A.12 Heisen­berg pic­ture

This book fol­lows the for­mu­la­tion of quan­tum me­chan­ics as de­vel­oped by Schrö­din­ger. How­ever, there is an­other, ear­lier, for­mu­la­tion due to Heisen­berg. This sub­sec­tion gives a brief de­scrip­tion so that you are aware of it when you run into it in lit­er­a­ture.

In the Schrö­din­ger pic­ture, phys­i­cal ob­serv­ables like po­si­tion and mo­men­tum are rep­re­sented by time-in­de­pen­dent op­er­a­tors. The time de­pen­dence is in the wave func­tion. This is some­what coun­ter­in­tu­itive be­cause clas­si­cally po­si­tion and mo­men­tum are time de­pen­dent quan­ti­ties. The Heisen­berg pic­ture re­moves the time de­pen­dence from the wave func­tion and ab­sorbs it into the op­er­a­tor.

To see how that works out, con­sider first the gen­eral form of the wave func­tion. It can be writ­ten as

\begin{displaymath}
\Psi(\ldots;t) = e^{-{\rm i}H t/\hbar} \Psi(\ldots;0)
\end{displaymath} (A.37)

where the ex­po­nen­tial of an op­er­a­tor is de­fined through its Tay­lor se­ries:
\begin{displaymath}
e^{-{\rm i}H t/\hbar} = 1 - {\rm i}\frac{t}{\hbar} H
- \frac{t^2}{2!\hbar^2} H^2 + \ldots
\end{displaymath} (A.38)

(To check the above ex­pres­sion for the wave func­tion, take the ini­tial wave func­tion to be any en­ergy eigen­func­tion of en­ergy $E$. You get the cor­rect $e^{-{{\rm i}}Et/\hbar}$ time de­pen­dence, chap­ter 7.1.2. Every $H$ be­comes an $E$. And if the ex­pres­sion works for any eigen­func­tion, it works for all their com­bi­na­tions too. That means that it works for any wave func­tion, be­cause the eigen­func­tions are com­plete. To be sure, the above form of the wave func­tion ap­plies only if the Hamil­ton­ian is in­de­pen­dent of time. Even if it is not, the trans­for­ma­tion from the ini­tial wave func­tion $\Psi(\ldots;0)$ to a later one $\Psi(\ldots;t)$ still re­mains a “uni­tary” one; one that keeps the wave func­tion nor­mal­ized. But then you will need to use the Schrö­din­ger equa­tion di­rectly to fig­ure out the time de­pen­dence.)

Now con­sider an ar­bi­trary Schrö­din­ger op­er­a­tor $\widehat{A}$. The phys­i­cal ef­fects of the op­er­a­tor can be char­ac­ter­ized by in­ner prod­ucts, as in

\begin{displaymath}
\langle \Psi_1(\ldots;t)\vert\widehat A \Psi_2(\ldots;t)\rangle
\end{displaymath} (A.39)

Such a dot prod­uct tells you what amount of a wave func­tion $\Psi_1$ is pro­duced by ap­ply­ing the op­er­a­tor on a wave func­tion $\Psi_2$. Know­ing these in­ner prod­ucts for all wave func­tions is equiv­a­lent to know­ing the op­er­a­tor.

If the time-de­pen­dent ex­po­nen­tials are now peeled off $\Psi_1$ and $\Psi_2$ and ab­sorbed into the op­er­a­tor, you get the time-de­pen­dent Heisen­berg op­er­a­tor

\begin{displaymath}
\widetilde A \equiv e^{{\rm i}H t/\hbar} \widehat A e^{-{\rm i}H t/\hbar}
\end{displaymath} (A.40)

Heisen­berg op­er­a­tors will be in­di­cated with a tilde in­stead of a hat. Note that the ar­gu­ment of the first ex­po­nen­tial changed sign be­cause it was taken to the other side of the in­ner prod­uct.

The op­er­a­tor $\widetilde{A}$ de­pends on time. To see how it evolves, dif­fer­en­ti­ate the prod­uct with re­spect to time:

\begin{displaymath}
\frac{{\rm d}\widetilde A}{{\rm d}t} =
\frac{{\rm i}}{\hba...
...hbar} \widehat A e^{-{\rm i}H t/\hbar} \frac{{\rm i}}{\hbar} H
\end{displaymath}

The first and third terms can be rec­og­nized as a mul­ti­ple of the com­mu­ta­tor of $H$ and $\widetilde{A}$, while the mid­dle term is the Heisen­berg ver­sion of the time de­riv­a­tive of $\widehat{A}$, in case $\widehat{A}$ does hap­pen to de­pend on time. So the evo­lu­tion equa­tion for the Heisen­berg op­er­a­tor be­comes
\begin{displaymath}
\frac{{\rm d}\widetilde A}{{\rm d}t} = \frac{{\rm i}}{\hbar...
...}H t/\hbar}
\left[H,\widehat A\right] e^{-{\rm i}H t/\hbar} %
\end{displaymath} (A.41)

(Note that there is no dif­fer­ence be­tween the Hamil­to­ni­ans $\widehat{H}$ and $\widetilde{H}$ be­cause $H$ com­mutes with it­self, hence with its ex­po­nen­tials.)

