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Preparing: Quantum mechanical pictures

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• Schrödinger picture
• Heisenberg picture

Author: Vukics Andras <vukics@optics.szfki.kfki.hu> PDF: http://bird.szfki.kfki.hu/~vukics/physics/notes/pictures.pdf Placed into the public domain 2004-07-13.

Of quantum mechanical pictures

Schrödinger picture

In the Schrödinger picture the state of the system is described by the state vector $|\psi _{e}(t)\rangle$, for which the time development is determined by the Hamiltonian $H$ of the system through the Schrödinger equation:

<math> i\hbar\frac d{dt}\ket{\Psi(t)}=\Ha\ket{\Psi(t)}. </math>

The expectation value of an arbitrary observable $Ob$ reads:

\begin{equation}\atl{\Ob}=\bra{\Psi(t)}\Ob\ket{\Psi(t)}.\end{equation}

Heisenberg picture

It is easy to verify that the formal solution of Eq.~(\ref{eq:sch-eq}) in the case of time independent Hamiltonian (which will always be the case in this article) reads:

<math>\ket{\Psi(t)}=e^{-\frac i\hbar\Ha t}\ket{\Psi(0)}=:\Un_H(t)\ket{\Psi(0)}.</math>

In the Heisenberg picture the state of the system is kept time independent

<math>
\ket{\Psi_H}=\ket{\Psi(0)},
</math>

and the time evolution is transported to the operators $Ob_{H}(t)$ in such a way that the expectation value is unchanged:

<math>
\bra{\Psi_H}\Ob_H(t)\ket{\Psi_H}=\atl{\Ob}_H=
\atl{\Ob}=\bra{\Psi(t)}\Ob\ket{\Psi(t)}=\\=
\bra{\Psi(0)}\Un^\dag_H(t)\Ob\Un_H(t)\ket{\Psi(0)},
</math>

from which it is easy to derive that the Heisenberg-picture operators are connected to the Schrädinger-picture ones as

<math>\Ob_H(t)=\Un^\dag_H(t)\Ob\Un_H(t).</math>

From this it is easy to see that the Hamiltonian is unchanged:

<math>\Ha_H=\Ha.\end{equation}</math>

It is easy to verify that the differential equation for the time development of the observables (the Heisenberg equation) reads:

<math>
\frac{d\Ob_H(t)}{dt}=\frac i\hbar\left[\Ha,\Ob_H(t)\right]
+\Un^\dag_H(t)\frac{\partial\Ob}{\partial t}\Un_H(t).
\label{eq:H-eq}
</math>

It is worth noting that in the case when the Hamiltonian has the form \(\Ha=\Ha^{(0)}+\Ha^{(int)}\), though it is true that \(\Ha_H=\Ha\), it is in general \emph{not} true that \(\Ha^{(0)}_H=\Ha^{(0)}\) or \(\Ha^{(int)}_H=\Ha^{(int)}\). Only in the case when \(\left[\Ha^{(0)},\Ha^{(int)}\right]=0\) it is true that \begin{eqnarray} &\Ha^{(0)}_H=\Un^\dag_H(t)\Ha^{(0)}\Un_H(t)=\Ha^{(0)},\\ &\Ha^{(int)}_H=\Un^\dag_H(t)\Ha^{(int)}\Un_H(t)=\Ha^{(int)}. \end{eqnarray} However it is always true that \(\Ha^{(0)}\) and \(\Ha^{(0)}_H\) have the same \emph{form} if each one is written with operators of the corresponding picture.

Interaction pictures

We use interaction picture when we have a rapid free time development and a slow development due to interactions, and we would like to eliminate the rapid motion to concentrate on the interactions. It is also well adopted for perturbative calculations.

There exists two types of interaction pictures, the Schr\"odinger one and the Heisenberg one. The first one is used when before turning to interaction picture we are in Schr\"odinger picture, and after the transformation we would like to preserve the form of the Schr\"odinger equation. The second one, the Heisenberg interaction picture is used when before the transformation we are in Heisenberg picture and we would like to preserve the form of the Heisenberg equation.

The Schrödinger interaction picture

In this picture the state of the system is written like \begin{equation} \ket{\Psi_I(t)}=\Uni\ket{\Psi(t)}, \end{equation} and we request that \begin{equation} i\hbar\frac d{dt}\ket{\Psi_I(t)}=\Ha_I\ket{\Psi_I(t)}. \end{equation} Using these and Eq.~(\ref{eq:sch-eq}) it is easy to show that the corresponding interaction-picture Hamiltonian reads \begin{equation} \Ha_I=\Uni\Ha\Unid+i\hbar\dt{\Uni}\Unid. \label{eq:int-Ham} \end{equation} The rest of the time development is transformed to the interaction-picture observables \(\Ob_I(t)=\Uni\Ob\Unid\) as \begin{equation} \dt{\Ob_I(t)}=\left[\dt{\Uni}\Unid,\Ob_I(t)\right]. \end{equation} We see that the form of the Heisenberg equation is \emph{not} preserved because \(\dt{\Uni}\Unid\) is not the corresponding interaction-picture Hamiltonian (\ref{eq:int-Ham}). We can not preserve the forms of the Schr\"odinger equation and the Heisenberg equation at the same time.

In particular if the Hamiltonian has the form

The Heisenberg interaction picture

Here we use Heisenberg picture before the transformation and transform the Heisenberg-picture observables: \begin{equation} \Ob_I(t)=\Uni\Ob_H(t)\Unid, \end{equation} and request that the Heisenberg equation remains unchanged: \begin{equation} \dt{\Obi}=\frac{i}{\hbar}\left[\Ha_I,\Obi\right]. \end{equation} From these and from Eq.~(\ref{eq:H-eq}) we derive that the corresponding interaction-picture Hamiltonian reads \begin{equation} \Ha_I=\Uni\Ha\Unid-i\hbar\dt{\Uni}\Unid. \label{eq:int-Hamv} \end{equation} Note that the only difference from Eq.~(\ref{eq:int-Ham}) is the minus sign. In this picture the Schr\"odinger equation for the state vector \(\ket{\Psi_I(t)}=\Uni\ket{\Psi_H}\) reads \begin{equation} i\hbar\frac{d}{dt}\ket{\Psi_I(t)}=\lz\Ha_I+\Uni\Ha\Unid\rz\ket{\Psi_I(t)}. \end{equation}