When thinking about the two pictures, what I found to be strange was: I can write the postulate of time-evolution in the Schroedinger picture by:

\begin{align} i \hbar \frac{d}{dt} \lvert \Psi(t) \rangle = \hat{H} \lvert \Psi(t) \rangle \end{align}

with a suitable Hamilton-operator. We note that it can also depend on time, so writing $$\hat{H}(t) $$ and $$\partial_t \hat{H}(t)$$ makes sense.

Now we come to the Heisenberg picture, where the time dependence is described as a time dependence of the operators: \begin{align} \frac{d}{dt} \hat{A} = \frac{i}{\hbar} [\hat{H}, \hat{A}] + \partial_t A \end{align}

My first question: Can somebody define the quantity $\partial_t A$ in a sense full way? I somehow know what it means, but if we define $A(t)$ as a Function that maps times $t$ to operator acting on M, then this a function dependent of one variable $t$. There isn't a way to distinguish explicit and implicit dependency, because for that we would require A to depend on another variable $x$, that itself could then depend on $t$, so there would be a difference between $\frac{d}{dt} \hat{A}$ and $\partial_t A$, because $\frac{d}{dt} \hat{A} = \partial_t A + \partial_x A \cdot \dot{x}\;.$ At the moment, writing $\partial_t A$ doesn't make sense, and if it does, it's just the derivation of A after the time, and that's what you want to know initially, because in Heisenberg picture, time dependence is solely in the operators.

What I have seen to overcome this problem is some "definition of $\partial_t \hat{A} = \hat{U}(t)^{\dagger} \partial_t \hat{A}_S(t) \hat{U}(t) \;,$ that relies on the Schrödinger-Picture, since it uses the Schrödinger-version of the Operator A, and the time-evolution-Operator $\hat{U}$, that needs to be calculated first, using Schrödinger's equation.

Question 2: Is it possible to formulate the time dependency of operators in the Heisenberg-Picture, without any use of the depending operators in the Schrödinger picture (and without the time evolution Operator, that should then arise as a consequence of this equations) ? If not, can we say that Schrödinger's picture is more general than Heisenberg's Picture?


3 Answers 3


I personally find most notation in the Heisenberg picture totally unsatisfactory for the exact same reason as you, its hard to look at the notation and not want to see a time dependent operator as a function from time to a matrix (or to an operator on Hilbert Space, or such).

But it's similar to thermodynamics, a thermodynamic variable could be written as a function of many things.

So first keep in mind that you can have a Schrödinger, Heisenberg, or Dirac Picture completely independent of a position or momentum representation as a few examples.

So if you think of your operator as a function of other operators then you can imagine $\hat A=\hat A(\hat x,\hat p, t)$ and now there are lots of partials possible. This makes some amount of sense if you imagine a quantization of a classical Hamiltonian where the Hamiltonian itself isn't just the energy, it is specifically a function of the position and the corresponding conjugate canonical momentum (and maybe time).

But also, don't get too bothered. Any explicit time dependency if a classical Hamiltonian comes from not including the sources of the interactions into the Hamiltonian. They are like external forces. Include them and they aren't external any more, and the Hamiltonian no longer has a time dependency any more.

So all the partials are fictions anyway from treating something as external. So they are hacks to begin with.

  • $\begingroup$ Do you know wether one can write any physically meaningful Operator $\hat{A}$ as a "function" of position and momentum operator and time? That would be an equivalence to the classical picture, where the complete dynamic is described by the phase-space x and p, and it would fully answer my question. $\endgroup$ Commented Jan 19, 2016 at 9:30
  • $\begingroup$ @Quantumwhisp I think the fact truly a time partial is zero answers your question. The time partial only looks nonzero if you exclude part of the system and label it as external. But as for a dependence on x and p, you have an x1 and and x2 and so on and each has a conjugate canonical momentum. But if you want to include spin you either have to call it an x and then have additional xs or else the answer is no, since the Hamiltonian can defend on spin (think charged spin 1/2 particle with a magnetic moment). $\endgroup$
    – Timaeus
    Commented Jan 19, 2016 at 15:56

First off, the pictures are equivalent, because you can always map one picture to the other. For example, go to back from Heisenberg to Schrodinger operators (in the case of constant $H$) use $$A_S(t) = e^{iHt} A_H(t) e^{-iHt}.$$ Since we can always recover the Schrodinger operator time dependency, there's no information lost.

