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For an observable $A$ and a Hamiltonian $H$, Wikipedia gives the time evolution equation for $A(t) = e^{iHt/\hbar} A e^{-iHt/\hbar}$ in the Heisenberg picture as

$$\frac{d}{dt} A(t) = \frac{i}{\hbar} [H, A] + \frac{\partial A}{\partial t}.$$

From their derivation it sure looks like $\frac{\partial A}{\partial t}$ is supposed to be the derivative of the original operator $A$ with respect to $t$ and $\frac{d}{dt} A(t)$ is the derivative of the transformed operator. However, the Wikipedia derivation then goes on to say that $\frac{\partial A}{\partial t}$ is the derivative with respect to time of the transformed operator. But if that's true, then what does $\frac{d}{dt} A(t)$ mean? Or is that just a mistake?

(I need to know which term to get rid of if $A$ is time-independent in the Schrodinger picture. I think it's $\frac{\partial A}{\partial t}$ but you can never be too sure of these things.)

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@Qiaochu Yuan: Please link to the Wikipedia page, so we can improve the Wikipedia page if needed. –  Qmechanic Jun 19 '11 at 13:55
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@Qmechanic: thanks! It looks like the general issue is that we want to differentiate functions of time and other variables which are themselves functions of time. But here $A$, as an operator, is only (possibly) a function of time, yes? –  Qiaochu Yuan Jun 19 '11 at 13:58
    
@Qiaochu Yuan: Often, $\hat{A}(t)= \sum_r\sum_{i's} f_{i_1,\ldots i_r}(t) ~\hat{z}^{i_1}(t)\ldots \hat{z}^{i_r}(t)$, where $\hat{z}^{i}(t)$ are the fundamental phase space operators (positions and momenta). Explicit differentiation means differentiation of the $f_i$ coefficient functions. –  Qmechanic Jun 19 '11 at 14:21
    
@Qiaochu Yuan: The relation $\hat{A}(t)=e^{\frac{i}{\hbar}{\hat{H}}t}\hat{A}(0)e^{-\frac{i}{\hbar}\hat{H}t}$ is only true if $\hat{H}$ and $\hat{A}(t)$ do not depend explicitly on time $t$. (Here and above I have only addressed the Heisenberg picture.) –  Qmechanic Jun 19 '11 at 16:11

4 Answers 4

up vote 7 down vote accepted

There is no mistake on the Wikipedia page and all the equations and statements are consistent with each other. In $$A_{\rm Heis.}(t) = e^{iHt/\hbar} A e^{-iHt/\hbar}$$ the letter $A$ in the middle of the product represents the Schrödinger picture operator $A = A_{\rm Schr.}$ that is not evolving with time because in the Schrödinger picture, the dynamical evolution is guaranteed by the evolution of the state vector $|\psi\rangle$.

However, this doesn't mean that the time derivative $dA_{\rm Schr.}/dt=0$. Instead, we have $$ \frac{dA_{\rm Schr.}}{dt} = \frac{\partial A_{\rm Schr.}}{\partial t} $$ Here, $A_{\rm Schr.}$ is meant to be a function of $x_i, p_j$, and $t$. In most cases, there is no dependence of the Schrödinger picture operators on $t$ - which we call an "explicit dependence" - but it is possible to consider a more general case in which this explicit dependence does exist (some terms in the energy, e.g. the electrostatic energy in an external field, may be naturally time-dependent).

In Schrödinger's picture, $dx_{i,\rm Schr.}/dt=0$ and $dp_{j,\rm Schr.}/dt=0$ which is why the total derivative of $A_{\rm Schr.}$ with respect to time is given just by the partial derivative with respect to time. Imagine, for example, $$ A_{\rm Schr.}(t) = c_1 x^2 + c_2 p^2 + c_3 (t) (xp+px) $$ We would have $$ \frac{dA_{\rm Schr.}(t)}{dt} = \frac{\partial c_3(t)}{\partial t} (xp+px).$$ These Schrödinger's picture operators are called "untransformed" on that Wikipedia page. The transformed ones are the Heisenberg picture operators given by $$A_{\rm Heis.}(t) = e^{iHt/\hbar} A_{\rm Schr.}(t) e^{-iHt/\hbar}$$ Their time derivative, $dA_{\rm Heis.}(t)/dt$, is more complicated. An easy differentiation gives exactly the formula involving $[H,A_{\rm Heis.}]$ that you quoted as well. $$\frac{d}{dt} A_{\rm Heis.}(t) = \frac{i}{\hbar} [H, A_{\rm Heis.}(t)] + \frac{\partial A_{\rm Heis.}(t)}{\partial t}.$$ The two terms in the commutator arise from the $t$-derivatives of the two exponentials in the formula for the Heisenberg $A_{\rm Heis.}(t)$ while the partial derivative arises from $dA_{\rm Schr.}/dt$ we have always had. (These simple equations remain this simple even for a time-dependent $A_{\rm Schr.}$; however, we have to assume that the total $H$ is time-independent, otherwise all the equations would get more complicated.) The two exponentials on both sides never disappear by any kind of derivative, so obviously, all the appearances of $A$ in the differential equation above are $A_{\rm Heis.}$. The displayed equation above is the (only) dynamical equation for the Heisenberg picture so it is self-contained and doesn't include any objects from other pictures.

In the Heisenberg picture, it is no longer the case that $dx_{\rm Heis.}(t)/dt=0$ (not!) and the similar identity fails for $p_{\rm Heis.}(t)$ as well. $A_{\rm Heis.}(t)$ is a general function of all the basic operators $x_{i,\rm Heis.}(t)$ and $p_{j,\rm Heis.}(t)$, as well as time $t$.

