How can any quantum superposition get produced, given conservation of energy?

Question

Imagine I have a two-level quantum system $Q$ (motivating example: a trapped-ion qubit) where there is some energy difference between the states $\vert 0\rangle_Q$ and $\vert 1\rangle_Q$. Suppose I can drive it between the two states with a laser, where it transitions by either emitting or absorbing a photon.

If I send a number state of photons and measure it in the number state after it interacts with the qubit, this totally decoheres the qubit, since the environment/experimenter learns whether the number of photons was gained or lost.

So to actually drive the qubit we use a laser. A coherent laser state will have minimal entanglement with the qubit after this interaction, no matter what basis we measure it in. Basically this is the energy version of the momentum exchange in a beamsplitter, discussed here.

But if I tried to prepare a laser state from the vacuum state:

$$ \vert 0\rangle_L\mapsto \sum_{n=0}^\infty \frac{\alpha^{n/2}}{\sqrt{n!}}\vert n\rangle_L$$

(ignoring normalization) then this process clearly violates conservation of energy. The energy to send $n$ photons must come from somewhere: the laser source. So the laser beam will actually be in a state

$$ \sum_{n=0}^\infty \frac{\alpha^{n/2}}{\sqrt{n!}}\vert \psi_{-n}\rangle_S\vert n\rangle_L$$

where $\vert\psi_{-n}\rangle_S$ is the state of the laser source $S$ after giving $n$ quanta of energy to the laser mode $L$ (this paper discusses why this gives the usual mixed state representation of laser light; see also this question).

Because of the entanglement, if we measure both the laser source and the laser mode in the energy eigenbasis, we will once again completely decohere the qubit, unless the state $\vert \psi_{-n}\rangle$ was already in a coherent superposition of energy eigenstates.

But then we just keep pushing this argument back: where did the laser source get all this energy? From interaction with a different environmental system, which itself must have been in a coherent superposition. Conservation of energy continues to cause a problem: no system can spontaneously become such a superposition; we can only ever get a superposition of energy eigenstates in one system if it's entangled with another system that gained/lost the same amount of energy.

We could also avoid decoherence if we just measure the laser source and the laser mode in some basis that is not energy eigenstates. However: statistical quantum mechanics seems to postulate that quantum systems will tend to thermal equilibrium, which puts them in the Gibbs state. The Gibbs state is a mixture of energy eigenstates. In other words, this postulate means that thermalizing with the environment is effectively a perfect measurement in the energy eigenbasis.

But this would imply that once we drive this qubit transition, eventually the environment will cause it to decohere, no matter how well we isolate the qubit itself. Experimentally, this seems to be false: trapped ion quantum computers where everything but the ion is at room temperature can maintain coherent superposition for minutes or longer.

This seems to mean that actually, everyday classical systems are not in the Gibbs state, but must have some superposition of energy. I haven't found anything that describes such a state, though!

Can anyone provide/reference such a description, or point out where my reasoning has gone wrong?

I'm not entirely sure I understand your question but is the Wigner-Araki-Yanase theorem the simplest expression of what you're describing? — , Feb 25 '22 at 23:58
part of your problem is that driving with a coherent state does not result in a 2-level system. The Jaynes-Cumming model is closer to what you want for the 2-level system. — ZeroTheHero, Feb 27 '22 at 01:44

Jagerber48 · Answer 1 · 2022-02-25T11:39:00.463

2

Short answer, but you said it yourself. If the laser is in a coherent state then, because of the unique properties of the coherent state, it is NOT entangled with the qubit after interacting with the qubit. As you say, the laser must become entangled with the environment during the turning on of the laser. But since the laser/qubit interaction doesn’t result in entanglement the qubit is not entangled with the environment so there are no loss of coherence concerns.

Well, there’s more to say. You stipulate that if we measure the environment in the energy eigenbasis we will decohere the qubit. Yes this is true. But we don’t typically measure the environment in the energy eigenbasis. One way to think of it is this. When you use a laser to put a qubit in a coherent superposition case it is actually the case the whole room is in a crazy superposition state of having one more or less quanta. But the room basically looks the same in either case. That single quanta is distributed among trillions of trillions of degrees of freedom. It would require an unthinkably precise measurement/interaction to observe the difference between the two situations. Can try to say more to clarify later.

