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Suppose Stanford Research Systems starts selling a two-level atom factory. Your grad student pushes a button, and bang, he gets a two level atom. Half the time the atom is produced in the ground state, and half the time the atom is produced in the excited state, but other than that you get the exact same atom every time.

National Instruments sells a cheap knockoff two-level atom factory that looks the same, but doesn't have the same output. In the NI machine, if your grad student pushes a button, he gets the same two-level atom the SRS machine makes, but the atom is always in a 50/50 superposition of ground and excited states with a random relative phase between the two states.

The "random relative phase between the two states" of the NI knockoff varies from atom to atom, and is unknown to the device's user.

Are these two machines distinguishable? What experiment would you do to distinguish their outputs?

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I asked this question because I think there's a good followup question about indistinguishability in QM. However, every time I try to form this followup question, it comes out hopelessly vague and mystical-sounding. I'll keep rephrasing 'til I get someting concrete. –  Andrew Apr 4 '11 at 21:08
    
I think this is a great question, and plays out on some fundamental notions of how we describe what is knowable in QM and how much redundancy is in our description –  lurscher Apr 4 '11 at 21:39
    
I see you marked an accepted answer. I find this question fascinating and would like to see it discussed further. I really was hoping people would add more analysis. Since no one did, I added what at initial glance (may be wrong) appears to be a counter example to hopefully at least get more discussion going. –  Edward Apr 5 '11 at 1:31
    
@Edward This question is essentially answered, but I'd like to see this discussed further too to clear up why two non-overlapping sections of Hilbert space can't be distinguished here. So I asked a new question physics.stackexchange.com/questions/8123/… –  Ginsberg Apr 6 '11 at 0:08

7 Answers 7

up vote 28 down vote accepted

These systems are not distiguishable. The average density matrix is the same, and the probability distribution obtained by performing any measurement depends only on the average density matrix.

For the first system, the density matrix is $$\frac{1}{2} \left[\left(\begin{array}{cc}1&0\cr 0&0\end{array}\right)+ \left(\begin{array}{cc}0&0\cr 0&1\end{array}\right)\right].$$

For the second system, the density matrix is $$\frac{1}{2\pi} \int_\theta \frac{1}{2}\left(\begin{array}{cc}1&e^{-i\theta}\cr e^{i \theta}&1\end{array}\right) d \theta.$$

It is easily checked that these are the same.

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It boils down to whether the phase is "truly random" or not, right? Perhaps there might be some experimental systematics in the cheap knock-off that show up in the random relative phase. As @Peter Morgan notes in his answer if the phase does average to zero then its hardly a cheap knock-off. –  user346 Apr 5 '11 at 4:59
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More precisely, it looks like we need to assume the "random phase" is uniformly distributed in the ensemble. For that is implicitly assumed in how the integral is written there. –  Ginsberg Apr 5 '11 at 22:40

Perhaps the machine described here can actually be built. Let me propose a heated flask containing a 50-50 mixture of gaseous, monatomic carbon-14 and nitrogen-14. When you push a button, a pinhole opens up and allows exactly one atom to escape. Is it: either a carbon atom or a nitrogen atom, with 50% probability, or is it an atom in a 50-50 superposition of the carbon/nitrogen states?

The weight of expert opinion in the answers posted thus far seems to indicate that these two descriptions are experimentally indistinguishable. I suspect this is correct, although it's a funny conclusion that flies in the face of the common-sense belief that an atom is either carbon or nitrogen, but not both. However I think two stipulations ought to be noted:

  1. I can see no reason why a given atom ought not to appear in an 80/20 superposition, so long as the long-term average is 50/50.

  2. I don't believe the machine is actually constructible because I don't think there is a mechanism which can reliably produce exactly one atom at a time. You never quite know just how many atoms you've let out, and that introduces enough uncertainty in the measurement to avoid any of the glaring contradictions that seem to be present.

EDIT: When Andrew posted this question, he promised a follow-up question. It's six months later and I haven't seen the follow-up. So here's what I think the follow up was going to be:

Suppose you have a gas in equilibrium. According to thermodynamics, the probability of an atom being in a given state is given by an exponential function of the energy. So, according to Copenhagen, we have atoms in different energy eigenstates which make random transitions from one state to another, emitting or absorbing photons when they make transitions. Question: Is there a way to experimentally distinguish this model from an alternate model where all the atoms are in continuously varying superpositions of states, radiating or absorbing continuously as the charge distribution of those superpositions oscillates like tiny antennas?

If Andrew is out there, I wonder if this was his follow-up question?

