I have been reading up on the Schrödinger picture of quantum mechanics. In this picture it is often said (e.g. here) that the operators are time-independent unless they depend explicitly on time. I cannot see what this is telling us since it seems to be saying that operators are time-independent unless they are not. Is this not exactly the same for the Heisenberg picture which is said to have time-dependent operators? i.e. what does the above saying actually mean and how does it distinguish the time-dependence of operators in the Schrödinger and Heisenberg pictures?

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    $\begingroup$ In Schrodinger picture, you can have a time-varying Hamiltonian $H(t)$ if, say, you flip an external field on and off. $\endgroup$
    – knzhou
    Sep 17, 2016 at 6:56

3 Answers 3


The definition of constant operator (please, see the disclaimer at the end of this post) differs in the two pictures. Specifically (here $U(t,t_0)$ is the time evolution operator):

Schrödinger picture: an operator $A_S (t)$ is constant if $$A_S(t)= A_S(t_0) \qquad $$ for all $t$.

Heisenberg picture: an operator $A_H (t)$ is constant if: $$U(t,t_0)A_H(t)U^{\dagger}(t,t_0)=A_H (t_0)$$ for all $t$.

Since the two pictures are connected by: $$A_H (t)=U^{\dagger}(t,t_0)A_S(t)U(t,t_0),$$ the constancy of an operator can be checked in either picture.

The point is that a non-constant operator is an operator whose definition is changing with time.

Consider for example a non-relativistic free particle. The Schrodinger operator: $$x (t) = x + \frac{p}{m}t$$ is an operator whose definition is changing with time. Note that I have declared that $x(t)$ is a Schrodinger operator or, more precisely, that the time dependence in $x(t)$ is that of the Schrodinger picture. If, on the other hand, I had told you that $x(t)$ was a Heisenberg operator, then it would have been a constant operator, namely the constant operator: $$x_H(t)=e^{iHt}xe^{-iHt}.$$

When one introduces an operator valued function $t\mapsto A(t)$, it must always be clear to what dynamical scheme he is referring to.

Disclaimer. I've just noticed that my above definition of “constant” operator (which, to be clear, is just my personal jargon) may cause some confusion with what is usually called a conserved operator. A conserved operator is by definition one whose matrix elements $$\langle \alpha (t)\vert A(t)\vert \beta (t)\rangle$$ are constant, where the kets and the operators are referred to any scheme. This translates, respectively, to the requirement: $$U(t,t_0)A_S (t)U(t,t_0)=A_S(t_0)$$ in the Schrodinger scheme and: $$A_H (t)=A_H (t_0)\,\,(\,=A_S(t_0)!)$$in the Heisenberg scheme.

For example, the Heisenberg operator: $$\mathbf K (t) = \dfrac{\intop\text d ^3 \mathbf x \mathcal H (\mathbf x ,t ) \mathbf x }{\intop\text d ^3 \mathbf x \mathcal H (\mathbf x ,t )}-\mathbf P t,$$ where $\mathcal H $ and $\mathbf P$ are the hamiltonian density and momentum of the free Klein-Gordon field, is a conserved operator but not a constant operator.