Sorry this is going to be more rambling than an answer.. but also, a laser is an out-of-equilibrium system. So you can't apply stuff like "However: statistical quantum mechanics seems to postulate that quantum systems will tend to thermal equilibrium, which puts them in the Gibbs state.".

edited Feb 25 '22 at 11:39

answered Feb 25 '22 at 10:43

Jagerber48

13,887

A coherent laser state will not entangle with the qubit, but I don't know if coherent laser states actually exist. If the laser itself is entangled with the laser source, then it will get entangled with the qubit when it interacts (which can be seen because measuring both the source and the laser in the energy eigenbasis would decohere the qubit). – Sam Jaques Feb 25 '22 at 13:37
About measuring a single quanta: Can you clarify what you mean by "observe the difference"? Definitely no measuring device could observe the difference in a way that a human experimenter could detect, but we just need "the environment" to measure the energy quanta for the decoherence to occur. I'm not sure whether the environment does perfectly measure the energy quanta (I assume not because quantum computers work), but evolving to a Gibbs state would be effectively a precise measurement of single energy quanta, right? – Sam Jaques Feb 25 '22 at 13:40
For the last point: The non-equilibrium laser might resolve things. But what if I start from thermal equilibrium, then start the laser and evolve the qubit, then wait until the laser and environment reach thermal equilibrium again? If thermal equilibrium is always a mixture over energy eigenstates, the problem still happens. – Sam Jaques Feb 25 '22 at 13:43
You misunderstand decoherence. Just because one system gets entangled with another doesn’t mean coherence between other systems is lost. Suppose we use one laser to excite the atom and then another to read it out. It is coherence between the two lasers and the atom that determines if an interference signal is observed. No matter if any of those are entangled with a larger environment. – Jagerber48 Feb 25 '22 at 15:36
I might be using "decoherence" incorrectly. What I mean is this: In the notation of my question, if the qubit was originally in state $|0\rangle_Q+|1\rangle_Q$, then after interaction with a laser pulse, the source+laser+qubit system is now in the state $\sum_n |\psi_{-n}\rangle | n -1 \rangle_L |1\rangle_Q + \sum_n |\psi_{-n}\rangle |n+1\rangle_L |0\rangle_Q$. Now the $|0\rangle_Q$ and $|1\rangle_Q$ states can no longer interfere with each other because they are entangled with a larger system. – Sam Jaques Feb 25 '22 at 15:58
you are right about the presence of entanglement. But, just because a system is entangled with another system doesn’t mean it can’t exhibit interference with a third system. – Jagerber48 Feb 25 '22 at 17:50
If $\langle \psi_{-n}|\psi_{-m}\rangle=\delta_{nm}$, then if you trace out the laser and source, the qubit is no longer in the $|0\rangle_Q+|1\rangle_Q$ state, but a mixture of those two states. Hence, the two states cannot interfere with each other anymore. What would it mean for them to interfere with another system? – Sam Jaques Feb 26 '22 at 08:53
@SamJaques You need to ask your question a little more clearly. If you measure anything in the system in the number basis then yes, you would probably need control over all environmental degrees of freedom to be able to observe quantum interference. Most quantum experiments end when a number basis measurement is performed, like photon number.
You need to specify more clearly a particular quantum experiment that you are not able to understand on your interpretation and then we could discuss more directly.
– Jagerber48 Feb 27 '22 at 22:53
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For example, you propose that the qubit begins in $|0\rangle_Q + |1\rangle_Q$. This state already breaks conservation of energy so it's hard to tell what you're driving at. Suggestion to specifically spell out, in detail, a quantum experiment you're having a hard time understanding. – Jagerber48 Feb 27 '22 at 22:54
Good point! Necessarily if $|0\rangle_Q$ has different energy than $|1\rangle_Q$, they must be entangled (at least a little bit) with an external system. The only way it looks remotely like a product state is if that external system already had a coherent superposition.
My imagined experiment is: 1) Set up a perfectly isolated qubit in the $|0\rangle_Q+|1\rangle_Q$ state that can be driven with a laser pulse. 2) Let the apparatus thermally interact with the environment for a long time. 3) Apply an $X$ gate via laser. 4) Allow the apparatus to thermally equilibriate again for a long time.
– Sam Jaques Feb 28 '22 at 10:06
Measure the qubit in the $|+\rangle_Q,|-\rangle_Q$ basis. The result I expect is that the measurement result is always $|+\rangle$. But this only happens if the environment does not measure the laser and source in the number basis. Since the environment will effectively measure the laser and source in some basis, what is that measurement?

Sam Jaques

Feb 28 '22 at 10:07

I'm afraid I don't follow. In step 1) you say the qubit is perfectly isolated. This means step 2) has no effect on the qubit. The qubit state you specify is an eigenstate of the $X$ gates so step 3) also has no effect on the qubit (i.e. doesn't even generate entanglement between qubit and laser). Again, because the qubit is perfectly isolated step 4) has no effect on anything. Finally, the qubit began in $|+\rangle_Q$, so yes, when you measure it in the $|+\rangle_Q$ $|-\rangle_Q$ basis we will always see it in the $|+\rangle_Q$ state. – Jagerber48 Mar 01 '22 at 05:39

I think you're not being rigorous in whatever you mean by "the environment will effectively measure..." – Jagerber48 Mar 01 '22 at 05:39

user34722 · Accepted Answer · 2022-03-02T02:40:13.500

Gibbs states are equilibrium states that are also stable under realistic measurement/decoherence, but we are talking about a highly non-equilibrium setting. What is the most common non-equilibrium state that is stable under decoherence? A coherent state!