For ex­am­ple, con­sider the Schrö­din­ger ${\widehat x}$ po­si­tion and ${\widehat p}_x$ lin­ear mo­men­tum op­er­a­tors of a par­ti­cle. These do not de­pend on time. Us­ing the com­mu­ta­tors as fig­ured out in chap­ter 7.2.1, the cor­re­spond­ing Heisen­berg op­er­a­tors evolve as:

\begin{displaymath}
\frac{{\rm d}\widetilde x}{{\rm d}t} = \frac{1}{m} \widetil...
...p}_x}{{\rm d}t} =
- \widetilde{\frac{\partial V}{\partial x}}
\end{displaymath}

Those have the ex­act same form as the equa­tions for the clas­si­cal po­si­tion and mo­men­tum of the par­ti­cle.

In fact, the equiv­a­lent of the gen­eral equa­tion (A.41) is also found in clas­si­cal physics: it is de­rived in ad­vanced me­chan­ics, with the so-called “Pois­son bracket” tak­ing the place of the com­mu­ta­tor. As a sim­ple ex­am­ple, con­sider one-di­men­sion­al mo­tion of a par­ti­cle. Any vari­able $a$ that de­pends on the po­si­tion and lin­ear mo­men­tum of the par­ti­cle, and maybe also ex­plic­itly on time, has a time de­riv­a­tive given by

\begin{displaymath}
\frac{{\rm d}a}{{\rm d}t} =
\frac{\partial a}{\partial x} ...
...} \frac{{\rm d}p_x}{{\rm d}t} +
\frac{\partial a}{\partial t}
\end{displaymath}

ac­cord­ing to the to­tal dif­fer­en­tial of cal­cu­lus. And from the clas­si­cal Hamil­ton­ian

\begin{displaymath}
H = \frac{p_x^2}{2m} + V
\end{displaymath}

it is seen that the time de­riv­a­tives of po­si­tion and mo­men­tum obey the clas­si­cal “Hamil­ton­ian dy­nam­ics”

\begin{displaymath}
\frac{{\rm d}x}{{\rm d}t} = \frac{\partial H}{\partial p_x}...
... \frac{{\rm d}p_x}{{\rm d}t} = - \frac{\partial H}{\partial x}
\end{displaymath}

Sub­sti­tut­ing this into the time de­riv­a­tive of $a$ gives

\begin{displaymath}
\frac{{\rm d}a}{{\rm d}t} =
\frac{\partial a}{\partial x} ...
...\frac{\partial H}{\partial x} +
\frac{\partial a}{\partial t}
\end{displaymath}

The first two terms in the right hand side are by de­f­i­n­i­tion mi­nus the Pois­son bracket $\{H,a\}_{\rm {P}}$, so

\begin{displaymath}
\frac{{\rm d}a}{{\rm d}t} = - \{H,a\}_{\rm {P}} + \frac{\pa...
... \frac{\partial a}{\partial x} \frac{\partial H}{\partial p_x}
\end{displaymath}

Note that the Pois­son bracket, like the com­mu­ta­tor, is an­ti­sym­met­ric un­der ex­change of $H$ and $a$. Ap­par­ently, for­mally iden­ti­fy­ing the Pois­son bracket with the com­mu­ta­tor di­vided by ${\rm i}\hbar$ brings you from clas­si­cal me­chan­ics to Heisen­berg’s quan­tum me­chan­ics.

More gen­er­ally, the clas­si­cal Hamil­ton­ian can de­pend on mul­ti­ple and non-Carte­sian co­or­di­nates, gener­i­cally called “gen­er­al­ized co­or­di­nates.” In that case, in the Pois­son bracket you must sum over all gen­er­al­ized co­or­di­nates and their as­so­ci­ated so-called “canon­i­cal” mo­menta. For a Carte­sian po­si­tion co­or­di­nate, the canon­i­cal mo­men­tum is the cor­re­spond­ing lin­ear mo­men­tum. For an an­gu­lar co­or­di­nate, it is the cor­re­spond­ing an­gu­lar mo­men­tum. In gen­eral, us­ing the so-called La­grangian for­mu­la­tion usu­ally cov­ered in an en­gi­neer­ing ed­u­ca­tion, and oth­er­wise found in ad­den­dum {A.1}, the canon­i­cal mo­men­tum is the de­riv­a­tive of the La­grangian with re­spect to the time de­riv­a­tive of the co­or­di­nate.

The bot­tom line is that the Heisen­berg equa­tions are usu­ally not easy to solve un­less you re­turn to the Schrö­din­ger pic­ture by peel­ing off the time de­pen­dence. In rel­a­tivis­tic ap­pli­ca­tions how­ever, time joins space as an ad­di­tional co­or­di­nate, and the Heisen­berg pic­ture be­comes more help­ful. It can also make it eas­ier to iden­tify the cor­re­spon­dence be­tween clas­si­cal equa­tions and the cor­re­spond­ing quan­tum op­er­a­tors.


Key Points
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In the Heisen­berg pic­ture, op­er­a­tors evolve in time just like their phys­i­cal vari­ables do in clas­si­cal physics.