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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 initial state $|\psi\rangle$. What is the work done on the system? Is it well-defined?

Also posted at https://www.physicsforums.com/threads/work-done-on-a-quantum-state.976256/

Edit: Adding more details for clarity.

The question has been inspired by the attempt to understand Jarzynski's equality for quantum systems. There is a lot of controversy about the definition of work done in literature.

Here are 2 papers which highlight this observation. No-go theorem for the characterisation of work fluctuations in coherent quantum systems Fluctuation theorems: Work is not an observable

Arnold Neumaier stated on physics forums that

"the dynamics is unitary, and decoherence or measurement don't figure. Work is undefined in this generality.".

I agree with this viewpoint.

It is my understanding that the back-reaction on the probe that creates the driving field is important to understand the definition of work in a general setting.

My goal is to define work in a way that can be motivated from experimental considerations. It seems clear to me that the 2 point measurement scheme where energy is measured before and after the evolution does not take into account coherence of the state into account in defining work done. There are other proposals, and none(to the extent I have evaluated them) seemed satisfactory.

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    $\begingroup$ Sorry for the d/v but your rep would imply that you are aware that you need to show your own thoughts and work. $\endgroup$ – StudyStudy Aug 19 at 19:10
  • $\begingroup$ @StudyStudy The question stands independent of my own research as it is a well posed question. However since you asked, one protocol that is used is as follows "An established protocol to measure work involves a double projective measurement of the energy of the system at the beginning and at the end of the evolution. Such a two-measurement protocol (TMP) can be described in terms of classical conditional probabilities" ref:arxiv.org/abs/1504.01574 . However there is a doubt about whether it can be used in this context. $\endgroup$ – Prathyush Aug 19 at 19:22
  • $\begingroup$ Why can't two projective measurement be used to define work in this context? Are you worried about initial coherences? $\endgroup$ – Sunyam Aug 20 at 15:10
  • $\begingroup$ @Sunyam Yes i am worried about initial coherences. If you look at the discussions physics forums Arnold says if evolution is unitary then work cannot be defined. I think the correct answer to this question is that work cannot be defined in a context independent way, and one has to consider the back reaction on the object that is responsible for the change in the Hamiltonian. Its an actively debated topic in literature. I may eventually write an answer based on my literature survey eventually. $\endgroup$ – Prathyush Aug 20 at 16:40
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    $\begingroup$ Your bounty description expands on the question a lot. Without it, it would be unclear what you are asking. I recommend editing your question and adding this additional information. (Then, it would be a quite interesting question imo.) $\endgroup$ – Noiralef Aug 24 at 10:30
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What is the work done on the system? Is it well-defined?

We can choose a definition that behaves the way we want it to in certain situations, but whatever definition we choose, it won't be perfect (Examples 1-4). Accounting for back-reaction doesn't fix this. Instead, it highlights a deeper ambiguity (Example 5).

That's okay, because the concept(s) of "work" is not one of the foundational concepts in our current understanding of nature. Rather than trying to perfect the definition of "work," we can do things the other way around. For any experiment that we would traditionally describe in terms of "work," we can try to describe the experiment in more fundamental terms instead.

The rest of this answer consists of examples illustrating why I think an unambiguous unified definition of "work" is neither feasible nor necessary. Contents:

  • Example 1: Charged particle in a uniform electric field

  • Example 2: A contrived superposition

  • Example 3: Statistical mechanics

  • Example 4: Jarzynski's equality (quantum version)

  • Example 5: Accounting for back-reaction

Example 1: Charged particle in a uniform electric field

This is a special case of the general setup described in the OP. In this example, we would normally say that work is being done, even though $H(t)$ is independent of $t$.

Consider a charged particle in an external potential whose gradient represents an external electric field. In the usual representation, the Hamiltonian is $$ H = \frac{-(\hbar \nabla)^2}{2m}+V(x), $$ using one-dimensional space for simplicity. Take the electric field (proportional to $\nabla V$) to be uniform in some large region of space, and take the particle's initial state to be a gaussian wavepacket with zero net momentum, localized somewhere near the middle of that region.

