Example of Quantum Dynamics
An electron is in an magnetic field in the z + z+ z + state
∣ ψ ( 0 ) ⟩ = ( cos θ 0 / 2 e i ϕ 0 sin θ 0 / 2 ) \left\lvert \psi(0) \right\rangle=\begin{pmatrix}\cos\theta_0/2 \\ e^{i\phi_0}\sin\theta_0/2\end{pmatrix} ∣ ψ ( 0 ) ⟩ = ( cos θ 0 /2 e i ϕ 0 sin θ 0 /2 ) Find out the expected value of S z , S x , S y S_z, S_x, S_y S z , S x , S y t t t
Classical charged particle has
μ ⃗ = q 2 m s ⃗ \vec{\mu}=\frac{q}{2m}\vec{s} μ = 2 m q s A quantum one also has a g g g
μ ⃗ = g q 2 m s ⃗ \vec{\mu}=g\frac{q}{2m}\vec{s} μ = g 2 m q s In the case of an electron, q = e , m = m e , g ≈ 2 q=e, m=m_e, g\approx 2 q = e , m = m e , g ≈ 2
μ ⃗ = g e ℏ 2 m e ⏟ μ B 1 ℏ s ⃗ \vec{\mu}=g\underbrace{\frac{e\hbar}{2m_e}}_{\mu_B}\frac{1}{\hbar}\vec{s} μ = g μ B 2 m e e ℏ ℏ 1 s Where μ B \mu_B μ B
The spin in magnetic field B ⃗ \vec{B} B
Clasically, the energy of a magnetic dipole in a B field is
U = − μ ⃗ ⋅ B ⃗ U=-\vec{\mu}\cdot\vec{B} U = − μ ⋅ B In quantum mechanics, the energy is the Hamiltonian operator H ^ \hat{H} H ^
γ = g μ B ℏ \gamma = \frac{g\mu_B}{\hbar} γ = ℏ g μ B So
H ^ = − μ ⃗ ⋅ B ⃗ = − γ s ⃗ ⋅ B ⃗ \hat{H}=-\vec{\mu}\cdot\vec{B}=-\gamma\vec{s}\cdot\vec{B} H ^ = − μ ⋅ B = − γ s ⋅ B Since B field is in the z + z+ z + B ⃗ = B S z ^ \vec{B}=B\hat{S_z} B = B S z ^
H ^ = − γ B ℏ 2 ( 1 0 0 − 1 ) = − γ B S ^ z \hat{H}=-\gamma B \frac{\hbar}{2}\begin{pmatrix}1 & 0 \\ 0 & -1\end{pmatrix}=-\gamma B \hat{S}_z H ^ = − γ B 2 ℏ ( 1 0 0 − 1 ) = − γ B S ^ z To do it the Schrödinger equation way, take the eigenvalues and Eigenstate of H H H
H ∣ z ± ⟩ = ∓ γ B ℏ 2 ∣ z ± ⟩ H\left\lvert z\pm \right\rangle=\mp\gamma B \frac{\hbar}{2}\left\lvert z\pm \right\rangle H ∣ z ± ⟩ = ∓ γ B 2 ℏ ∣ z ± ⟩ Because Hamiltonian is time-independent, we can use the Uniform dynamics solution to find U ( t , t 0 ) U(t,t_0) U ( t , t 0 )
U ( t , 0 ) ∣ z ± ⟩ = e − i H t / ℏ ∣ z ± ⟩ = e ± i γ B t / 2 ∣ z ± ⟩ U(t,0)\left\lvert z\pm \right\rangle=e^{-iHt/\hbar}\left\lvert z\pm \right\rangle=e^{\pm i\gamma B t/2}\left\lvert z\pm \right\rangle U ( t , 0 ) ∣ z ± ⟩ = e − i H t /ℏ ∣ z ± ⟩ = e ± iγ B t /2 ∣ z ± ⟩ Looking at the state ∣ ψ ( 0 ) ⟩ \left\lvert \psi(0) \right\rangle ∣ ψ ( 0 ) ⟩ H H H
