# I What is a time-dependent operator?

Tags:
1. Jul 12, 2016

### weezy

While studying Ehrenfest's theorem I came across this formula for time-derivatives of expectation values. What I can't understand is why is position/momentum operator time-independent? What does it mean to be a time-dependent operator? Since position/momentum of a particle may change continuously I believe they should be time dependent. I know that I'm wrong with my reasoning. Where am I confusing?

#### Attached Files:

• ###### Screen Shot 2016-07-12 at 3.56.37 PM.png
File size:
3.6 KB
Views:
52
2. Jul 12, 2016

### vanhees71

The fundamental operators (for the usual QM 1 systems these are $x$ and $p$) are time-independent by definition (as in classical Hamiltonian mechanics; they become time dependent only through the Hamilton equations of motion if you want to determine the actual trajectory of the system in phase space). The partial time derivative is acting only on the explicit time dependence of an (operator valued) function $A(x,p,t)$. If $A$ is not explicitly time dependent this partial time derivative doesn't occur.

3. Jul 12, 2016

### Jilang

The operator is not time dependent but what it acts on could be.

4. Jul 12, 2016

### dextercioby

Time dependence means dynamics. You have three ways in QM (called pictures) to encode dynamics: in the states (Schrödinger), in the observables (Heisenberg) or in both (Dirac - Tomonaga - Schwinger).

5. Jul 13, 2016

### weezy

Especially regarding Heisenberg's & Schrodinger's picture: All operators are time dependent and states are time independent. The opposite is true for Schrodinger's picture. But how are these two equivalent?

6. Jul 13, 2016

### vanhees71

There's a time-dependent unitary transformation between the two. Take an example where the Hamiltonian is not explicitly time dependent.

Schrödinger picture:
The state ket fulfills the EoM
$$\mathrm{i} \frac{\mathrm{d}}{\mathrm{d} t}|\psi_S(t) \rangle = \hat{H} |\psi_S(t) \rangle.$$
This is solved by the time-evolution operator
$$|\psi_S(t) \rangle=\exp(-\mathrm{i} \hat{H} t) |\psi_S(0) \rangle=\hat{U}_S(t) |\psi_S(0) \rangle.$$
The fundamental operators (say $x$ and $p$ for a particle moving along a line as the most simple example) are time-independent
$$\hat{x}_S(t)=\hat{x}_S(0), \quad \hat{p}_S(t)=\hat{p}_S(0).$$
Thus also the corresponding (generalized) eigenvectors are time-independent in the Schrödinger picture:
$$|u_{Sx}(t) \rangle = |u_{Sx}(0) \rangle, \quad u_{Sp}(t) \rangle=|u_{Sp}(0) \rangle.$$
The picture independent wave function is in this case given by
$$\psi(t,x)=\langle u_{Sx}(t)|\psi_S(t) \rangle=\langle u_{Sx}(0)|\hat{U}(t) \psi_S(0) \rangle.$$

Heisenberg picture:

The state kets are time independent
$$|\psi_H(t) \rangle=|\psi_H(0) \rangle.$$

The fundamental operators get a time dependence by the full Hamiltonian via the EoM
$$\dot{\hat{x}_H}=\frac{1}{\mathrm{i}} [\hat{x}_H,\hat{H}] \; \Rightarrow \; \hat{x}_H(t)=\exp(\mathrm{i} \hat{H} t) \hat{x}_H(0) \exp(-\mathrm{i} \hat{H} t) = \hat{U}^{-1}(t) \hat{x}_H(0) \hat{U}(t).$$
The same holds for $\hat{p}$. The eigenvectors thus fulfill
$$|u_{Hx}(t) \rangle = \hat{U}^{-1}(t) |u_{Hx}(0) \rangle.$$
Assuming that the two pictures of time evolution agree at the initial time, you get
$$|\psi_H(t) \rangle=|\psi_H(0) \rangle=|\psi_S(0) \rangle=\hat{U}^{-1} |\psi_S(t) \rangle.$$
This implies that the wave function is picture independent, as it must be, since
$$\psi(t,x)=\langle u_{Hx}(t) | \psi_H(t) \rangle = \langle \hat{U}^{-1}(t) u_{Sx}(t)|\hat{U}^{-1}(t) \psi_S(t) \rangle = \langle u_{Sx}(t)|\psi_S(t) \rangle.$$
This is so, because you switch all states and operators from one picture to the other with the unitary transformation $\hat{U}(t)$. This means that all the physically relevant outcome of any calculation cannot depend on in which picture of time evolution it is calculated.

All this formalism can be generalized to transformations between general pictures (Dirac pictures), where the time dependence of state kets and fundamental operators is quite arbitrarily chosen by splitting the Hamiltonian in two additive parts with the one part governing the time evolution of the state kets and the other the time evolution of the fundamental operators. This splitting is arbitrary and chosing different splittings leads to unitary equivalent time evolutions such that the final outcomes for physical quantities are still picture independent.

This holds true only in non-relativistic quantum theory in the 1st-quantization formalism. In relativistic QFTs it's wrong, as is proven by Haag's theorem, and a lot of trouble (which is hitherto still unsolved for realistic models like the Standard Model of elementary particle physics) with relativistic QFT originates from this.