We already know what happens: the particle accelerates. More accurately, the expectation value of the position operator accelerates. By analogy with classical mechanics, we would say that the electric field is doing "work" on the particle.

In this case, "work" refers to a change in (the expectation value of) part of the system's energy, namely the particle's kinetic energy. The keyword is part, a theme that will be revisited in Example 5.

Example 2: A contrived superposition

Consider the single-charged-particle model again, with a Hamiltonian of the form shown above. This time, take the field to be uniform in two large regions of space, $R$ and $L$, with the electric field vector directed to the right in $R$ and to the left in $L$. Take the initial state to be a superposition of two right-moving gaussian wavepackets, one in the middle of $R$ and one in the middle of $L$.

Again, we already know what happens in this situation: the wavepacket in one region accelerates, and the wavepacket in the other region decelerates. But we have just one particle, whose state is a quantum superposition of these two wavepackets, one accelerating and one decelerating, and we can choose the magnitudes of the field in each region so that the expectation value of the particle's momentum doesn't change at all.

How should we quantify the "work" done on the particle in this case? Is "work" even a useful concept here?

Example 3: Statistical mechanics

Consider the first law of thermodynamics: $T\,dS=dE+p\,dV$. We typically describe the overall change in the system's energy (the $dE$ term) as being partly due to "work" (the $p\,dV$ term) and partly due to heat transfer (the $T\,dS$ term).

In statistical mechanics, we interpret $S(E,V)$ as the log of the number of microstates compatible with the given constraints on the total energy $E$ and the total volume $V$, and the first law is simply the definitions of $T$ and $p$.

In a system with an enormous number of particles, most of the microstates compatible with the volume constraint $V$ are states in which the particles are distributed throughout the whole available volume. States in which a significant part of the available volume is empty hardly affect the number $S(E,V)$ at all.

But in the opposite extreme of a system with just one particle localized somewhere inside a volume of astronomical proportions, as far as laboratory-scale experiments with that particle are concerned, counting energy eigenstates is no longer a useful thing to do, and defining "work" in terms of changes in that volume is likewise practically useless. The definition of "work" used in Example 1 is more useful in this extreme, at least if we exclude situations like the one in Example 2.

The message here is that different definitions of work are useful under different situations.

Example 4: Jarzynski's equality (quantum version)

For the purpose of highlighting how it defines "work," here's a quick review of the quantum version of Jarzynski's equality. Consider a system with time-dependent Hamiltonian $H(t)$. Use the Schrödinger picture, and let $\rho(t)$ denote the density matrix representing the state at time $t$. Then $$ \rho(t) =u(t)\rho(0) u^\dagger(t) \tag{1} $$ with $$ \frac{d}{dt}u(t)=-iH(t)u(t) \hskip2cm u(0)=1. \tag{2} $$ (I'm using a lowercase $u$ here for consistency with Example 5.) For each $t$, let $\mathcal{E}(t)$ denote a complete set of orthonormal eigenstates of $H(t)$, and define $$ E(t,k) := \langle k|H(t)|k\rangle \hskip1cm \text{for each} \hskip1cm |k\rangle\in\mathcal{E}(t) \tag{3} $$ and $$ Z(t) := \text{trace} \left(e^{-\beta H(t)} \right) = \sum_{|k\rangle\in\mathcal{E}(t)} e^{-\beta E(t,k)}. \tag{4} $$ For simplicity, I'll keep the inverse temperature $\beta$ fixed. If the initial state is $$ \rho(0)=\frac{e^{-\beta H(0)}}{Z(0)}, \tag{5} $$ and if $H(t)$ is measured at $t=0$ and again at $t=1$, then the joint probability of getting the outcomes $|j\rangle$ and $|k\rangle$, respectively, is $$ p(j,k) = \big|\langle k|u(1)|j\rangle\big|^2 \frac{e^{-\beta E(0,j)}}{Z(0)} \tag{6} $$ with $|j\rangle\in\mathcal{E}(0)$ and $|k\rangle\in\mathcal{E}(1)$. The quantum version of Jarzynski's equality says $$ \sum_{j,k}p(j,k) e^{-\beta W(j,k)} = \frac{Z(1)}{Z(0)} \tag{7} $$ with $W(j,k) := E(1,k)-E(0,j)$. The proof is easy: the factors $e^{\pm\beta E(0,j)}$ cancel each other, and then evaluating the sum over $j$ eliminates the dependence on $u(1)$. Equation (7) is a special case of equation (2.7) in Jarzynski Relations for Quantum Systems and Some Applications.

