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Back in 2011, this question asked about the Wikipedia version of the Heisenberg equation of motion for an operator $A$:

\begin{equation*} \frac{d}{dt} A(t) = \frac{i}{\hbar} \left[ H, A(t) \right] + \frac{\partial A}{\partial t} \end{equation*}

The accepted response asserted that "there is no mistake on the Wikipedia page", and indeed the formula is the same today as it was then.

But it's not correct, I think.

The first two occurrences of $A$ refer to $A_H$, "in the Heisenberg picture", where

\begin{equation*} A_H(t) = U^\dagger(t) A_S(t) U(t) \end{equation*}

where $U(t)$ is the unitary time-evolution operator and $A_S(t)$ is the operator in the Schrodinger picture, which may or may not be time-dependent.

So far so good. However, the last term is not

\begin{equation*} \frac{\partial A_H}{\partial t} \end{equation*}

but rather \begin{equation*} \left(\frac{dA_S}{dt} \right)_H = U^\dagger(t) \frac{dA_S}{dt} U(t) \end{equation*}

In fact, if one wanted an equation of motion using exclusively $A_H$, I think one would write: \begin{equation*} \frac{d}{dt} A_H(t) = \frac{i}{\hbar} \left[ H_H(t), A_H(t) \right] + U^\dagger(t) \frac{d}{dt} \left(U(t) A_H(t) U^\dagger(t) \right) U(t) \end{equation*}

But no one ever does, as far as I can see. Is it wrong, or too ugly for prime time, or???

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  • $\begingroup$ Just a small point, don't you have the commutator round the wrong way? $\endgroup$ – Ruvi Lecamwasam Mar 27 '15 at 8:13
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You may see the idea on why the formula is written like that by a comparison with the classical phase space formula.

In classical phase space you have observables of the type $a(x,p,t)$, or more precisely $a(x(t),p(t),t)$; where $x(t)$, $p(t)$ and $t$ are considered independent variables. The equation of motion for $a$ is given by: $$\frac{d}{dt}a(x(t),p(t),t)=\{h,a\}+\partial_t a\; ;$$ where $h$ is the classical hamiltonian, $\{\cdot,\cdot\}$ the Poisson bracket and the partial derivative in time is intended as deriving the functional form $a(\cdot,\cdot,\cdot)$ above w.r.t the third variable only (i.e. disregarding the fact that the first two argument may depend on the third). This is quite usual when dealing with function of many variables, and is a widely accepted convention (only the explicit dependence is taken into account when taking partial derivatives).

In quantizing, a formula with the same flavour is obtained replacing functions with operators (with suitable ordering, and in their Heisenberg version)---for me it will be replacing small letters with capital ones---and Poisson brackets with $\frac{i}{\hbar}[\cdot,\cdot]$. This is just a naïve procedure, and not an exact derivation of the formula; nevertheless you get: $$\frac{d}{dt}A(X(t),P(t),t)=\frac{i}{\hbar}[H,A]+\partial_t A\; .$$ Again the meaning remains unchanged: the partial derivative is intended as the derivative of the functional form $A(\cdot,\cdot,\cdot)$ with respect to the third variable only.

Note (related to the OP comment in the other answer): As a disclaimer, I will not consider in the following any problem of domains of operators, convergence of series and so on; nevertheless also these problems are to be taken into account. The statements have to be intended as true on suitable domains and with suitably regular functions.

Let $A(X,P,t)$ be a ploynomial function (regardless of the ordering), whith $[X,P]\neq 0$, and $t$ commutes with everything. Let also $U(t)=U(X,P,t)$ be a unitary operator (in particular we will use $U^*(t)U(t)=U(t)U^*(t)=id$, where $id$ is the identity operator) such that $X(t):= U^*(t)XU(t)$, and $P(t)=U^*(t)PU(t)$. It is then clear that, using the property of $U$ above, $$U^*(t)A(X,P,t)U(t)=A(X(t),P(t),t)\; .$$ This can be extended, roughly speaking, to any function that admits a series expansion.

Now given this equality, if you interpret the partial derivative as acting only explicitly on $t$, you see that again you obtain for polynomials/functions with series expansion $$\partial_t A(X(t),P(t),t)=U^*(t)\partial_tA(X,P,t)U(t)\; ,$$ for the functional form of $\partial_t A(\cdot,\cdot,t)$ is always the same, and the "replacement" of $X,P$ with $X(t),P(t)$ (by means of $U$) can be done either before or after the derivative without changing the functional form (just changing $X$ in $X(t)$ and $P$ in $P(t)$).