∣ ψ ( 0 ) ⟩ = cos θ 0 / 2 ∣ z + ⟩ + e i ϕ 0 sin θ 0 / 2 ∣ z − ⟩ \left\lvert \psi(0) \right\rangle=\cos\theta_0/2\left\lvert z+ \right\rangle+e^{i\phi_0}\sin\theta_0/2\left\lvert z- \right\rangle ∣ ψ ( 0 ) ⟩ = cos θ 0 /2 ∣ z + ⟩ + e i ϕ 0 sin θ 0 /2 ∣ z − ⟩ ∣ ψ ( t ) ⟩ = U ( t ) ∣ ψ ( 0 ) ⟩ \left\lvert \psi(t) \right\rangle=U(t)\left\lvert \psi(0) \right\rangle ∣ ψ ( t ) ⟩ = U ( t ) ∣ ψ ( 0 ) ⟩ ∣ ψ ( t ) ⟩ = ( e i γ B t / 2 ) cos θ 0 / 2 ∣ z + ⟩ + ( e − i γ B t / 2 ) e i ϕ 0 sin θ 0 / 2 ∣ z − ⟩ \left\lvert \psi(t) \right\rangle=(e^{i\gamma B t/2})\cos\theta_0/2\left\lvert z+ \right\rangle+(e^{-i\gamma B t/2})e^{i\phi_0}\sin\theta_0/2\left\lvert z- \right\rangle ∣ ψ ( t ) ⟩ = ( e iγ B t /2 ) cos θ 0 /2 ∣ z + ⟩ + ( e − iγ B t /2 ) e i ϕ 0 sin θ 0 /2 ∣ z − ⟩ Global phase should be separated out
∣ ψ ( t ) ⟩ = e i γ B t / 2 [ cos θ 0 / 2 ∣ z + ⟩ + ( e i ( − γ B t + ϕ 0 ) ) sin θ 0 / 2 ∣ z − ⟩ ] \left\lvert \psi(t) \right\rangle=e^{i\gamma B t/2}\left[\cos\theta_0/2\left\lvert z+ \right\rangle+(e^{i(-\gamma B t+\phi_0)})\sin\theta_0/2\left\lvert z- \right\rangle\right] ∣ ψ ( t ) ⟩ = e iγ B t /2 [ cos θ 0 /2 ∣ z + ⟩ + ( e i ( − γ B t + ϕ 0 ) ) sin θ 0 /2 ∣ z − ⟩ ] Hence
⟨ ψ ( t ) ∣ S ^ z ∣ ψ ( t ) ⟩ = ℏ 2 ( ψ + ∗ ψ − ∗ ) ( 1 0 0 − 1 ) ( ψ + ψ − ) \langle\psi(t)|\hat{S}_z|\psi(t)\rangle = \frac{\hbar}{2}\begin{pmatrix}\psi_+^* & \psi_-^*\end{pmatrix}\begin{pmatrix}1 & 0 \\ 0 & -1\end{pmatrix}\begin{pmatrix}\psi_+ \\ \psi_-\end{pmatrix} ⟨ ψ ( t ) ∣ S ^ z ∣ ψ ( t )⟩ = 2 ℏ ( ψ + ∗ ψ − ∗ ) ( 1 0 0 − 1 ) ( ψ + ψ − ) = ℏ 2 ( ψ + ∗ ψ − ∗ ) ( ψ + − ψ − ) = \frac{\hbar}{2}\begin{pmatrix}\psi_+^* & \psi_-^*\end{pmatrix}\begin{pmatrix}\psi_+ \\ -\psi_-\end{pmatrix} = 2 ℏ ( ψ + ∗ ψ − ∗ ) ( ψ + − ψ − ) = ℏ 2 ( ψ + ∗ ψ + − ψ − ∗ ψ − ) = ℏ 2 ( ∣ ψ + ∣ 2 − ∣ ψ − ∣ 2 ) = \frac{\hbar}{2}\left(\psi_+^*\psi_+ - \psi_-^*\psi_-\right) = \frac{\hbar}{2}\left(|\psi_+|^2 - |\psi_-|^2\right) = 2 ℏ ( ψ + ∗ ψ + − ψ − ∗ ψ − ) = 2 ℏ ( ∣ ψ + ∣ 2 − ∣ ψ − ∣ 2 ) = ℏ 2 ( cos 2 θ 0 2 − sin 2 θ 0 2 ) = \frac{\hbar}{2}\left(\cos^2\frac{\theta_0}{2} - \sin^2\frac{\theta_0}{2}\right) = 2 ℏ ( cos 2 2 θ 0 − sin 2 2 θ 0 ) = ℏ 2 cos θ 0 (Note: unchanging in time) = \frac{\hbar}{2}\cos\theta_0 \quad \text{(Note: unchanging in time)} = 2 ℏ cos θ 0 (Note: unchanging in time) ⟨ ψ ( t ) ∣ S ^ x ∣ ψ ( t ) ⟩ = ℏ 2 ( ψ + ∗ ψ − ∗ ) ( 0 1 1 0 ) ( ψ + ψ − ) \langle\psi(t)|\hat{S}_x|\psi(t)\rangle = \frac{\hbar}{2}\begin{pmatrix}\psi_+^* & \psi_-^*\end{pmatrix}\begin{pmatrix}0 & 1 \\ 1 & 0\end{pmatrix}\begin{pmatrix}\psi_+ \\ \psi_-\end{pmatrix} ⟨ ψ ( t ) ∣ S ^ x ∣ ψ ( t )⟩ = 2 ℏ ( ψ + ∗ ψ − ∗ ) ( 0 1 1 0 ) ( ψ + ψ − ) = ℏ 2 ( ψ + ∗ ψ − ∗ ) ( ψ − ψ + ) = \frac{\hbar}{2}\begin{pmatrix}\psi_+^* & \psi_-^*\end{pmatrix}\begin{pmatrix}\psi_- \\ \psi_+\end{pmatrix} = 2 ℏ ( ψ + ∗ ψ − ∗ ) ( ψ − ψ + ) = ℏ 2 ( ψ + ∗ ψ − + ψ − ∗ ψ + ) = \frac{\hbar}{2}\left(\psi_+^*\psi_- + \psi_-^*\psi_+\right) = 2 ℏ ( ψ + ∗ ψ − + ψ − ∗ ψ + ) Plug in ψ + = e + i γ B t / 2 cos ( θ 0 / 2 ) \psi_+ = e^{+i\gamma B t/2}\cos(\theta_0/2) ψ + = e + iγ B t /2 cos ( θ 0 /2 ) ψ − = e − i γ B t / 2 e i ϕ 0 sin ( θ 0 / 2 ) \psi_- = e^{-i\gamma B t/2}e^{i\phi_0}\sin(\theta_0/2) ψ − = e − iγ B t /2 e i ϕ 0 sin ( θ 0 /2 )
ψ + ∗ ψ − = e − i γ B t / 2 cos θ 0 2 ⋅ e − i γ B t / 2 e i ϕ 0 sin θ 0 2 = e i ( ϕ 0 − γ B t ) cos θ 0 2 sin θ 0 2 \psi_+^*\psi_- = e^{-i\gamma B t/2}\cos\frac{\theta_0}{2} \cdot e^{-i\gamma B t/2}e^{i\phi_0}\sin\frac{\theta_0}{2} = e^{i(\phi_0 - \gamma B t)}\cos\frac{\theta_0}{2}\sin\frac{\theta_0}{2} ψ + ∗ ψ − = e − iγ B t /2 cos 2 θ 0 ⋅ e − iγ B t /2 e i ϕ 0 sin 2 θ 0 = e i ( ϕ 0 − γ B t ) cos 2 θ 0 sin 2 θ 0 ψ − ∗ ψ + = e − i ( ϕ 0 − γ B t ) cos θ 0 2 sin θ 0 2 \psi_-^*\psi_+ = e^{-i(\phi_0 - \gamma B t)}\cos\frac{\theta_0}{2}\sin\frac{\theta_0}{2} ψ − ∗ ψ + = e − i ( ϕ 0 − γ B t ) cos 2 θ 0 sin 2 θ 0 ψ + ∗ ψ − + ψ − ∗ ψ + = 2 cos θ 0 2 sin θ 0 2 cos ( ϕ 0 − γ B t ) = sin θ 0 cos ( ϕ 0 − γ B t ) \psi_+^*\psi_- + \psi_-^*\psi_+ = 2\cos\frac{\theta_0}{2}\sin\frac{\theta_0}{2}\cos(\phi_0 - \gamma B t) = \sin\theta_0 \cos(\phi_0 - \gamma B t) ψ + ∗ ψ − + ψ − ∗ ψ + = 2 cos 2 θ 0 sin 2 θ 0 cos ( ϕ 0 − γ B t ) = sin θ 0 cos ( ϕ 0 − γ B t ) ⇒ ⟨ ψ ( t ) ∣ S ^ x ∣ ψ ( t ) ⟩ = ℏ 2 sin θ 0 cos ( ϕ 0 − γ B t ) \Rightarrow\langle\psi(t)|\hat{S}_x|\psi(t)\rangle = \frac{\hbar}{2}\sin\theta_0 \cos(\phi_0 - \gamma B t) ⇒ ⟨ ψ ( t ) ∣ S ^ x ∣ ψ ( t )⟩ = 2 ℏ sin θ 0 cos ( ϕ 0 − γ B t ) ⟨ ψ ( t ) ∣ S ^ y ∣ ψ ( t ) ⟩ = ℏ 