Even though the proof is easy, equation (7) is interesting because the left-hand side involves $u(1)$, which says something about how the system gets from $t=0$ to $t=1$, whereas the right-hand side only involves the initial and final Hamiltonians $H(0)$ and $H(1)$, regardless of what happens in between. We can make it sound even more interesting by referring to the quantity $W(j,k)$ as "work."

In Example 1, where $H(t)$ is independent of $t$, this definition of "work" would give $$ \sum_{j,k}p(j,k)f\big(W(j,k)\big) = \sum_{j}p(j,j)f\big(W(j,j)\big)= f(0), $$ for any function $f$ and for any initial state (not just for a thermal state), so this definition of work is not equivalent to the definition we would normally use in a context like Example 1. Again, different definitions of work are useful in different situations.

Example 5: Accounting for back-reaction

The OP considers a time-dependent Hamiltonian $H(t)$, which we often do as a way of accounting for how external devices affect the system of interest when back-reaction is negligible. In the paper Fluctuation relations and strong inequalities for thermally isolated systems, Jarzynski acknowledges the limitations of such a model:

It is hard to imagine an experimental situation in which a macroscopic quantum system is so utterly isolated from its environment that it evolves unitarily as an external parameter is varied from one value to another –- surely a wayward photon or gas molecule will scatter off the system, spoiling unitarity. ... These considerations highlight the idealizations that are made (and should always be kept in mind) when choosing specific dynamical equations of motion to model the evolution of a many-particle system.

To account for back-reaction, we can use a more realistic model that includes the microscopic quantum dynamics of whatever devices are affecting the subsystem of interest, together with the subsystem of interest itself, all as part of one big truly-closed system with an overall time-independent Hamiltonian $H$. Such a model automatically includes back-reaction: $H$ is time-independent, so the total energy is conserved. However, doing this doesn't make "work" unambiguous. Instead, it exposes a deeper ambiguity.

One example of such a model is QED + QCD (quantum electrodynamics combined with quantum chromodynamics), which is rich enough to account for both the molecular and nuclear dynamics of most of the laboratory-scale experiments that might be associated with Jarzynski's equality. In QED+QCD, what part of the total system's energy would we use to define the work done on the subsystem of interest? The answer depends on the state, because the state defines what objects are present, what lab equipment is present, and how everything is configured, both macroscopically and microscopically. Even the context of a given state, the concept of a subsystem's "energy" is fundamentally ambiguous. The Hamiltonian of QED+QCD involves one electron field, one up-quark field, and so on, and all of the objects and lab equipment and air molecules are described in terms of these same quantum fields. In QED+QCD, there is no general or perfect way to separate the energy of a subsystem from the energy of the rest of the world.

The difficulty is not limited to QED+QCD. To formulate it in general terms, consider any quantum system with a time-independent Hamiltonian $H$. In the Schrödinger picture, which I'm using here, the model is defined by its set $\Omega$ of time-independent observables, together with the Hamiltonian $H$ that generates time-evolution. A subsystem is a simply subset $\omega\subset\Omega$.

(A less general but more popular definition of "subsystem" involves writing the Hilbert space $\cal{H}$ as a tensor product $\cal{H}=\cal{H}_1\otimes\cal{H}_2$ and taking $\omega$ to be all of the observables in $\Omega$ that act non-trivially only on one factor $\cal{H}_1$.)