As I said, this may not be true with more general functions, anyways this type of Heisenberg evolution formulas, written like that, are valid only in very special situations, if you want to be precise and consider domain problems and the proper definition of functions of unbounded operators.

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  • $\begingroup$ Thank you for this answer. I certainly have no problems with the classical equation of motion; it's as rigorous as physics gets, I think. For the quantum version, I've only seen derivations that start with the Schrodinger picture; your "direct replacement" procedure is intriguing. Forgive my skepticism, but do you have any references for it? Thanks. $\endgroup$ – Art Brown Mar 27 '15 at 5:28
  • $\begingroup$ @ArtBrown The direct replacement procedure is not a rigorous equality (as I said it was a naïve procedure, and I have modified the text accordingly to make it clearer): the problem is that the quantization of a poisson bracket is not, for polynomials of order equal or higher than three, the commutator of the quantized classical functions (there are additional terms). Nevertheless the heuristic replacement gives you an idea on how to consider the variables as independent in the formula you are interested in. $\endgroup$ – yuggib Mar 27 '15 at 7:55
  • $\begingroup$ Your edit is very helpful, thanks. Note: it turns out I introduced a sign error in the commutator of the equation I was trying to copy from Wikipedia. It's now fixed, but my typo may have leaked into your answer. $\endgroup$ – Art Brown Mar 29 '15 at 5:25
  • $\begingroup$ @ArtBrown You're welcome :) Also I corrected the sign to be coherent with your edit. $\endgroup$ – yuggib Mar 29 '15 at 7:03
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Let $A_S := A(X,P,t)$ be expressible as a polynomial in $X,P,t$ then: $$ A_H := U^* \ A_S \ U = U^* \ A(X,P,t) \ U = A(X_H,P_H,t) $$ Let $B_S := B(X,P,T) = \partial_t A_S = \partial_t A(X,P,t) $, then $B$ is also expressible as a polynomial in $X,P,t$. Therefore: $$ B_H := U^* \ B_S \ U = U^* \ B(X,P,t) \ U = B(X_H,P_H,t) = \partial_t A(X_H,P_H,t) = \partial_t A_H$$ Thus $$ \left( \partial_t A_S \right)_H = \partial_t A_H $$ since the Heisenberg-picture only transforms like $(X,P,t)\rightarrow(X_H,P_H,t)$ and leaves the functional dependance on time $t$ untouched (if A is expressible as a polynomial). To be more clear one shoud really write this like \begin{align} (X,P,t)&\rightarrow(X_H,P_H,t') \\ \\ \left( \partial_t A_S \right)_H & = \partial_t' A_H \end{align}

$$$$

For clarification: If $A(X,t) = X \cdot t$ then $$ \partial_t A(X,t) = X $$ and similar $$ \partial_t A(X_H,t) = X_H $$ because the first parameter of A is thought to be an independent variable, even tough in general $$ \partial_t X_H(t) \ne 0 $$ This is actually the same situation as in classical mechanics when one moves from generalized coordinates $q,\dot q,t$ to $q,p,t$. One really has to take care whether $t$ is an independent variable or parametrization.

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  • $\begingroup$ Thank you. I'm having trouble distinguishing your answer from yuggib's edit (which has time priority). Is the $t$ vs. $t'$ distinction important? Also, in your last equation, do you mean the full (convective?) time derivative instead of the partial derivative? $\endgroup$ – Art Brown Mar 29 '15 at 5:18
  • $\begingroup$ @ArtBrown: The distinction between $t$ and $t'$ is important insofar that one emphasizes the change of the coordinate system. Often, one does not explicitly use this notation since $t=t'$. Now, $X_H(t)$ is a path on some manifold with coordinates $(X_H, t)$. It just happens to be that one has chosen $t$ as a parametrization of that path. In this sense $\partial_t X_H(t)$ shall denote the change of $X_H(t)$ along the path. In differential geometry one would write this as $dX_H$ independant of parametrization. $\endgroup$ – image Mar 29 '15 at 12:51
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Back in 2011, [this question][1] asked about the [Wikipedia version][2] of the Heisenberg equation of motion for an operator $A$:

\begin{equation*} \frac{d}{dt} A(t) = \frac{i}{\hbar} \left[ A(t) ,H\right] + \frac{\partial A}{\partial t} \end{equation*}

The accepted response asserted that "there is no mistake on the Wikipedia page", and indeed the formula is the same today as it was then.

