Heisenberg equation of motion -- Partial derivative question

[LagrangeEuler]

Heisenberg equation of motion for operators are given by
i\hbar\frac{d\hat{A}}{dt}=i\hbar\frac{\partial \hat{A}}{\partial t}+[\hat{A},\hat{H}].
Almost always $\frac{\partial \hat{A}}{\partial t}=0$. When that is not the case?

 

[vanhees71]

The logic is usually as follows: You have some fundamental operators that generate the algebra of observables. E.g., in non-relativistic QM for a single particle without spin you have $\hat{\vec{x}}$ and $\hat{\vec{p}}$, from which you can build all the operators describing other observables and particularly also the Hamiltonian. By definition $\hat{\vec{x}}$ and $\hat{\vec{p}}$ are not explicitly time-dependent, i.e., in the Schrödinger picture they don't depend on time at all. In the Heisenberg picture they "move" with the full Hamiltonian, i.e., fulfilling
$$\mathrm{i} \hbar \mathrm{d}_t \hat{\vec{x}}(t)=[\hat{\vec{x}},\hat{H}], \quad \mathrm{i} \hbar \mathrm{d}_t \hat{\vec{p}}(t)=[\hat{\vec{p}},\hat{H}].$$
Now any other observable is a function of $\hat{\vec{x}}$, $\hat{\vec{p}}$ and (maybe) also explicitly of time: $\hat{A}(\hat{\vec{x}},\hat{\vec{p}},t)$. Then in the Heisenberg picture you have a time-dependence due to the time-dependence of $\hat{\vec{x}}$ and $\hat{\vec{p}}$ as well as the explicit time dependence. In your formula the commutator takes care of the former and the partial time derivative of the latter time dependence of $\hat{A}$. Note that in the Schrödinger picture in this case $\hat{A}$ has only the explicit time dependence and thus
$$\mathrm{d}_t \hat{A}_{\text{S}}=\partial_t \hat{A}_{\text{S}}.$$

 

[hilbert2]

Heisenberg equation of motion for operators are given by
i\hbar\frac{d\hat{A}}{dt}=i\hbar\frac{\partial \hat{A}}{\partial t}+[\hat{A},\hat{H}].
Almost always $\frac{\partial \hat{A}}{\partial t}=0$. When that is not the case?

If an explicit time dependence is needed in the Hamiltonian, it means that the effect of something is not modeled properly by including the particles causing it, but in an "effective" way as a time-varying potential energy function. One example is the Rabi oscillation model of the interaction between atoms and electromagnetic waves. It would be much more difficult to calculate the equivalent results if treating the electromagnetic field as an actual quantum system along with the atom or molecule.

 

[samalkhaiat]

Almost always $\frac{\partial \hat{A}}{\partial t}=0$. When that is not the case?

In relativistic QFT, we have an opposite example. That is, there is a very important conserved operator (i.e., \frac{dA}{dt} = 0) which depends explicitly on time (i.e., \frac{\partial A}{\partial t} \neq 0). Do you recognise that operator? Hint: no relativistic QFT can be studied without it.

 

[Demystifier]

In relativistic QFT, we have an opposite example. That is, there is a very important conserved operator (i.e., \frac{dA}{dt} = 0) which depends explicitly on time (i.e., \frac{\partial A}{\partial t} \neq 0). Do you recognise that operator? Hint: no relativistic QFT can be studied without it.

Further hint. The angular momentum ${\bf L}={\bf r}\times{\bf p}$ is not Lorentz covariant.

 

[LagrangeEuler]

Just to try to understand. If Hamiltonian is time-independent then there is no explicit time dependence for operators in the Hamiltonian. Is that the correct statement?

 

[vanhees71]

Yes. If the Hamiltonian is not explicitly time dependent it's time-independent in both the Schrödinger and the Heisenberg picture.

Of course $\vec{L}=\vec{r} \times \vec{p}=-\vec{p} \times \vec{r}$!

 

[LagrangeEuler]

Thanks. For me, Heisenberg picture is nonintuitive. Commutator of the operator of the coordinate
[x_H(t),x_H(t')] \propto \sin\omega(t'-t)
I can not understand where can I used this. And why this is important? What I can get from it?

 

[Demystifier]

Of course $\vec{L}=\vec{r} \times \vec{p}=-\vec{p} \times \vec{r}$!

Thanks for the sign correction (I've edited it now), I never find the orientation of pseudovectors intuitive.

 

[vanhees71]

It's just convention.

 

[vanhees71]

Thanks. For me, Heisenberg picture is nonintuitive. Commutator of the operator of the coordinate
[x_H(t),x_H(t')] \propto \sin\omega(t'-t)
I can not understand where can I used this. And why this is important? What I can get from it?

Where did you get this formula from, and what's $\omega$?

For a free particle you have (check it, that's a very intuitive example :-)):
$$\hat{x}(t)=\hat{x}_0 + \frac{1}{m} \hat{p}_0 t, \quad \hat{p}(t)=\hat{p}_0.$$
This gives
$$[\hat{x}(t),\hat{x}(t')]=[\hat{x}_0+\frac{1}{m} \hat{p}_0 t, \hat{x}_0+\frac{1}{m} \hat{p}_0 t']=\frac{\mathrm{i} \hbar}{m} (t'-t).$$

 

[hilbert2]

Where did you get this formula from, and what's $\omega$?

I think it's the solution for the harmonic oscillator, as in this document: https://ocw.mit.edu/courses/physics...all-2013/lecture-notes/MIT8_05F13_Chap_06.pdf .

 

[vanhees71]

Sure, that makes sense. It's one of the few problems which can be immediately solved as in classical mechanics, because the EoM is linear (the same holds of course for the free particle and a particle moving under influence of a constastant force with $V(\hat{x})=-F \hat{x}$, with $F=\text{const}$).

 

[hilbert2]

Sure, that makes sense. It's one of the few problems which can be immediately solved as in classical mechanics, because the EoM is linear (the same holds of course for the free particle and a particle moving under influence of a constastant force with $V(\hat{x})=-F \hat{x}$, with $F=\text{const}$).

That made me try to find information on whether the quantum counterparts of other isochronous classical systems (time-periodic with energy-independent period) also have a classical-like Heisenberg picture time evolution, but didn't find the answer. That kind of systems sometimes can in some cases have equally spaced energy eigenvalues when quantized, just like the simple harmonic oscillator.

https://iopscience.iop.org/article/10.1088/1742-6596/87/1/012007/pdf

Edit: Looking at Fig. 1 of this paper, it seems that the classical trajectory of an isochronous system doesn't necessarily coincide with the time evolution of expectation value $\langle x\rangle$ in the equivalent quantum version (but isn't completely different either): https://www.sciencedirect.com/science/article/abs/pii/037596017990197X

 

[LagrangeEuler]

Yes, it is for LHO. But I am not sure what to do with this. As in your example. When I calculate this what to do with that result?

 

[hilbert2]

Yes, it is for LHO. But I am not sure what to do with this. As in your example. When I calculate this what to do with that result?

It just shows how similar QM is with classical mechanics in the special case of LHO. And most likely you can also use it to calculate some practical results about a system that can be described with the model of the harmonic oscillator, e.g. a vibrating chemical bond between atoms.