Defining the state of the subsystem is trivial. Let $$ \psi(\cdots,t) := \frac{ \big\langle\psi(t)\,\big|\cdots\big|\,\psi(t)\big\rangle }{ \big\langle\psi(t)\,\big|\,\psi(t)\big\rangle} \tag{8} $$ be the state of the complete system at time $t$. The dots "$\cdots$" are a placeholder for an operator: if $X$ is an observable, then $\psi(X,t)$ is the expectation value of $X$ at time $t$. The Hamiltonian $H$ enters via the relationship $$ \psi(\cdots,t) = \psi\big(U^\dagger(t)\cdots U(t),0\big) \tag{9} $$ with the unitary operators $U(t)$ defined by $$ \frac{d}{dt} U(t)=-iH\,U(t). \tag{10} $$ The state of the subsystem $\omega\subset\Omega$ at time $t$ is simply the restriction of the function $\psi(\cdots,t)$ to observables in $\omega$. There is no need to factorize the Hilbert space, no need to take a "partial trace," or anything like that. Those can be useful tools for calculations, but they're not needed conceptually.

Equation (7) refers to the subsystem's Hamiltonian. What does that mean in the general framework? When does a subsystem act like it has a (possibly time-dependent) Hamiltonian of its own? That's probably a big enough question to fill several research careers, so I won't try to answer it here, but I'll suggest a way to formulate the question. The question asks about the feasibility of an approximation $$ \psi(X,t)\approx\psi\big(u^\dagger(t)X u(t),0\big) \hskip1cm \text{for all } X\in\omega\subset\Omega, \tag{11} $$ with unitary operators $u(t)$ given by (2) for some time-dependent Hamiltonian $H(t)$ that commutes with everything that commutes with everything in $\omega$. In other words, we want $H(t)$ to belong to the double commutant of $\omega$. In other other words, we want $H(t)$ to belong to the von Neumann algebra generated by $\omega$.

If we can find such an $H(t)$, then we can use it to define the energy of the subsystem, which is a prerequisite for defining "work." In general, we have no guarantee that such an $H(t)$ exists, but at least we have a mathematical formulation of the question.

Equation (7) assumes that the system is initially in a thermal state (5). How can we express an initial condition like (5) in the general framework? One way would be to use the approximation (11) to justify reverting to a model with no back-reaction, but that would defeat the purpose of using the more general framework.

Here's another way: Suppose that $H(t)$ satisfies the conditions shown above, and to simplify the language and notation, suppose that its spectrum is discrete. Let $P(t,k)$ be the projection operator onto the eigenvector $|k\rangle$ of $H(t)$ with eigenvalue $E(t,k)$. Then an initial condition like (5) can be expressed like this: $$ \psi\big(P(0,k),0\big) \propto e^{-\beta E(0,k)} \tag{12} $$ with a proportionality factor that is independent of $k$. By construction, the projection operators $P(t,k)$ belong to the von Neumann algebra generated by $\omega$, so the condition (12) describes the state of the subsystem without assuming anything further about the state of the complete system.

Equations (8)-(11) would provide a general approach to accounting for back-reaction, but it all relies on an identification of the subsystem of interest as a subset $\omega\subset\Omega$ of the system's observables. Thinking about the QED+QCD example makes it clear that in general, a state-independent definition of "subsystem" does not exist. Nevermind the ambiguities about how to define the subsystem's energy, which of course is a prerequisite for defining "work." Here we have a much deeper ambiguity: the very concept of "subsystem of interest" is state-dependent, because the existence and configuration of the whole experiment is state-dependent. In relativistic quantum field theory, observables are not tied to particles, because particles are just phenomena. Observables are tied to regions of space (or spacetime in the Heisenberg picture). This is a deep obstacle to the existence of any definition of "work" that behaves as desired in arbitrary quantum systems.

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