But it's not correct, I think.

The first two occurrences of $A$ refer to $A_H$, "in the Heisenberg picture", where

\begin{equation*} A_H(t) = U^\dagger(t) A_S(t) U(t) \end{equation*}

where $U(t)$ is the unitary time-evolution operator and $A_S(t)$ is the operator in the Schrodinger picture, which may or may not be time-dependent.

This is all fine and dandy...

So far so good. However, the last term is not

\begin{equation*} \frac{\partial A_H}{\partial t} \end{equation*}

Well, you are sort of correct... I.e., you are not even wrong in Pauli's sense of the phrase. What you have written above does not really make sense to write. Or rather, you seem to be re-interpreting what you wrote above as some kind of total derivative again...

but rather \begin{equation*} \left(\frac{dA_S}{dt} \right)_H =U^\dagger(t) \frac{dA_S}{dt} U(t) \end{equation*}

Yes, that is the correct way to interpret the meaning of $$ \frac{\partial A}{\partial t}\;. $$

Who ever said anything otherwise... You seem to be misinterpreting the meaning of the symbols $\frac{\partial A_H}{\partial t}$ and then blaming Wikipedia for your misinterpretation of the symbols...

So. Just to be clear. The following is true. $$ \frac{dA_H}{dt}=i[H,A_H]+U^\dagger \frac{\partial A_S}{\partial t}U $$ And there's no harm in writing either $dA_S/dt$ or $\partial A_S/\partial t$, both mean the same thing here.

I think that an example will help clear everything up and will help you understand why the authors of the Wikipedia article chose the notation that they did.

Suppose that $$ A_S=xp+x^2p^2t^2\;, $$ which is an operator with some explicit time dependence. Here $x$ and $p$ are the position and momentum Schrodinger operators.

Then $$ \frac{\partial A_S}{\partial t}=2x^2p^2t $$ and $$ U^\dagger\frac{\partial A_S}{\partial t}U=2x_H^2p_H^2t\;. $$ See how on the RHS of the above equation all of the operators are in the Heisenberg picture? I think this is the fact that the authors of the Wikipedia article are trying to impart with their notation. I.e., you don't really keep the $U^\dagger$ and $U$ separated out on the edges of the partial derivative, you incorporate them into the term by actually inserting $U^\dagger U$ to be able to write all the operators in the Heisenberg picture. I.e., $$ U^\dagger 2x^2p^2t U=2U^\dagger x UU^\dagger x UU^\dagger p UU^\dagger p Ut=2x_h^2p_H^2t $$

So, anyways, we end up with $$ \frac{dA_H}{dt}=i[H,x_Hp_H+x_H^2p_H^2t^2]+2x_H^2p_H^2t $$

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  • $\begingroup$ Good answer, but I thought it was Pauli who coined the phrase "not even wrong" $\endgroup$ – Sean Mar 26 '15 at 10:03
  • $\begingroup$ Good catch. I knew it was one of those guys ending with "i". $\endgroup$ – hft Mar 26 '15 at 13:59
  • $\begingroup$ Thank you for this answer. Unfortunately, I'm a bit confused by it. 1) On the one hand, your "true" equation looks the same as mine, and different from Wikipedia's. 2) On the other, in your example, it looks like $\partial A_H / \partial t = (dA_S/dt)_H $, which would validate the interpretation that you said "did not really make sense to write". In fact, if this equality holds in general, I would have to concede Wikipedia has it right. Thoughts? $\endgroup$ – Art Brown Mar 27 '15 at 5:41
  • $\begingroup$ $\frac{d A_H}{d t}$ does not equal $(\frac{dA_S}{dt})_H$. One has the $U^\dagger$ and $U$ being acted on by the derivative and one does not. I'm not sure exactly what you are expressing by $\frac{\partial A_H}{\partial t}$. I think you are confused about the meaning of the symbols and I'm trying to explain this to you in the answer I gave. $\endgroup$ – hft Mar 27 '15 at 6:10
  • $\begingroup$ Are you just being contrarian? Admittedly physicists play fairly fast and loose with symbols. Do you really not understand what is meant by the symbols or are you just being very picky? $\endgroup$ – hft Mar 27 '15 at 6:15

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