However you don't need coherent states to make coherent states, and that's not how lasers make them. The short answer is that measurement in a basis that does not commute with photon number produces coherence between different energy states, and plenty of realistic measurements (like electric field) fit the bill.

Mommy, where do coherent states come from?

According to this answer and the textbook it cites, truly ideal lasers actually emit a classical mixture of Fock states (states of definite photon number). Unlike coherent states with their well-defined phases, this mixed state is rotationally symmetric and does not have any coherence. Hence the mystery.

You can understand a lot of the physics of a laser by considering a charged mass on a spring. If the mass is displaced from the origin (i.e. it is initialized in a coherent state), then the resulting oscillating charge will radiate a coherent state. No surprises so far. You want to understand how we can get a coherent state when the mass starts in a state of definite energy, i.e. a Fock state*. In the phase space of the oscillator, this Fock state looks like a ring, and this ring contracts toward the origin as the system radiates. If the initial excitation number $N_0$ is large, then the probability distribution of emitted photons is Poissonian, just like a coherent state. The joint state of the oscillator + radiation is the entangled state

$$e^{-|\alpha(t)|^2/2}\sum_n \frac{\alpha(t)^{n}}{\sqrt{n!}} |N_0 - n\rangle |n\rangle$$

just like in the statement of your question. These entangled 'rings' are highly non-classical. Specifically, they are highly unstable to realistic measurement and decoherence. To see what happens under various measurements, consider writing the density matrix for the light mode in two different ways

$$\rho_\text{light} \propto \sum_n \frac{\alpha^{2n}}{n!} |n\rangle\langle n| \propto \int_0^{2\pi} |\alpha e^{i\phi}\rangle\langle\alpha e^{i\phi}|d\phi$$

Observers measuring energy ($n$) won't disturb the state too much, because the $n$ distribution $\alpha^{2n}/n!$ is sharply peaked around $n=|\alpha|^2$. However even a very course measurement of electric field will distinguish between $|+\alpha\rangle$ and $|-\alpha\rangle$ if $\alpha$ is large!** For instance, if we measure the electric field of the radiation by placing a very light charge and measuring its velocity at a later time, the mass will move up or down depending on the measurement outcome.*** In short, electric field measurements, which create superposition between different $|n\rangle$ states, cannot be ignored macroscopically. Note that this argument doesn't apply to thermal states, because they don't contain macroscopically different values of $E$ (and therefore an imprecise measurement of $E$ leaves a Gibbs state unchanged).

For a broader context, states that are stable under decoherence are referred to as pointer states, which are discussed in the context of quantum Darwinism. At high frequencies, where you can't play the trick of measuring the electric field with a point mass, some have argued that coherent states actually don't exist in practice, but are rather merely a 'convenient fiction.' See this paper.

*Initializing in a Fock state is somewhat analogous to having population inversion in a real laser. Things still work out of you prepare a mixture of Fock states, as long as the variance isn't too large. Initializing in a thermal state won't do, because then we will just emit thermal radiation!

**In fact the highly unstable state $|\alpha\rangle + |-\alpha\rangle$ is called a cat states, in reference to Schrodinger's cat.

***This resolves the apparent paradox in your original question. Qubits made of atoms are essentially electric or magnetic field sensors, because they couple to the electromagnetic field via $(a+a^\dagger)\sigma_x$. Thus if we interact a bunch of qubits with a 'ring state,' the ensemble has effectively measured the electric field and collapsed it onto a coherent state. Upon observation of the whole system, the physics will be basically consistent with it having been in a coherent state to begin with.

