Heisenberg picture

Equivalence of Heisenberg's equation to the Schrödinger equation

For the sake of pedagogy, the Heisenberg picture is introduced here from the subsequent, but more familiar, Schrödinger picture.

According to Schrödinger's equation, the quantum state at time ${\displaystyle t}$ is ${\displaystyle |\psi (t)⟩=U(t)|\psi (0)⟩}$, where ${\displaystyle U(t)=T{e}^{-\frac{i}{\hslash }{\int }_0^{t}ds{H}_{\mathrm{S}}(s)}}$ is the time-evolution operator induced by a Hamiltonian ${\displaystyle H_{\mathrm{S}}(t)}$ that could depend on time, and ${\displaystyle |\psi (0)⟩}$ is the initial state. ${\displaystyle T}$ refers to time-ordering, ħ is the reduced Planck constant, and i is the imaginary unit. Given an observable ${\displaystyle A_{\mathrm{S}}(t)}$ in the Schrödinger picture, which is a Hermitian linear operator that could also be time-dependent, in the state ${\displaystyle |\psi (t)⟩}$, its expectation value is given by $${\displaystyle ⟨A{⟩}_t=⟨\psi (t)|A_{\mathrm{S}}(t)|\psi (t)⟩.}$$

In the Heisenberg picture, the quantum state is assumed to remain constant at its initial value ${\displaystyle |\psi (0)⟩}$, whereas operators evolve with time according to the definition $${\displaystyle A_{\mathrm{H}}(t):=U^{†}(t)A_{\mathrm{S}}(t)U(t).}$$ This readily implies ${\displaystyle ⟨A{⟩}_t=⟨\psi (0)|A_{\mathrm{H}}(t)|\psi (0)⟩}$, so the same expectation value can be obtained by working in either picture. The Schrödinger equation for the time-evolution operator is $${\displaystyle \frac{d}{dt}U(t)=-\frac{i}{\hslash }{H}_{\mathrm{S}}(t)U(t).}$$ It follows that $${\displaystyle \begin{array}{cc}\frac{d}{dt}{A}_{\mathrm{H}}(t)& =\left(\frac{d}{dt}{U}^{†}(t)\right)A_{\mathrm{S}}(t)U(t)+U^{†}(t)A_{\mathrm{S}}(t)\left(\frac{d}{dt}U(t)\right)+U^{†}(t)\left(\frac{\mathrm{\partial }{A}_{\mathrm{S}}}{\mathrm{\partial }t}\right)U(t)\\ & =\frac{i}{\hslash }{U}^{†}(t)H_{\mathrm{S}}(t)A_{\mathrm{S}}(t)U(t)-\frac{i}{\hslash }{U}^{†}(t)A_{\mathrm{S}}(t)H_{\mathrm{S}}(t)U(t)+U^{†}(t)\left(\frac{\mathrm{\partial }{A}_{\mathrm{S}}}{\mathrm{\partial }t}\right)U(t)\\ & =\frac{i}{\hslash }{U}^{†}(t)H_{\mathrm{S}}(t)U(t)U^{†}(t)A_{\mathrm{S}}(t)U(t)-\frac{i}{\hslash }{U}^{†}(t)A_{\mathrm{S}}(t)U(t)U^{†}(t)H_{\mathrm{S}}(t)U(t)+{\left(\frac{\mathrm{\partial }{A}_{\mathrm{S}}}{\mathrm{\partial }t}\right)}_{\mathrm{H}}\\ & =\frac{i}{\hslash }[H_{\mathrm{H}}(t),A_{\mathrm{H}}(t)]+{\left(\frac{\mathrm{\partial }{A}_{\mathrm{S}}}{\mathrm{\partial }t}\right)}_{\mathrm{H}},\end{array}}$$ where differentiation was carried out according to the product rule. This is Heisenberg's equation of motion. Note that the Hamiltonian that appears in the final line above is the Heisenberg Hamiltonian ${\displaystyle H_{\mathrm{H}}(t)}$, which may differ from the Schrödinger Hamiltonian ${\displaystyle H_{\mathrm{S}}(t)}$.