7. Jul 13, 2016

### kith

As an application of what vanhees71 wrote, have a look at how the time-dependency can be shifted for the expectation value of an observable $A$:
$$\begin{eqnarray*} \langle A \rangle(t) &=& \langle \psi_S(t) | A_S | \psi_S(t) \rangle \\ &=& \langle \psi_S(0) | U^\dagger(t) A_S U(t)| \psi_S(0) \rangle \\ &=& \langle \psi_H | A_H(t) | \psi_H \rangle \end{eqnarray*}$$

8. Jul 13, 2016

### weezy

Thanks. I'd also like to know if the time-derivative can be regarded as a time dependent operator or not?

9. Jul 13, 2016

### kith

You may be interested in this recent thread which is about $\frac{d}{d t}$ in the context of relativistic QM. Generally, time is a parameter and not an observable in the sense of QM.

10. Jul 14, 2016

### weezy

I couldn't follow most of the discussion as I haven't studied so far. Could you tell me if the energy operator $E=i \hbar \frac{\partial }{\partial t}$ is time dependent or not? As $\frac{dE}{dt}$ is not zero?

11. Jul 14, 2016

### PeroK

I think you're missing the fundamental point. The operator itself may be time-dependent or not; but this is not the same as the observables being time dependent.

Take, as an example, the potential $V$. If $V$ is time-independent, then it's the same operator at any time. E.g.:

$V(x) = \frac{1}{2}m \omega x^2$

Now, if the expected value of $x$ changes with time, then the expected value of $V(x)$ changes with time. The dynamic quanity is time-dependent, but the operator is not.

A simple example of a time-dependent operator would be:

$V(x) = 0$ when $(t < 0)$
$V(x) = \frac{1}{2}m \omega x^2$ when $(0 < t < T)$
$V(x) = 0$ when $(t > T)$

Now, the operator itself is time-dependent: the potential is only switched on for a finite period $(0, T)$.

Or, you could have:

$V(x, t) = \frac{1}{2}m \omega x^2 e^{-t}$

Where the potential is reducing exponentially over time.

12. Jul 14, 2016

### vanhees71

This is wrong! $\mathrm{i} \hbar \partial_t$ is \textbf{not} the energy operator. The Hamiltonian represents the energy, and it is a function of the fundamental operators and perhaps explicitly of time.

I think, it's important for you do first get a good grasp of classical mechanics in the Hamiltonian formulation with Poisson brackets. Otherwise, it's very hard to make sense out of quantum theory.

13. Jul 14, 2016

### weezy

Isn't $V(x) = \frac{1}{2}m \omega x^2$ always time dependent? A free particle is always under the influence of the potential hence constantly in motion?

14. Jul 14, 2016

### PeroK

It's nothing to do with the motion of any particle. It's the definition of the operator. The definition of $V$ does not depend on time.

15. Jul 14, 2016

### weezy

No I was worried about the $x^2$ term. Since x varies with time so I inferred that V must vary with time too

16. Jul 14, 2016

### PeroK

If you drew a graph of $V(x)$ then that graph would be time-independent: it's the same function, hence the same operator, at any time.

But, if you tried to draw the graph of $V(x, t) = \frac{1}{2}m \omega x^2 e^{-t}$, then you'd have to draw a different graph for each time $t$. The potential itself (regardless of what any particle is doing) is changing with time. That is what is meant by a time-dependent potential.

You should look again at the two examples of time-dependent potentials in post #11.

17. Jul 14, 2016

### weezy

Okay so even though x is a function of t, V isn't.. I guess this is what is meant by 'explicit time dependence' isn't it?

18. Jul 14, 2016

### PeroK

Yes. The potential itself isn't changing but the expected value of a particle's potential energy is likely to be time-dependent.

19. Jul 15, 2016

### vanhees71

Again, I can only recommend to learn classical Hamiltonian mechanics first.

Your potential is NOT explicitly time dependent since time doesn't occur explicitly. The Hamiltonian (and all other operators describing observables) are functions of the fundamental observables (in your case these are $x$ and $p$ building the Heisenberg algebra of observables for a spinless non-relativistic particle) and perhaps explicitly on time. Only in the latter case you have an explicit time dependence.

In classical mechanics the Hamilton canonical equations of motion define trajectories in phase space, and the solutions are described as functions $x(t)$ and $p(t)$, but this time dependence is not what's meant when writing down a phase-space variable as function of $x$ and $p$ like $H(x,p,t)$, where the three variables occuring as arguments are taken as independent variables. Of course, it can make sense to look at an observable along the trajectory of the particle in phase space, i.e., $H[x(t),p(t),t]$, but that's only relevant in such specific cases, and then you make it explicit by mentioning that you mean to look at the observable along the particle's trajectory.

In quantum mechanics the time dependence of operators that represent observables and states (represented either by state vectors in the case of pure states or more generally the statistical operator) is pretty arbitrary. That's called the "choice of the picture of time evolution", and this choice is just for convenience to solve a given problem. The outcome of the physical observables (probabilities, expectation values, correlation functions,...) are of course independent of this choice.