# Schrödinger vs. Heisenberg Picture

In this article we discuss two alternative ways of looking at the time development. We already discussed that the time development of the state vector $|\psi(t) \rangle$ is decribed by the abstract Schrödinger Equation which we introduced as a postulate. Recall the Schrödinger Equation \begin{align} i \, \hbar \frac{d}{dt} |\psi(t) \rangle & = H \, |\psi(t) \rangle\\ \end{align} has the solution \begin{align} |\psi(t) \rangle & = e^{- \frac{i H }{\hbar}\,(t-t_0)} |\psi(t_0) \, \rangle\\ \end{align} where \begin{align} U(t,t_0)=e^{- \frac{i H }{\hbar}\,(t-t_0)} \end{align} is the unitary evolution operator. An operator is called unitary if \begin{align} U^{\dagger}=U^{-1} \quad \text{and} \quad UU^{\dagger}=U^{\dagger}U=\mathbb{I} \end{align} which is obviously the case for the time development operator. Unitary operations preserve the norm since \begin{align} \langle \psi(t)|\psi(t) \rangle & = \langle \psi(t_0)| e^{ \frac{i H^{\dagger} }{\hbar}\,(t-t_0)} e^{- \frac{i H }{\hbar}\,(t-t_0)} |\psi(t_0) \, \rangle\\ \end{align} the Hamiltonian is hermitian $H^{\dagger}=H$, so we have \begin{align} \langle \psi(t)|\psi(t) \rangle & = \langle \psi(t_0)| e^{ \frac{i H }{\hbar}\,(t-t_0)} e^{- \frac{i H }{\hbar}\,(t-t_0)} |\psi(t_0) \, \rangle\\ \end{align} since the Hamiltonian commutes with itself $[H,H]=0$ \begin{align} e^{ \frac{i H }{\hbar}\,(t-t_0)} e^{- \frac{i H }{\hbar}\,(t-t_0)} = \mathbb{I} \end{align} is just the unit operator. But notice that in general the product \begin{align} e^{A} e^{B} \not= e^{A+B} \quad \text{unless} \quad [A,B]=0 \end{align} because in general \begin{align} e^{A} e^{B} \not= e^{A+B}e^{[A,B]/2} \end{align} However in this case we find \begin{align} \langle \psi(t)|\psi(t) \rangle & = \langle \psi(t_0)| \psi(t_0) \, \rangle\\ \end{align} This means the length of the state vector $|\psi(t) \rangle$ in Hilbert Space does not change with time. This is a general property of unitary transformations. One can visualise the time development as movement of the state vector on the unit sphere.

Let us now determine the mean value of the oberservable $A$ and how $\langle A \rangle$ changes with time. The mean value of $A$ is \begin{align} \langle A \rangle (t)=\langle \psi(t)|A|\psi(t) \rangle \end{align} but we know $|\psi(t) \rangle$ so we can simply plug in \begin{align} \langle A \rangle (t)=\langle \psi(0)|e^{ \frac{i H }{\hbar}\,(t)} A e^{- \frac{i H }{\hbar}\,(t)} |\psi(0) \, \rangle \end{align} in this new expression, there are two ways of looking at the equation.

The time development operators can eather act on the ket and the bra, or on the operator $A$. In the first case the state vector changes with time - this is called the Schrödinger picture - and in the second case the observable is time dependent - we call this the Heiseberg picture.

\begin{align} \langle A \rangle (t) = \langle \psi_S(t)| A_S | \psi_S(t) \, \rangle \end{align} \begin{align} \langle A \rangle (t) = \langle \phi_H| A_H(t) | \phi_H \, \rangle \end{align} the state vector in the Heisenberg picture are denoted $|\psi_S(t_0)=| \phi_H \, \rangle$ to underline that they are fixed vectors independent of time.

Up to now we only considered time dependence of the state vector, but somtimes it is convenient to work in the Heisenberg Picture.

