i H t / ℏ = e − i c 0 t / ℏ [ cos ( c t ℏ ) I − i sin ( c t ℏ ) c ^ ⋅ σ ] . \begin{aligned}
U(t)
&=
e^{-iHt/\hbar}
\\
&=
e^{-ic_0t/\hbar}
\left[
\cos\!\left(\frac{ct}{\hbar}\right)I
-
i\sin\!\left(\frac{ct}{\hbar}\right)
\widehat{\boldsymbol c}\cdot\boldsymbol\sigma
\right].
\end{aligned} U ( t ) = e − i H t /ℏ = e − i c 0 t /ℏ [ cos ( ℏ c t ) I − i sin ( ℏ c t ) c ⋅ σ ] . The scalar c 0 I c_0I c 0 I c \boldsymbol c c 2 c 2c 2 c
For a physical magnetic moment μ = γ S \boldsymbol\mu=\gamma\mathbf S μ = γ S
H = − μ ⋅ B = − ℏ γ 2 B ⋅ σ . H
=
-\boldsymbol\mu\cdot\mathbf B
=
-\frac{\hbar\gamma}{2}
\mathbf B\cdot\boldsymbol\sigma. H = − μ ⋅ B = − 2 ℏ γ B ⋅ σ . The sign of the gyromagnetic ratio γ \gamma γ
The dimensionless matrices
σ + = σ x + i σ y 2 = ( 0 1 0 0 ) , σ − = σ x − i σ y 2 = ( 0 0 1 0 ) \sigma_+
=
\frac{\sigma_x+i\sigma_y}{2}
=
\begin{pmatrix}0&1\\0&0\end{pmatrix},
\qquad
\sigma_-
=
\frac{\sigma_x-i\sigma_y}{2}
=
\begin{pmatrix}0&0\\1&0\end{pmatrix} σ + = 2 σ x + i σ y = ( 0 0 1 0 ) , σ − = 2 σ x − i σ y = ( 0 1 0 0 ) are related to physical spin ladders by
S + = ℏ σ + , S − = ℏ σ − . S_+=\hbar\sigma_+,
\qquad
S_-=\hbar\sigma_-. S + = ℏ σ + , S − = ℏ σ − . The general spin-j j j Ladder-Operator Action .
Write the ordered basis before writing a matrix.
Decide whether the symbols are dimensionless Pauli matrices or physical
spin operators carrying ℏ \hbar ℏ
Convert a measurement axis to n ⋅ σ \boldsymbol n\cdot\boldsymbol\sigma n ⋅ σ
Convert a state to its Bloch vector when several directional measurements
are needed.
For rotations or exponentials, use
( n ⋅ σ ) 2 = I (\boldsymbol n\cdot\boldsymbol\sigma)^2=I ( n ⋅ σ ) 2 = I
For a generic Hermitian matrix, separate the trace term c 0 I c_0I c 0 I
Check Hermiticity, normalization, and limiting cases such as
U ( 0 ) = I U(0)=I U ( 0 ) = I
Forgetting the factor ℏ / 2 \hbar/2 ℏ/2 σ i \sigma_i σ i S i S_i S i
Reversing the sign of the upper-right entry of σ y \sigma_y σ y
Copying matrices from a source with the opposite basis order.
Treating a generic two-level pseudospin as literal spatial angular momentum.
Dropping relative phase when constructing the Bloch vector.
Using cos 2 γ \cos^2\gamma cos 2 γ cos 2 ( γ / 2 ) \cos^2(\gamma/2) cos 2 ( γ /2 )
Confusing active U O U † UOU^\dagger U O U † U † O U U^\dagger OU U † O U
Treating the 2 π 2\pi 2 π
Using σ ± \sigma_\pm σ ± S ± S_\pm S ±
Assuming the basis displayed here is automatically the energy basis.
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 , World
Scientific, 1998.
M. A. Nielsen and I. L. Chuang, Quantum Computation and Quantum
Information , 10th anniversary ed., Cambridge University Press, 2010.
