Graduate Can Work Be Defined for a Quantum System with a Time-Dependent Hamiltonian?

[Prathyush]

I have a Hamiltonian $H_{\lambda(t)}$, where $\lambda (t)$ characterizes a time dependent path in parameter space. The parameter is changed in finite time from $\lambda (t_i)$ to $\lambda(t_f)$ . At $t = t_i$ the system is in the intial state $|\Psi>$. What is the work done on the system? Is it well defined?

 

[thephystudent]

In first instance, just the difference between the energy before and after the time-evolution as eg. in this paper ,
further subtleties arise when you want to consider dissipation ('Heat') as well, in this case you will need to make kind of a subjective distinction between which energy is considered useful or tractable, and which not...

 

[Prathyush]

@thephystudent Very interesting paper.

I was studying this question in the context of understanding Jarjinski's equality for quantum systems. From my preliminary scan of the literature, I have not see an very a compelling argument for how the work must be defined yet.

After a quick read of the paper you linked, He uses expectation value of the Energy as you said. The protocol described uses individual pulses. I don't think it is correct to think about work done as the paper describes atleast in the context that I want to think about. Each bragg pulse can be though of as doing work on the system individually. So one does not need more than one bragg pulse to discuss this question. After the bragg pulse passes he assumes the system undergoes decoherence, by neglecting the diagonal terms. I think this assumption can be more carefully examined to understand this question better.

It also relates to the question, does measurement of work cause decoherence in the system. I think it is intuitive to expect this. So it is natural to ask the question how does one measure the work? Does measurement of work decohere the system?

 

[Prathyush]

@vanhees71 @A. Neumaier Please can you look into this question? It appears to me that one if one does a measurement of work one inevitably decoheres the system. So if one does a continuous work measurement the evolution will be not be unitary. However if one only changes the parameter without actually measuring the work i.e ignoring the back reaction on the apparatus that is used to do the work, the evolution can be described by a unitary evolution, but in that case one cannot define the work done.

 

[A. Neumaier]

@vanhees71 @A. Neumaier Please can you look into this question? It appears to me that one if one does a measurement of work one inevitably decoheres the system. So if one does a continuous work measurement the evolution will be not be unitary. However if one only changes the parameter without actually measuring the work i.e ignoring the back reaction on the apparatus that is used to do the work, the evolution can be described by a unitary evolution, but in that case one cannot define the work done.

In the text of #1, the dynamics is unitary, and decoherence or measurement don't figure. Work is undefined in this generality.

I haven't looked at the paper since I am on holiday.

 

[Prathyush]

In the text of #1, the dynamics is unitary, and decoherence or measurement don't figure. Work is undefined in this generality.

Exactly what I was thinking work done should not be well defined quantity if one does unitary evolution under a fixed time varying external field.

The standard protocol to define work is the Two time measurement protocol. Where a measurement of energy of system is done before and after the process. This is a not satisfactory method because it destroys interference.

I am trying to see if one can construction a system where I can both measure the work done on the system while allowing it to evolve under an external field(although not a precisely as external field). This can be done by coupling the system to a external probe particle. The initial wavefunction of the probe can be made in such a way that it produces a response nearly identical to an external field. However the back reaction on the probe cannot be neglected. At some later time using some suitable method the back reaction of the probe i.e the change in the energy of the probe can be measured to give us the work done. The back reaction and eventual measurement cause a decoherence of the system.

 

[thephystudent]

Hi, actually the paper I referred to was my Master's thesis, but I haven't touched the subject afterwards so I don't remember all details (though I may be a bit biased because I see everything through this lense ;) ). I believe the difference between work and heat is that work is reversible ; unitary dynamics are always reversible so there is no heat.

The dephasing between the Bragg pulses is not unitary, I believe it can be explicitly written in Lindblad form and generates heat. I believe this Point of view is the same as (among others) the papers of Allahverdyan and Van Nieuwenhuizen cited, but I don't know how far it applies to your work.