You already know that a total time derivative $dx/dt$ is not the same thing as a partial time derivative $\partial x / \partial t$, but it might be confusing in this new context. As an example, consider a time dependent Schrodinger operator of the form $V_S(x_S, t)$. The Heisenberg operator is $$V_H(x, t) = e^{-iHt} V_S(x_S, t) e^{iHt} = V_S(x_H(t), t).$$ The partial time derivative comes from how the function $V_S(x, t)$ depends on $t$. The total time derivative also includes how $x_H(t)$ evolves. So we can write Heisenberg operators purely in terms of other Heisenberg operators just fine, and we can still tell apart the two kinds of time dependence.

  • $\begingroup$ But you have to use the schrödinger-Operator to express the implicit and explicit time dependence of the Operator, isn't that right? $\endgroup$ Commented Jan 19, 2016 at 10:00
  • $\begingroup$ Not really... like, if you wanted an operator of the form $\hat{x} t$ (in Schrodinger picture), then the Heisenberg operator is $\hat{x}_H(t) t$. There's no Schrodinger operator in there at all. $\endgroup$
    – knzhou
    Commented Jan 19, 2016 at 15:56

This is something that I've also noticed but haven't seen anyone else mention, but yes, the Schrödinger equation is stronger than the Heisenberg equation, even in the time independent case. To see this, consider the two equations:

\begin{align} i \hbar \frac{d}{dt} \Psi(t) = \hat{H} \Psi(t) \end{align}

\begin{align} \frac{d}{dt} \hat{A} = \frac{i}{\hbar} [\hat{H}, \hat{A}] \end{align}

For well-behaved Hamiltonians, the general solution to Schrödinger's equation is $$\Psi(t) = e^{-i\hat{H}t/\hbar} \Psi_0$$ for initial condition $\Psi(0) = \Psi_0$, and the general solution to Heisenberg's equation is $$\hat{A}(t) = e^{i[\hat{H}, \cdot]t/\hbar} \hat{A}_0 = e^{i\hat{H}t/\hbar} \hat{A}_0 e^{-i\hat{H}t/\hbar}$$ for initial condition $\hat{A}(0) = \hat{A}_0$.

While it is trivial to get the general solution of the Heisenberg equation from the Schrödinger equation, the converse is not true. I.e., if you have a concrete expression for $e^{-i\hat{H}t/\hbar}$, then it is very easy to find $e^{i[\hat{H}, \cdot]t/\hbar}$, but the converse is not easy. In most cases though, you can't find the general solution to either equations (well, if you don't count the Taylor series solution, which is impractical to use).

This means that finding a solution to Heisenberg's equation won't give you a solution to Schrödinger's equation, nor make it easier to find, in general. And finding the solution to Schrödinger's equation is usually better, since it gives you the evolution of the probability density. Let me give you a concrete example. In the case of the harmonic oscillator, solving Heisenberg's equation for the position and momentum operator (see this article) will give you this result

\begin{equation} \hat{x}_H(t) = \hat{x} cos(\omega t) + \frac{\hat{p}}{m\omega} sin(\omega t) \end{equation}

\begin{equation} \hat{p}_H(t) = \hat{p} cos(\omega t) - m\omega \hat{x} sin(\omega t) \end{equation}

which is just classical mechanics in disguise. This doesn't help with finding a solution to Schrödinger's equation one bit.

Edit: what I mean by stronger, or more general, here is not rigorously mathematical, though it's something you see in practice: solving Heisenberg's equation, in practice, won't in general help you find a particular solution to Schrödinger's equation. I guess someone might come here and prove by mathematical means that the two pictures are just as strong, but these proofs of equivalence always beg the question; implicitly, the proofs always use the general solution to Schrödinger's equation (i.e. it uses $e^{-i\hat{H}t/\hbar}$) to show the pictures are equivalent. I.e., in order to use Heisenberg to get Schrodinger, you already need the solution to Schrodinger's equation.


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