Construct the spin-up eigenstate along
n = ( sin θ , 0 , cos θ ) \boldsymbol n=(\sin\theta,0,\cos\theta) n = ( sin θ , 0 , cos θ )
Solution Here ϕ = 0 \phi=0 ϕ = 0
∣ n , + ⟩ = ( cos ( θ / 2 ) sin ( θ / 2 ) ) . \lvert\boldsymbol n,+\rangle
=
\begin{pmatrix}
\cos(\theta/2)\\
\sin(\theta/2)
\end{pmatrix}. ∣ n , + ⟩ = ( cos ( θ /2 ) sin ( θ /2 ) ) . The directional matrix is
σ n = ( cos θ sin θ sin θ − cos θ ) . \sigma_{\boldsymbol n}
=
\begin{pmatrix}
\cos\theta&\sin\theta\\
\sin\theta&-\cos\theta
\end{pmatrix}. σ n = ( cos θ sin θ sin θ − cos θ ) . Using angle-addition identities in the matrix product gives
σ n ∣ n , + ⟩ = ∣ n , + ⟩ . \sigma_{\boldsymbol n}
\lvert\boldsymbol n,+\rangle
=
\lvert\boldsymbol n,+\rangle. σ n ∣ n , + ⟩ = ∣ n , + ⟩ . Therefore the physical spin eigenvalue is + ℏ / 2 +\hbar/2 + ℏ/2
A beam prepared in ∣ + z ⟩ \lvert+z\rangle ∣ + z ⟩ γ \gamma γ z z z
Solution The prepared Bloch vector is r = z ^ \boldsymbol r=\hat{\mathbf z} r = z ^
p + = 1 + cos γ 2 = cos 2 γ 2 , p_+
=
\frac{1+\cos\gamma}{2}
=
\cos^2\frac{\gamma}{2}, p + = 2 1 + cos γ = cos 2 2 γ , and
p − = 1 − cos γ 2 = sin 2 γ 2 . p_-
=
\frac{1-\cos\gamma}{2}
=
\sin^2\frac{\gamma}{2}. p − = 2 1 − cos γ = sin 2 2 γ . The variance is
( Δ S n ) 2 = ℏ 2 4 ( 1 − cos 2 γ ) = ℏ 2 4 sin 2 γ . (\Delta S_{\boldsymbol n})^2
=
\frac{\hbar^2}{4}
\left(1-\cos^2\gamma\right)
=
\frac{\hbar^2}{4}\sin^2\gamma. ( Δ S n ) 2 = 4 ℏ 2 ( 1 − cos 2 γ ) = 4 ℏ 2 sin 2 γ .
Apply an active rotation by angle π \pi π x x x ∣ + z ⟩ \lvert+z\rangle ∣ + z ⟩
Solution The rotation operator is
U x ( π ) = cos π 2 I − i sin π 2 σ x = − i σ x . U_x(\pi)
=
\cos\frac{\pi}{2}I
-
i\sin\frac{\pi}{2}\sigma_x
=
-i\sigma_x. U x ( π ) = cos 2 π I − i sin 2 π σ x = − i σ x . Since σ x ∣ + z ⟩ = ∣ − z ⟩ \sigma_x\lvert+z\rangle=\lvert-z\rangle σ x ∣ + z ⟩ = ∣ − z ⟩
U x ( π ) ∣ + z ⟩ = − i ∣ − z ⟩ . U_x(\pi)\lvert+z\rangle
=
-i\lvert-z\rangle. U x ( π ) ∣ + z ⟩ = − i ∣ − z ⟩ . The phase − i -i − i − z -z − z
For
H = ( Δ g g − Δ ) , Δ , g ∈ R , H
=
\begin{pmatrix}
\Delta&g\\
g&-\Delta
\end{pmatrix},
\qquad
\Delta,g\in\mathbb R, H = ( Δ g g − Δ ) , Δ , g ∈ R , find the energies and the exact propagator.
Solution Write
H = g σ x + Δ σ z . H
=
g\sigma_x+\Delta\sigma_z. H = g σ x + Δ σ z . Thus c 0 = 0 c_0=0 c 0 = 0 c = ( g , 0 , Δ ) \boldsymbol c=(g,0,\Delta) c = ( g , 0 , Δ )
Ω = g 2 + Δ 2 . \Omega
=
\sqrt{g^2+\Delta^2}. Ω = g 2 + Δ 2 . The energies are
E ± = ± Ω . E_\pm=\pm\Omega. E ± = ± Ω. The propagator is
U ( t ) = cos ( Ω t ℏ ) I − i sin ( Ω t ℏ ) g σ x + Δ σ z Ω . U(t)
=
\cos\!\left(\frac{\Omega t}{\hbar}\right)I
-
i\sin\!\left(\frac{\Omega t}{\hbar}\right)
\frac{g\sigma_x+\Delta\sigma_z}{\Omega}. U ( t ) = cos ( ℏ Ω t ) I − i sin ( ℏ Ω t ) Ω g σ x + Δ σ z . At g = 0 g=0 g = 0 Δ = 0 \Delta=0 Δ = 0 x x x