 

[Prathyush]

@thephystudent Why do you say the dephasing between bragg pulses is not unitary? In the paper they use a well defined Hamiltonian evolution to describe bragg pulses. That is a unitary evolution.

EDIT: I am sorry I misunderstood what you wrote, Indeed the dephasing after the interaction is a nonunitary evolution. I did not know it can be associated with a heat transfer. That would be very interesting to see explicitly.

I think arnold is correct when he says you cannot define work done for a unitary evolution, without introducing measurements somewhere. One has to use the back reaction effect on the probe to define work a few modern papers seem to agree with this viewpoint.

 

[vanhees71]

I've only looked cross reading on the above cited paper by Verstraelen et al. The key point is that you deal with an explicitly time-dependent Hamiltonian and thus energy is not conserved here. That's also the case in classical mechanics: If you have forces derived from a time-dependent potential, you can still use the Hamilton principle but total energy is not conserved.

This is easily seen in the quantum case using the fundamental rule that the time-derivative of an observable (represented by a self-adjoint operator $\hat{O}$) is represented by the operator
$$\mathring{\hat{O}}=\frac{1}{\mathrm{i} \hbar} [\hat{O},\hat{H}]+\partial_t \hat{O},$$
where the partial time derivative refers to the explicit time-dependence of $\hat{O}$ only (i.e., the only time-dependence at all in the Schrödinger picture of time evolution). Now set $\hat{O}=\hat{H}$. Then you have
$$\mathring{\hat{H}}=\partial_t \hat{H},$$
i.e., energy is conserved if and only if the Hamiltonian is not explicitly time-dependent.

Of course the time evolution is still unitary. In the Schrödinger picture the states evolve in time with the unitary operator $\hat{U}(t,t_0)$ definied by the "operator Schrödinger equation",
$$\mathrm{i} \hbar \partial_t \hat{U}(t,t_0)=\hat{H}(t) \hat{U}(t,t_0).$$
This time-evolution operator now is a function of $t$ and $t_0$ separately, not only on $(t-t_0)$ as in the case of a time-independent Hamiltonian. The formal solution of the operator Schrödinger equation is given, using the time-ordering symbol, by
$$\hat{U}(t,t')=\mathcal{T}_c \exp \left [-\frac{\mathrm{i}}{\hbar} \int_{t_0}^t \mathrm{d} t' \hat{H}(t') \right].$$
The timeordering symbol tells you that in the expansion of the operator exponential you have to time order the Hamiltonians in the occurring multiple integrals always such the the time arguments are ordered as increasing from right to left.

Now the paper deals with the interesting situation that you of course also do not have the usual equilibrium situation since the canonical (or grand canonical) stat. op. $\exp[-\hat{H}/(k_B T)]$ is not stationary since $\hat{H}$ depends explicitly ontime, but that you still can define a generalized "Gibbs ensemble" by using the maximum-entropy method under appropriate constraints.

 

[Prathyush]

@vanhees71 Yes. I agree, but you did not address the issue about work done.
Perhaps you can look at the following papers available on arxiv

No-go theorem for the characterisation of work fluctuations in coherent quantum systems
Fluctuation theorems: Work is not an observable

 

[Prathyush]

In the text of #1, the dynamics is unitary, and decoherence or measurement don't figure. Work is undefined in this generality.

@A. Neumaier

In regard to what you said, under what circumstances can the concept of work be well defined? Because classically we have good definition for work.

I am trying to explore this question about work done on quantum system where do you think I can begin?

Thank you for your time.

 

[A. Neumaier]

In regard to what you said, under what circumstances can the concept of work be well defined? Because classically we have good definition for work.

I am trying to explore this question about work done on quantum system where do you think I can begin?

Your context is such that it is difficult to say anything constructive. To be able to talk about work one needs a (mechanica, electrical, thermodynamic, etc.) force acting on the system, typically created through a gradient of some potential of the respective kind. A change of parameter in a Hamiltonian is nothing of this sort. So there is nothing I can say.