I might be using the wrong terminology when I say "coherent state". YYour oscillator+radiation state has the same form as my source+laser state, where $|\psi_{-n}\rangle = |N_0-n\rangle$. But in your case all the source states at different excitation numbers are definitely orthogonal to each other. Thus the joint system is not an eigenstate of $I\otimes a$ (where $a$ is the annihilator on the radiation system). — Sam Jaques, Feb 26 '22 at 09:10
But measuring the electric field would settle my confusion, sort of! If atoms really measure $a+a^\dagger$, rather than energy, then coherent states of photon number would happen all the time. However, that seems inconsistent with an assertion that thermal equilibrium is a mixture of energy eigenstates, because when and how do the atoms switch from measuring electric field eigenstates to energy eigenstates? I'm happy to believe that the Gibbs state is also a convenient fiction but I'd like to understand, if possible, why we use it. — Sam Jaques, Feb 26 '22 at 09:12
Yes your first comment here is exactly my point (maybe I should have stuck with your notation). An ideal laser emits the state from your question. Only measurement of some kind can break the symmetry to give you a definite phase. — user34722, Feb 26 '22 at 17:53
This measurement changes the state macroscopically because it is highly non-equilibrium, because electric field takes on a macroscopically large expectation value. So the fact that atoms measure E is relevant, even if the measurement isn't very precise. It's usually irrelevant for thermal states, because the thermal fluctuations are tiny and a course measurement of E gives 0 (which microscopically means that your averaging over measurement outcomes, and hence no coherence). This is all related to the quantum to classical transition. — user34722, Feb 26 '22 at 18:07
I'll try to paraphrase your last comment, please let me know if I misrepresent it: At the scale I'm thinking of (individual photons), atoms measure the electric field and hence produce coherent states. This means the actual state of atoms in an electric field is not technically a Gibbs state, but that's fine because at the macroscopic scale of thermodynamics, the energy superpositions induced by measuring E are relatively small. Moreover, because each atom is producing its own tiny superposition, the averaged result ends up being a very tight band of energy thanks to the central limit theorem. — Sam Jaques, Feb 28 '22 at 10:00
Yes it's important that the atoms (or at least someone) measures E, but the more important issue is whether that measurement outcome is macroscopically relevant (and therefore cannot be averaged over and ignored). If I measure E but average over measurement outcomes, the average state does not have coherence. It's only when we condition on the measurement outcome that we randomly pick out a specific phase, which then yields a coherent state. I modified the answer to emphasize this point, have a look. — user34722, Mar 02 '22 at 02:33
Rereading your question, I think that this also shows where your original logic breaks down. It's true that thermalization/decoherence eventually destroys coherence between energy eigenstates (coherent states will eventually damp down to become thermal again), but that doesn't mean that measurement in other bases isn't relevant to our perception of the macroscopic world, particularly in non-equilibrium settings. — user34722, Mar 02 '22 at 02:46
I'm still a bit confused but this has done the most for clarifying things for me. The mixed state you wrote is not disturbed at all by measuring photon number/energy $\hat{n}$ because its a mixture of eigenstates of that. The pure state above would be very disturbed by that measurement, no matter how sharply it's peaked. But if the measurement of $\hat{n}$ is approximated by measuring $a+a^\dagger$, then I can imagine (though I don't immediately see the math) that measuring $a+a^\dagger$ might give you some information on $n$ that could function as a coarse measurement of $n$? — Sam Jaques, Mar 02 '22 at 16:20
Glad to hear it! Two things: 1. There might be a confusion about measurement vs. decoherence. Perfect (strong) measurement of n will project you into a specific Fock state $|n\rangle$. That will certainly change the state, no matter how I write it. You can think of decoherence as measuring and then averaging over measurement outcomes. That will not change a diagonal density matrix. 2. Imperfect (weak) measurement of n would not disturbed the pure state above very much, because it is sharply peaked. The measurement would need to be able to distinguish tiny differences in energy. — user34722, Mar 03 '22 at 05:06
Precise measurements of energy that can distinguish tiny differences in energy take a long time, which is sort of a manifestation of the time-energy uncertainty relation $\Delta E\Delta t>\hbar$. A realistic environment will indeed cause decoherence in the energy eigenbasis, but this is a slow process (because of 2). This decoherence is precisely what causes the oscillations to damp! — user34722, Mar 03 '22 at 05:13

Ofer Naaman · Answer 3 · 2022-02-26T04:48:22.763

The laser is in a coherent state. If the intensity of the laser corresponds to average of $n$ photons, the state will have fluctuations in that number that go like $\sqrt{n}$.

If $n$ is large enough, the fluctuations completely overwhelm the 1 photon that may have been absorbed by the qubit. In other words, there is fundamentally no way to derive any information about the state of the qubit by measuring the coherent state before and after it has interacted with the qubit. Therefore, since no information can be extracted, no decoherence is necessary.

That's why we want to couple the qubit very weakly to the control drive (laser or microwaves): weak coupling means that we need many photons in the coherent field needed to affect a pi pulse, which means the fluctuations will be large, and therefore the control drive cannot be used to extract information.

To answer your question "what about conservation of energy" more directly - if someone was to keep track meticulously of all energy quanta, they could not, even in principle, notice that you skimmed off one photon from the laser to excite the qubit.

How can any quantum superposition get produced, given conservation of energy?

3 Answers3

Mommy, where do coherent states come from?

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