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This is such a nice philosophical question with such a neat resolution that I can't resist dropping a comment. The reduced density matrices of the atom are the same for Stanford and National, but quantum mechanics is irreducibly holistic. The wave function describes the entire universe. If the atom was prepared by Stanford, it will be entangled with traces of the environmental record in Stanford in a particular way, but if it was prepared at National, it will be entangled with traces at National in a different but still specific way. Holistically, there is indeed a difference. To suppose the atom can be considered in isolation from the rest of the world is a major fallacy in quantum mechanics.

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The density matrices in both cases are identical. If quantum mechanics is exactly linear, both states ought to be indistinguishable. But if there are some slight nonlinearities in the time evolution, we ought to be able to distinguish between them in principle. But you have to realize nonlinearities in quantum mechanics lead to all sorts of problems, which is why most people assume quantum mechanics is exactly linear.

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Let me give some reference that might be useful to make things clear.
It's Landau-Lifshitz, book 5, chapter 5:

The averaging by means of the statisitcal matrix ... has a twofold nature. It comprises both the averaging due to the probalistic nature of the quantum description (even when as complete as possible) and the statistical averaging necessiated by the incompleteness of our information concerning the object considered.... It must be borne in mind, however, that these constituents cannot be separated; the whole averaging procedure is carried out as a single operation, and cannot be represented as the result of succesive averagings, one purely quantum-mechanical and the other purely statistical.

This "twofold averaging" is exactly the reason why the two states cannot be distinguished in any way.

Let me add another nice citation:

It must be emphasised that the averaging over various $\psi$ states, which we have used in order to illustrate the transition from a complete to an incomplete quantum-mechanical description has only a very formal significance. In particular, it would be quite incorrect to suppose that the description by means of the density matrix signifies that the subsystem can be in various $\psi$ states with various probabilities and that the average is over these probabilities. Such a treatment would be in conflict with the basic pronciples of quantum mechanics.

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EDIT: Reread it some hours later and found my error. I figured I was doing something wrong. I was applying operations out of order when calculating the conditional probability. It is 1/2 in each case. I'll leave the commentary untouched.


I think the answer is Yes, or at least I'm not entirely convinced the answer is no.

I will provide an example below, but I don't find it very convincing since I just ad-hoc approached it, and don't have a nice "overarching" principle to take away from this. Basically, consider this more as a comment to get discussion going, than a full fledged answer.

The other answers show that the expectation value of measuring the system to be in a particular state is the same. Basically the density matrix of the ensemble is the same, but the density matrix of the first machine only has two possible outputs while the second has an infinite number. Focusing immediately on the ensemble average seems to be throwing away any possibility we have of distinguishing them.


Here's an attempt at distinguishing them:

Machine 1 possible output, only pure states
$|0\rangle$
$|1\rangle$

Machine 2 possible output, any state
$\frac{1}{\sqrt{2}}(|0\rangle + p |1\rangle)$
where $p = e^{i\theta}$ with $0 \le \theta < 2\pi$

Now take some other qubit B (it doesn't matter here physically what it is) of prepared state $\frac{1}{\sqrt{2}}(|0\rangle+|1\rangle)$ to get the product states:

machine 1
$\frac{1}{\sqrt{2}}(|0\rangle+|1\rangle)|0\rangle = \frac{1}{\sqrt{2}}(|00\rangle +|10\rangle)$
$\frac{1}{\sqrt{2}}(|0\rangle+|1\rangle)|1\rangle = \frac{1}{\sqrt{2}}(|01\rangle +|11\rangle)$

machine 2
$\frac{1}{2}(|0\rangle+|1\rangle)(|0\rangle + p |1\rangle) = \frac{1}{2}(|00\rangle+p|01\rangle + |10\rangle + p |11\rangle)$

Now let's introduce an interaction which can cause some interference:
$|00\rangle \rightarrow |00\rangle$
$|01\rangle \rightarrow \frac{1}{\sqrt{2}}(|01\rangle + |10\rangle)$
$|10\rangle \rightarrow \frac{1}{\sqrt{2}}(|01\rangle - |10\rangle)$
$|11\rangle \rightarrow |11\rangle$

now we have
machine 1
$\frac{1}{\sqrt{2}}(|00\rangle +|10\rangle) \rightarrow \frac{1}{\sqrt{2}}|00\rangle + \frac{1}{2}(|01\rangle-|10\rangle)$
$\frac{1}{\sqrt{2}}(|01\rangle +|11\rangle) \rightarrow \frac{1}{\sqrt{2}}|11\rangle + \frac{1}{2}(|01\rangle+|10\rangle)$
machine 2
$\frac{1}{2}(|00\rangle+p|01\rangle + |10\rangle + p |11\rangle) \rightarrow \frac{1}{2}(|00\rangle + p\frac{1}{\sqrt{2}}(|01\rangle+|10\rangle) + \frac{1}{\sqrt{2}}(|01\rangle-|10\rangle) + p |11\rangle)$
$ \ \ \ = \frac{1}{2}(|00\rangle + (p+1)\frac{1}{\sqrt{2}}|01\rangle+(p-1)\frac{1}{\sqrt{2}}|10\rangle + p |11\rangle)$

Now let's do two measurements. First measure the state of B to be 0 or 1, then measure the sate of the atom to be 0 or 1.