2 ( ψ + ∗ ψ − ∗ ) ( 0 − i i 0 ) ( ψ + ψ − ) \langle\psi(t)|\hat{S}_y|\psi(t)\rangle = \frac{\hbar}{2}\begin{pmatrix}\psi_+^* & \psi_-^*\end{pmatrix}\begin{pmatrix}0 & -i \\ i & 0\end{pmatrix}\begin{pmatrix}\psi_+ \\ \psi_-\end{pmatrix} ⟨ ψ ( t ) ∣ S ^ y ∣ ψ ( t )⟩ = 2 ℏ ( ψ + ∗ ψ − ∗ ) ( 0 i − i 0 ) ( ψ + ψ − ) = ℏ 2 ( ψ + ∗ ψ − ∗ ) ( − i ψ − i ψ + ) = \frac{\hbar}{2}\begin{pmatrix}\psi_+^* & \psi_-^*\end{pmatrix}\begin{pmatrix}-i\psi_- \\ i\psi_+\end{pmatrix} = 2 ℏ ( ψ + ∗ ψ − ∗ ) ( − i ψ − i ψ + ) = ℏ 2 ( − i ψ + ∗ ψ − + i ψ − ∗ ψ + ) = i ℏ 2 ( ψ − ∗ ψ + − ψ + ∗ ψ − ) = \frac{\hbar}{2}\left(-i\psi_+^*\psi_- + i\psi_-^*\psi_+\right) = \frac{i\hbar}{2}\left(\psi_-^*\psi_+ - \psi_+^*\psi_-\right) = 2 ℏ ( − i ψ + ∗ ψ − + i ψ − ∗ ψ + ) = 2 i ℏ ( ψ − ∗ ψ + − ψ + ∗ ψ − ) Using the same products as above, with e i x − e − i x = 2 i sin x e^{ix} - e^{-ix} = 2i\sin x e i x − e − i x = 2 i sin x
ψ − ∗ ψ + − ψ + ∗ ψ − = − 2 i cos θ 0 2 sin θ 0 2 sin ( ϕ 0 − γ B t ) = − i sin θ 0 sin ( ϕ 0 − γ B t ) \psi_-^*\psi_+ - \psi_+^*\psi_- = -2i\cos\frac{\theta_0}{2}\sin\frac{\theta_0}{2}\sin(\phi_0 - \gamma B t) = -i\sin\theta_0\sin(\phi_0 - \gamma B t) ψ − ∗ ψ + − ψ + ∗ ψ − = − 2 i cos 2 θ 0 sin 2 θ 0 sin ( ϕ 0 − γ B t ) = − i sin θ 0 sin ( ϕ 0 − γ B t ) ⇒ ⟨ ψ ( t ) ∣ S ^ y ∣ ψ ( t ) ⟩ = i ℏ 2 ⋅ ( − i ) sin θ 0 sin ( ϕ 0 − γ B t ) = ℏ 2 sin θ 0 sin ( ϕ 0 − γ B t ) \Rightarrow\langle\psi(t)|\hat{S}_y|\psi(t)\rangle = \frac{i\hbar}{2}\cdot(-i)\sin\theta_0\sin(\phi_0 - \gamma B t) = \frac{\hbar}{2}\sin\theta_0\sin(\phi_0 - \gamma B t) ⇒ ⟨ ψ ( t ) ∣ S ^ y ∣ ψ ( t )⟩ = 2 i ℏ ⋅ ( − i ) sin θ 0 sin ( ϕ 0 − γ B t ) = 2 ℏ sin θ 0 sin ( ϕ 0 − γ B t ) Another way to do it is to use the Heisenberg Picture (Heisenberg, 1925)
d d t A ^ ( t ) = i ℏ [ H ^ , A ^ ] \frac{d}{dt}\hat{A}(t) = \frac{i}{\hbar}[\hat{H}, \hat{A}] d t d A ^ ( t ) = ℏ i [ H ^ , A ^ ] Recalling the Pauli Matrices and their commutators, we can derive it for spin operators S z , S x , S y S_z,S_x,S_y S z , S x , S y
[ S ^ x , S ^ y ] = i ℏ S ^ z , [ S ^ y , S ^ z ] = i ℏ S ^ x , [ S ^ z , S ^ x ] = i ℏ S ^ y [\hat{S}_x, \hat{S}_y] = i\hbar\hat{S}_z, \quad [\hat{S}_y, \hat{S}_z] = i\hbar\hat{S}_x, \quad [\hat{S}_z, \hat{S}_x] = i\hbar\hat{S}_y [ S ^ x , S ^ y ] = i ℏ S ^ z , [ S ^ y , S ^ z ] = i ℏ S ^ x , [ S ^ z , S ^ x ] = i ℏ S ^ y For our Hamiltonian H ^ = − γ B S ^ z \hat{H} = -\gamma B \hat{S}_z H ^ = − γ B S ^ z
