Hamiltonian on a Composite System

Non-interacting Hamiltonian

H(total)=H(A)+H(B)H^{(total)}=H^{(A)}+H^{(B)}

Where H(A),H(B)H^{(A)}, H^{(B)} are non interacting.

Two spin-1/2 particles exist where

H(total)=S^x(total)+S^x(A)+S^x(B)H^{(total)}=\hat{S}^{(total)}_x+ \hat{S}^{(A)}_x + \hat{S}^{(B)}_x =S^x⊗I+I⊗S^x=\hat{S}_x\otimes I + I \otimes \hat{S}_x

Let there be a state where

∣ψ⟩=a11∣+,+⟩+a12∣+,−⟩+a21∣−,+⟩+a22∣−,−⟩\left\lvert \psi \right\rangle=a_{11}\left\lvert +,+ \right\rangle+a_{12}\left\lvert +,- \right\rangle+a_{21}\left\lvert -,+ \right\rangle+a_{22}\left\lvert -,- \right\rangle

Recall what a S^x\hat{S}_x operator does here.

Sx≜ℏ2(0110)S_x\triangleq\frac{\hbar}{2}\begin{pmatrix}0 & 1\\ 1 & 0\end{pmatrix}\quad

basically,

Sx^(A)=ℏ2(0110)⊗(1001)\hat{S_x}^{(A)}=\frac{\hbar}{2}\begin{pmatrix}0 & 1\\ 1 & 0\end{pmatrix}\otimes \begin{pmatrix}1 & 0\\ 0 & 1\end{pmatrix} =ℏ2(0⋅I1⋅I1⋅I0⋅I)= \frac{\hbar}{2}\begin{pmatrix} 0 \cdot I & 1 \cdot I \\ 1 \cdot I & 0 \cdot I \end{pmatrix} =ℏ2(0010000110000100)= \frac{\hbar}{2}\begin{pmatrix} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \end{pmatrix}

So

S^x(A)∣ψ⟩=ℏ2(a11∣+,+⟩+a12∣+,−⟩−a21∣−,+⟩−a22∣−,−⟩)\hat{S}_x^{(A)}\left\lvert \psi \right\rangle=\frac{\hbar}{2}(a_{11}\left\lvert +,+ \right\rangle+a_{12}\left\lvert +,- \right\rangle-a_{21}\left\lvert -,+ \right\rangle-a_{22}\left\lvert -,- \right\rangle) S^x(B)∣ψ⟩=ℏ2(a11∣+,+⟩−a12∣+,−⟩+a21∣−,+⟩−a22∣−,−⟩)\hat{S}_x^{(B)}\left\lvert \psi \right\rangle=\frac{\hbar}{2}(a_{11}\left\lvert +,+ \right\rangle-a_{12}\left\lvert +,- \right\rangle+a_{21}\left\lvert -,+ \right\rangle-a_{22}\left\lvert -,- \right\rangle) S^x(total)∣ψ⟩=ℏ(a11∣+,+⟩−a12∣−,−⟩)\hat{S}_x^{(total)}\left\lvert \psi \right\rangle=\hbar(a_{11}\left\lvert +,+ \right\rangle-a_{12}\left\lvert -,- \right\rangle)
Eigenstate of S^x(tot)\hat{S}_x^{(\text{tot})}Eigenvalue
$+,+\rangle$
$+,-\rangle, \
$-,-\rangle$

Looks like a spin-1

Time Evolution of Non-interacting Composite Systems

If H=S^x(total)H=\hat{S}_x^{(total)} then

e−itS^x(total)=e−it(S^x(A)+S^x(B))e^{-it\hat{S}_x^{(total)}}=e^{-it\left(\hat{S}_x^{(A)}+\hat{S}_x^{(B)}\right)} =e−itS^x(A)e−itS^x(B)=e^{-it\hat{S}_x^{(A)}}e^{-it\hat{S}_x^{(B)}} =(e−itS^⊗I)(I⊗e−itS^)=(e^{-it\hat{S}}\otimes I)(I \otimes e^{-it\hat{S}}) =e−itS^x⊗e−itS^x=e^{-it\hat{S}_x}\otimes e^{-it\hat{S}_x}

Note that

H(tot)=H1(A)+H2(B)  ⟹  U(tot)=U1(A)U2(B)=U1⊗U2H^{(\text{tot})} = H_1^{(A)} + H_2^{(B)} \implies U^{(\text{tot})} = U_1^{(A)} U_2^{(B)} = U_1 \otimes U_2 U(tot)(∣v⟩⊗∣w⟩)=(U1∣v⟩)⊗(U2∣w⟩)U^{(\text{tot})}(|v\rangle \otimes |w\rangle) = (U_1 |v\rangle) \otimes (U_2 |w\rangle) product state⟶product state\text{product state} \longrightarrow \text{product state}

Unitary evolution under non-interacting Hamiltonian cannot generate an entangled state.