  • $\begingroup$ So..., the Schrödinger operator $x(t)$ is not an observable ? $\endgroup$
    – Our
    Jan 26, 2019 at 8:55
  • $\begingroup$ @onurcanbektas why would you say that? $\endgroup$
    – pppqqq
    Jan 26, 2019 at 12:51
  • $\begingroup$ Because in Schrödinger's picture, observables are time-independent. In particular, the position and momentum operators, see physics.stackexchange.com/questions/456724/… $\endgroup$
    – Our
    Jan 26, 2019 at 12:52
  • $\begingroup$ @onurcanbektas here I was not addressing the question whether an operator is an observable or not. Usually, in QM, hermitian operators are assumed to represent observable quantities (apart from density matrices, which are hermitian but used to characterize the state of a system, instead). Anyway, in the Schrodinger picture the $x$ operator would be time-independent, $x(t)=x(t_0)$, and yes it represent an observable quantity. $\endgroup$
    – pppqqq
    Jan 26, 2019 at 12:57
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    $\begingroup$ @onurcanbektas OK, there is some confusion here, let me restate my point. If a Schrodinger operator $\xi_S(t)$ is defined as $\xi _S (t)\equiv x+pt/m$, it has a different physical intepretation from the Heisenberg position operator $x_H(t)$, even though they share the same mathematical expression as linear operators on the Hilbert space: $\xi_S(t)=x_H(t)$. On the other hand, if we define $x_S(t)=x$, then $x_S(t)$ and $x_H(t)$ have different mathematical expressions, but the same physical interpretation. In any case, I'm not implying that $\xi _S$ is not a physical observable. $\endgroup$
    – pppqqq
    Jan 26, 2019 at 13:46

the operators are time-independent unless they depend explicitly on time. I cannot see what this is telling us...

The confusion is due to unfortunate terminology. There is a difference between "explicitly does not depend on time" and "time-independent".

Operator does not explicitly depend on time:

It is assumed that every operator is, in general, a function of coordinate and momenta operators $x_i$ and $p_i$ and time parameter $t$. If the operator is special in the sense it can be expressed, at all times, as value of the same function of coordinates and momenta, it is said that it does not explicitly depend on time. For example, the operator given by the function $U(x,p,t)=kx^2/2$ does not explicitly depend on time but the operator given by the function $U_1(x,p,t)= kx^2/2-qEx\cos\Omega t$ does.

Operator is time-independent:

Operator $\hat{A}_t$ is time-independent when it maps any vector $\psi_0$ to the same image counterpart vector $\psi_0'$ at all times:

$$ \forall t: \hat{A}_t \psi_0 = \psi_0' $$

An operator that does not explicitly depend on time, such as $k\hat{x}^2/2$, can be time-independent, but not necessarily.

If $\hat{x},\hat{p}$ are time-independent, as in the Schroedinger picture, $k\hat{x}^2/2$ is time-independent as well.

However, if $\hat{x}$ or $\hat{p}$ is time-dependent, as in the Heisenberg picture, $k\hat{x}^2/2$ may be time-dependent, even if it does not depend explicitly on time.

  • $\begingroup$ I thought $\hat x, \hat p$ are assumed to be time-independent operators, see physics.stackexchange.com/questions/456724/… $\endgroup$
    – Our
    Jan 26, 2019 at 9:01
  • $\begingroup$ Those symbols do not have universal meaning. It depends on the "picture" - choice of formalism. In "Schroedinger picture" they are time-independent. In "Heisenberg picture", they evolve in time. $\endgroup$ Jan 26, 2019 at 9:31
  • $\begingroup$ I implicitly assumed that when you write $U(x,p,t)=kx^2/2$, $x$ was corresponding to the "normal" form i.e from Schrödinger's point of view, of the operator. $\endgroup$
    – Our
    Jan 26, 2019 at 9:33
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    $\begingroup$ It may help to note that dependence of those symbols on time, explicit or not, is a general mathematical property, it is good to understand the difference without mixing in quantum theory specifics such as the choice of Schroedinger or Heisenberg picture. $\endgroup$ Jan 26, 2019 at 9:41
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    $\begingroup$ What Heisenberg picture does is to strip $\psi$ of time dependence and put it into operators. The operator may or may not have been time-dependent before the change of picture. $\endgroup$ Jan 26, 2019 at 11:10

Time dependence comes in many ways. Not just explicit time dependence. What Schrödinger picture says is that the state of a system evolves with time but the operators which are used to measure the position for example, they are fixed. So if you measure the position at two different times they have different probability density because the state has evolved.

On the other hand Heisenberg picture says that a unique system is determined by a unique state vector. But if you now want to measure the position in two different times you'll need two different operators.


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