d d t S ^ z ( t ) = i ℏ [ − γ B S ^ z ( t ) , S ^ z ( t ) ] = − i γ B ℏ [ S ^ z ( t ) , S ^ z ( t ) ] = 0 \frac{d}{dt}\hat{S}_z(t) = \frac{i}{\hbar}[-\gamma B \hat{S}_z(t), \hat{S}_z(t)] = -\frac{i\gamma B}{\hbar}[\hat{S}_z(t), \hat{S}_z(t)] = 0 d t d S ^ z ( t ) = ℏ i [ − γ B S ^ z ( t ) , S ^ z ( t )] = − ℏ iγ B [ S ^ z ( t ) , S ^ z ( t )] = 0 d d t S ^ x ( t ) = i ℏ [ − γ B S ^ z ( t ) , S ^ x ( t ) ] \frac{d}{dt}\hat{S}_x(t) = \frac{i}{\hbar}[-\gamma B \hat{S}_z(t), \hat{S}_x(t)] d t d S ^ x ( t ) = ℏ i [ − γ B S ^ z ( t ) , S ^ x ( t )] = − i γ B ℏ [ S ^ z ( t ) , S ^ x ( t ) ] = -\frac{i\gamma B}{\hbar}[\hat{S}_z(t), \hat{S}_x(t)] = − ℏ iγ B [ S ^ z ( t ) , S ^ x ( t )] = − i γ B ℏ ⋅ i ℏ S ^ y ( t ) = γ B S ^ y ( t ) = -\frac{i\gamma B}{\hbar}\cdot i\hbar \hat{S}_y(t) = \gamma B \hat{S}_y(t) = − ℏ iγ B ⋅ i ℏ S ^ y ( t ) = γ B S ^ y ( t ) d d t S ^ y ( t ) = i ℏ [ − γ B S ^ z ( t ) , S ^ y ( t ) ] \frac{d}{dt}\hat{S}_y(t) = \frac{i}{\hbar}[-\gamma B \hat{S}_z(t), \hat{S}_y(t)] d t d S ^ y ( t ) = ℏ i [ − γ B S ^ z ( t ) , S ^ y ( t )] = − i γ B ℏ [ S ^ z ( t ) , S ^ y ( t ) ] = -\frac{i\gamma B}{\hbar}[\hat{S}_z(t), \hat{S}_y(t)] = − ℏ iγ B [ S ^ z ( t ) , S ^ y ( t )] = − i γ B ℏ ⋅ ( − i ℏ S ^ x ( t ) ) = − γ B S ^ x ( t ) = -\frac{i\gamma B}{\hbar}\cdot(-i\hbar \hat{S}_x(t)) = -\gamma B \hat{S}_x(t) = − ℏ iγ B ⋅ ( − i ℏ S ^ x ( t )) = − γ B S ^ x ( t ) Collecting all three:
d d t ( S ^ x , S ^ y , S ^ z ) = γ B ( S ^ y , − S ^ x , 0 ) \frac{d}{dt}(\hat{S}_x, \hat{S}_y, \hat{S}_z) = \gamma B(\hat{S}_y, -\hat{S}_x, 0) d t d ( S ^ x , S ^ y , S ^ z ) = γ B ( S ^ y , − S ^ x , 0 ) Using Cross Product formula:
= − γ ( 0 , 0 , B ) × ( S ^ x , S ^ y , S ^ z ) = -\gamma(0, 0, B)\times(\hat{S}_x, \hat{S}_y, \hat{S}_z) = − γ ( 0 , 0 , B ) × ( S ^ x , S ^ y , S ^ z ) d S ⃗ d t = − γ B ⃗ × S ⃗ (Larmor precession) \boxed{\frac{d\vec{S}}{dt} = -\gamma \vec{B}\times\vec{S}} \quad \text{(Larmor precession)} d t d S = − γ B × S (Larmor precession) Hamiltonian Rotation Quirk