Interacting Hamiltonian

H=H(A)⊗I+I⊗H(B)⏟local part+Hint⏟interaction termH = \underbrace{H^{(A)} \otimes I + I \otimes H^{(B)}}_{\text{local part}} + \underbrace{H_{\text{int}}}_{\text{interaction term}}

Where HintH_{int} is any operator on HA⊗HBH_A\otimes H_B.

Example

Note σx\sigma_x is defined in Pauli matrix.

H(total)=ℏωσx(A)σx(B)=ℏω(σx⊗σx)H^{(total)}=\hbar \omega \sigma_x^{(A)}\sigma_x^{(B)}=\hbar\omega (\sigma_x\otimes \sigma_x)

Note that σx⊗σx\sigma_x\otimes \sigma_x cannot be written as σx⊗I+I⊗σx\sigma_x\otimes I + I \otimes \sigma_x hence this belongs in the HintH_{int} term. We observe that the unitary evolution is

U(total)=e−itH/ℏ=e−iωt(σx⊗σx)U^{(total)}=e^{-itH/\hbar}=e^{-i\omega t(\sigma_x\otimes \sigma_x)}

Suppose ∣ψ(0)⟩=∣0⟩⊗∣0⟩\left\lvert \psi(0) \right\rangle = \left\lvert 0 \right\rangle \otimes \left\lvert 0 \right\rangle

∣ψ(0)⟩=∣+⟩+∣−⟩2⊗∣+⟩+∣−⟩2\left\lvert \psi(0) \right\rangle = \frac{\left\lvert + \right\rangle + \left\lvert - \right\rangle}{\sqrt{2}} \otimes \frac{\left\lvert + \right\rangle + \left\lvert - \right\rangle}{\sqrt{2}} \\ =12(∣+,+⟩+∣+,−⟩+∣−,+⟩+∣−,−⟩)= \frac{1}{2}\left(\left\lvert +,+ \right\rangle + \left\lvert +,- \right\rangle + \left\lvert -,+ \right\rangle + \left\lvert -,- \right\rangle\right)

Then via the unitary time evolution we get

∣ψ(t)⟩=e−iωtσx⊗σx∣ψ(0)⟩=12(e−iωtσx⊗σx∣+,+⟩+e−iωtσx⊗σx∣+,−⟩+e−iωtσx⊗σx∣−,+⟩+e−iωtσx⊗σx∣−,−⟩)=12(e−iωt∣+,+⟩+eiωt∣+,−⟩+eiωt∣−,+⟩+e−iωt∣−,−⟩)\begin{align*} \left\lvert \psi(t) \right\rangle & = e^{-i\omega t \sigma_x \otimes \sigma_x}\left\lvert \psi(0) \right\rangle \\ & = \frac{1}{2}\left(e^{-i\omega t \sigma_x \otimes \sigma_x}\left\lvert +,+ \right\rangle + e^{-i\omega t \sigma_x \otimes \sigma_x}\left\lvert +,- \right\rangle + e^{-i\omega t \sigma_x \otimes \sigma_x}\left\lvert -,+ \right\rangle + e^{-i\omega t \sigma_x \otimes \sigma_x}\left\lvert -,- \right\rangle\right) \\ & = \frac{1}{2}\left(e^{-i\omega t}\left\lvert +,+ \right\rangle + e^{i\omega t}\left\lvert +,- \right\rangle + e^{i\omega t}\left\lvert -,+ \right\rangle + e^{-i\omega t}\left\lvert -,- \right\rangle\right) \end{align*}

Note

e−iωtσx⊗σx∣+,+⟩=e−iωt∣+,+⟩e^{-i\omega t \sigma_x \otimes \sigma_x} \left\lvert +,+ \right\rangle = e^{-i\omega t} \left\lvert +,+ \right\rangle e−iωtσx⊗σx∣+,−⟩=e+iωt∣+,−⟩e^{-i\omega t \sigma_x \otimes \sigma_x} \left\lvert +,- \right\rangle = e^{+i\omega t} \left\lvert +,- \right\rangle e−iωtσx⊗σx∣−,+⟩=e+iωt∣−,+⟩e^{-i\omega t \sigma_x \otimes \sigma_x} \left\lvert -,+ \right\rangle = e^{+i\omega t} \left\lvert -,+ \right\rangle e−iωtσx⊗σx∣−,−⟩=e−iωt∣−,−⟩e^{-i\omega t \sigma_x \otimes \sigma_x} \left\lvert -,- \right\rangle = e^{-i\omega t} \left\lvert -,- \right\rangle

So

det⁡12(e−iωteiωteiωte−iωt)=14(e−2iωt−e2iωt)=−i2sin⁡(2ωt)≠0\det \frac{1}{2}\begin{pmatrix} e^{-i\omega t} & e^{i\omega t} \\ e^{i\omega t} & e^{-i\omega t} \end{pmatrix} = \frac{1}{4}\left(e^{-2i\omega t} - e^{2i\omega t}\right) = -\frac{i}{2}\sin(2\omega t) \neq 0   ⟹  Entangled state for t>0,t≠kπω\implies \text{Entangled state for } t > 0, \quad t \neq \frac{k\pi}{\omega} Maximal when t=π4ω\text{Maximal when } t = \frac{\pi}{4\omega}