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This is terrible notation (on the part of the math / physics community as a whole, not this answer in particular). So if I'm reading this answer correctly, $\frac{\partial A}{\partial t}$ really means $\frac{\partial A_{\text{Schr}}(x, p, t)}{\partial t}$ (keeping $x$ and $p$ constant) whereas $\frac{d}{dt} A(t)$ means $\frac{d}{dt} \left( e^{iHt/\hbar} A_{\text{Schr}}(x(t), p(t), t) e^{-iHt/\hbar} \right)$? –  Qiaochu Yuan Jun 19 '11 at 15:21
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You make it sound contrived but there is nothing contrived about it. If $A_{Heis}(t) = e.A_{Schr}.e$, then obviously $d A_{Heis}(t)/dt = d(e.A_{Schr}.e)/dt$. This is called substitution and I am sure that most mathematicians would agree that this is true. Obviously, when we use symbols, we must know what they mean. The Wikipedia page assumes that the reader knows which $A(t)$ is the Schr. picture and which one is the Heis. picture. I just made this distinction more explicit. But when people know what the symbols mean, the notation of derivatives is unambiguous and standard. –  Luboš Motl Jun 19 '11 at 15:52
    
@Qiaochu Yuan: The point is that implicit and explicit dependences change their meaning when going back and forth between Schroedinger and Heisenberg picture because the basic operators get changed as well. –  Qmechanic Jun 19 '11 at 15:52
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Just to be sure, in the equation $dA/dt = i/\hbar [H,A] +\partial A/\partial t$, all the symbols $A$ are obviously still $A_{Heis}$, including one in the last term. This whole equation is in Heisenberg picture - it is the dynamical equation determining the time evolution of all of physics in the Heisenberg picture so of course that it cannot rely on other pictures. –  Luboš Motl Jun 19 '11 at 15:53
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Dear Qiaochu, even if $A_{Heis}$ meant nothing else, we still wrote its "definition" in terms of $A_{Schr}$, by the conjugation, so it is definitely not an undefined object. It is totally well-defined - if you decide that the conjugation is its definition - and all questions about it may be answered by pure thought. Yes, it is true that $\partial x_{Heis} / \partial t = 0$ and similarly for $p_{Heis}$. However, $dx_{Heis}/dt$ is not zero, and neither is $dp_{Heis}/dt$. –  Luboš Motl Jun 19 '11 at 16:12

It's easiest to derive this from the Schrödinger picture:

Let $B(t)$ be a time-dependent operator in the Schrödinger picture. The corresponding operator in the Heisenberg picture is $A(t) = e^{iHt/\hbar} B(t) e^{-iHt/\hbar}$. Differentiation with respect to $t$ gives

$$ \frac{d}{dt} A(t) = e^{iHt/\hbar} \left(\frac{i}{\hbar} H B(t) + \frac{\partial}{\partial t}B(t) - \frac{i}{\hbar} B(t) H) \right) e^{-iHt/\hbar} $$ $$ = e^{iHt/\hbar} \left(\frac{i}{\hbar} [H,B(t)] + \frac{\partial}{\partial t}B(t)\right) e^{-iHt/\hbar} = \frac{i}{\hbar} [H,A(t)] + \frac{\partial A}{\partial t} $$

In other words, the last partial derivative is to be understood in the sense that you take the operator $\frac{\partial B}{\partial t}$ and "evolve it in time" via the Schrödinger equation.


Useful non-example: the velocity operator $\vec v$. The velocity operator is the derivative of the position operator, but it's the total derivative as the system evolves. Hence,

$$ \vec v = \frac{i}{\hbar} [H,\vec r] .$$

In the Schrödinger picture, the position operator is, of course, time independent. Since $H$ is time independent as well, this is also the right velocity operator in the Schrödinger picture.

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As always in the Hamiltonian formulation of mechanics, whether classical or quantum, $$\partial A\over\partial t$$ means the way $A$ varies explicitly in time simply from the occurrence of $t$ explicitly in its formula.

But some of the other parts of the formula of $A$ might change with time also, thus contributing something to the total change in $A$ as time goes by, notated $$dA\over dt.$$

This is the same as the notation in the chain rule in several variables where $df=\frac{\partial f}{\partial x}dx + \frac{\partial f}{\partial t}dt$. The differential on the Left Hand Side is the « total differential » $df$ but it is the sum of two terms, only one of which is the explicit dependence of $f$ on $t$.

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The Heisenberg picture is defined as

$$A_{\mathrm{H}}(t) = e^{iHt/\hbar} A_{\mathrm{S}}(t) e^{-iHt/\hbar}$$

differentiating both sides we obtain

$$i\hbar \frac{\mathrm{d}}{\mathrm{d} t} A_{\mathrm{H}}(t) = [ A_{\mathrm{H}}(t), H] + i\hbar \left( \frac{\mathrm{d}}{\mathrm{d} t} A_{\mathrm{S}}(t) \right)_{\mathrm{H}} \>\>\>\>\>\>\>\>\>\>\>\>\>\> (1)$$

Some textbooks rewrite the last term using the notation [*]

$$\frac{\partial}{\partial t} A_{\mathrm{H}}(t) \equiv \left( \frac{\mathrm{d}}{\mathrm{d} t} A_{\mathrm{S}}(t) \right)_{\mathrm{H}}$$

[*] I agree on that this notation is awkward for mathematicians (it is not a true partial derivative) and the more rigorous physics textbooks use (1) with the total time derivative.

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