An important special case of the equation above is obtained if the Hamiltonian ${\displaystyle H_{\mathrm{S}}}$ does not vary with time. Then the time-evolution operator can be written as $${\displaystyle U(t)=e^{-\frac{i}{\hslash }t{H}_{\mathrm{S}}},}$$ and hence ${\displaystyle H_{\mathrm{H}}\equiv H_{\mathrm{S}}\equiv H}$ since ${\displaystyle U(t)}$ now commutes with ${\displaystyle H}$. Therefore, $${\displaystyle ⟨A{⟩}_t=⟨\psi (0)|e^{\frac{i}{\hslash }tH}{A}_{\mathrm{S}}(t)e^{-\frac{i}{\hslash }tH}|\psi (0)⟩}$$ and following the previous analyses, $${\displaystyle \begin{array}{cc}\frac{d}{dt}{A}_{\mathrm{H}}(t)& =\frac{i}{\hslash }[H,A_{\mathrm{H}}(t)]+e^{\frac{i}{\hslash }tH}\left(\frac{\mathrm{\partial }{A}_{\mathrm{S}}}{\mathrm{\partial }t}\right)e^{-\frac{i}{\hslash }tH}.\end{array}}$$

Furthermore, if ${\displaystyle A_{\mathrm{S}}\equiv A}$ is also time-independent, then the last term vanishes and $${\displaystyle \frac{d}{dt}{A}_{\mathrm{H}}(t)=\frac{i}{\hslash }[H,A_{\mathrm{H}}(t)],}$$ where ${\displaystyle A_{\mathrm{H}}(t)\equiv A(t)=e^{\frac{i}{\hslash }tH}A{e}^{-\frac{i}{\hslash }tH}}$ in this particular case. The equation is solved by use of the standard operator identity,$${\displaystyle e^{B}A{e}^{-B}=A+[B,A]+\frac{1}{2!}[B,[B,A]]+\frac{1}{3!}[B,[B,[B,A]]]+\cdots ,}$$ which implies $${\displaystyle A(t)=A+\frac{it}{\hslash }[H,A]+\frac{1}{2!}{\left(\frac{it}{\hslash }\right)}^2[H,[H,A]]+\frac{1}{3!}{\left(\frac{it}{\hslash }\right)}^3[H,[H,[H,A]]]+\cdots }$$

A similar relation also holds for classical mechanics, the classical limit of the above, given by the correspondence between Poisson brackets and commutators: $${\displaystyle \frac{[A,H]}{\hslash }⟷i\{A,H\}.}$$ In classical mechanics, for an A with no explicit time dependence, $${\displaystyle \{A,H\}=\frac{dA}{dt},}$$ so again the expression for A(t) is the Taylor expansion around t = 0.

In effect, the initial state of the quantum system has receded from view, and is only considered at the final step of taking specific expectation values or matrix elements of observables that evolved in time according to the Heisenberg equation of motion. A similar analysis applies if the initial state is mixed.

The time evolved state ${\displaystyle |\psi (t)⟩}$ in the Schrödinger picture is sometimes written as ${\displaystyle |{\psi }_{\mathrm{S}}(t)⟩}$ to differentiate it from the evolved state ${\displaystyle |{\psi }_{\mathrm{I}}(t)⟩}$ that appears in the different interaction picture.

Commutator relations

Commutator relations may look different than in the Schrödinger picture, because of the time dependence of operators. For example, consider the operators x(t₁), x(t₂), p(t₁) and p(t₂). The time evolution of those operators depends on the Hamiltonian of the system. Considering the one-dimensional harmonic oscillator, $${\displaystyle H=\frac{p^2}{2m}+\frac{m{\omega }^2{x}^2}{2},}$$ the evolution of the position and momentum operators is given by: $${\displaystyle \frac{d}{dt}x(t)=\frac{i}{\hslash }[H,x(t)]=\frac{p}{m},}$$ $${\displaystyle \frac{d}{dt}p(t)=\frac{i}{\hslash }[H,p(t)]=-m{\omega }^{2}x.}$$

Note that the Hamiltonian is time independent and hence x(t) and p(t) are the position and momentum operators in the Heisenberg picture. Differentiating both equations once more and solving for them with proper initial conditions, $${\displaystyle \dot{p}(0)=-m{\omega }^2{x}_0,}$$ $${\displaystyle \dot{x}(0)=\frac{p_0}{m},}$$ leads to $${\displaystyle x(t)=x_0\cos (\omega t)+\frac{p_0}{\omega m}\sin (\omega t),}$$ $${\displaystyle p(t)=p_0\cos (\omega t)-m\omega x_0\sin (\omega t).}$$

Direct computation yields the more general commutator relations, $${\displaystyle [x(t_1),x(t_2)]=\frac{i\hslash }{m\omega }\sin \left(\omega t_2-\omega t_1\right),}$$ $${\displaystyle [p(t_1),p(t_2)]=i\hslash m\omega \sin \left(\omega t_2-\omega t_1\right),}$$ $${\displaystyle [x(t_1),p(t_2)]=i\hslash \cos \left(\omega t_2-\omega t_1\right).}$$

For ${\displaystyle t_1=t_2}$, one simply recovers the standard canonical commutation relations valid in all pictures.

Summary comparison of evolution in all pictures

For a time-independent Hamiltonian H_S, where H_0,S is the free Hamiltonian,