To find the equivalent of Schrödinger's Equation called the Heisenberg equation of motion, we differentiate $\langle A \rangle$ \begin{align} \left\langle \frac{dA}{dt} \right\rangle=\langle \psi(0)|e^{ \frac{i H }{\hbar}\,t} A e^{- \frac{i H }{\hbar}\,t} |\psi(0) \, \rangle \end{align} differenciating the first factor $e^{\frac{i H}{\hbar}\, t}$ gives just a factor $\frac{i H }{\hbar}$ times the original equation. The second factor brings down $-\frac{i H }{\hbar}$ times the original equation. Us $A_H(t)$ for $e^{ \frac{i H }{\hbar}\,t} A e^{- \frac{i H }{\hbar}\,t}$ we have \begin{align} \left\langle \frac{dA}{dt} \right\rangle=\langle \psi(0)|\frac{i H }{\hbar} A_H(t) |\psi(0) \, \rangle - \langle \psi(0)|A_H(t)\frac{i H }{\hbar} |\psi(0) \, \rangle = \frac{1}{i \hbar} \langle \psi(0)|A_H(t) H |\psi(0) \, \rangle - \langle \psi(0)|H A_H(t) |\psi(0) \, \rangle=\frac{1}{i \hbar}\langle [A_H(t),H] \rangle \end{align} so we get \begin{align} \left\langle \frac{dA}{dt} \right\rangle=\frac{1}{i \hbar}\langle [A_H(t),H] \rangle \end{align} If $A_H(t)$ has explicit time dependence (not coming from the time development operators) then we need to include the partial derivative of $A_H(t)$ with respect to time. We could have also startet with the equation \begin{align} \frac{d}{dt}\langle A(t) \rangle=\frac{d}{dt}(\langle \psi(t)|A_S(t)|\psi(t) \rangle)=\frac{d}{dt}\langle \psi(t)|A_S(t)|\psi(t) \rangle+\langle \psi(t)|\frac{d}{dt}A_S(t)|\psi(t) \rangle + \langle \psi(t)|A_H(t)|\frac{d}{dt}|\psi(t) \rangle \end{align} where the following terms are familiar \begin{align} \frac{d}{dt}\langle \psi(t)|A_S(t)|\psi(t) \rangle+ \langle \psi(t)|A_S(t)|\frac{d}{dt}|\psi(t) \rangle =\frac{1}{i \hbar}\langle [A_H(t),H] \rangle \end{align} and \begin{align} \langle \psi(t)|\frac{d}{dt}A_S(t)|\psi(t) \rangle =\left\langle \frac{dA_S(t)}{dt} \right\rangle \end{align} To specify that $d/dt$ only acts on the explicit time dependence of $A_S(t)$ and $A_H(t)$ we write it as partial derivative \begin{align} \langle \psi(t)| \frac{\partial A_S(t)}{\partial t}|\psi(t) \rangle = \langle \psi_H(0)|\frac{d}{dt}A_H(t)|\psi_H(t_0) \rangle= \left\langle \frac{\partial A_H(t)}{\partial t} \right\rangle \end{align} Finally we arive at the general

Heisenberg Equation of Motion \begin{align} \frac{dA}{dt} =\frac{1}{i \hbar} [A_H(t),H]+ \frac{\partial A_H(t)}{\partial t} \end{align} The mean value of the oberservable coincides with the classical equations of motion, this fact is known as Ehrenfest Theorem \begin{align} \left\langle \frac{dA}{dt} \right\rangle=\frac{1}{i \hbar}\langle [A_H(t),H] \rangle+\left\langle \frac{\partial A_H(t)}{\partial t} \right\rangle \end{align}

The Heisenberg Equation of Motion is the analog of the classical canonical transformation, where some variable changes unter transformations generated by the Hamiltonian.

The evolution of the mean value of an observable in the Schrödinger picture $\frac{d}{dt}\langle A_S(t) \rangle$ is equal to the mean value in the Heisenberg picture $\left\langle \frac{dA}{dt} \right\rangle$ \begin{align} \left\langle \frac{dA}{dt} \right\rangle=\frac{d}{dt}\langle A(t) \rangle \end{align}

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