Conditional probability on the ensemble:
Given that we find B in state 1, what is the probability of finding the atom in state 0?
machine 1
(1/2) x 1 + (1/2) x (1/3) = 4/6

machine 2
$\frac{\frac{1}{2}(p-1)^2}{\frac{1}{2}(p-1)^2 + p^2} = \frac{\frac{1}{2}(2 - 2\cos\theta)}{\frac{1}{2}(2 - 2\cos\theta) + 1} = \frac{1 - \cos\theta}{2 - \cos\theta}$

Now averaging over $\theta$
$\mathrm{Prob} = \frac{1}{2\pi}\int_0^{2\pi}\frac{1 - \cos\theta}{2 - \cos\theta} d\theta = 1 - \frac{1}{\sqrt{3}}$


Now, it is quite possible I've made a mistake here. But my main point is that the other answers seem to be throwing away the useful information to obtain solely an average of the initial output states. As the answers stand now, they do not mathematically convince me that we can never obtain an effect by adding interactions and multiple measurements with conditional probability or maybe 'weak' measurements, since individually the states have much different density matrices. Hopefully I didn't make a mistake above, but even if I did, I'd still very much like to hear more in the other answers beyond what is currently written. This is a fascinating question, so I'm quite interested in discussing this further.

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+1 if you fix the answer so it's correct, instead of adding an edit notice. –  Sklivvz Apr 5 '11 at 10:19
    
Interesting! I'll read through carefully. –  Andrew Apr 5 '11 at 12:17
    
Hmm... it seems to be the problem is that was have an ensemble of states of form $|0\rangle + e^{i\theta}|1\rangle$ with uniform distribution in theta. What you seem to want is to interact with systems and then ask conditional statements (or trace over the other systems if we don't care about them in the end) to get $|0\rangle + e^{i\theta}|1\rangle \rightarrow |0\rangle + e^{i\psi}|1\rangle$ such that the ensemble has a non-uniform distribution in $\psi$. The question is then, can we prove this is impossible? –  Ginsberg Apr 5 '11 at 22:29
    
+1 For making me think. I also appreciate someone who can realize their own mistake AND admit it. –  Ginsberg Apr 5 '11 at 22:35

Case 1: $\frac{1}{2}\left[\left|0\right>\left<0\right|+\left|1\right>\left<1\right|\right]$.

Case 2, average over phases $0$ to $2\pi$: $$\frac{\int\left[(\left|0\right>+e^{i\theta}\left|1\right>) (\left<0\right|+e^{-i\theta}\left<1\right|)\right]d\theta} {\int\left[(\left<0\right|+e^{-i\theta}\left<1\right|) (\left|0\right>+e^{i\theta}\left|1\right>)\right]d\theta}.$$ The cross terms average to zero because $\int\limits_0^{2\pi} e^{i\theta}d\theta=0$, so it's the same density matrix. If this is really what the different manufacturers deliver, it's not a cheap knock-off.

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I got the "another Answer posted" message only seconds before I'd finished. When I saw how similar it was I almost didn't post it, but hey, the notation is different. –  Peter Morgan Apr 4 '11 at 20:22
    
interesting, so physically they are equivalent states (up to our knowledge) but described mathematically in different forms –  lurscher Apr 4 '11 at 20:46
    
@lurscher, it's picky, but I think of almost everything about this question as more Mathematics than Physics. An experimentalist would want the product datasheets to tell us what types of atoms are delivered, how close to thermal equilibrium the kinetic energies are, at what temperature, what atomic energy levels are present, what the distributions of delivery are over time, a lot more characterization of what I would get if I brought the products. To me, whether the NI machine delivers Caesium instead of Xenon, one every second instead of one every millisecond, or whatever, is the Physics. –  Peter Morgan Apr 4 '11 at 22:12
    
@Peter: is it? Sounds like engineering to me. Experimental physics at best. –  Marek Apr 4 '11 at 22:22
    
@Marek You're right that my response to @lurscher has its problems. Nonetheless, it is the ambition of every Physical Theory to be accurate enough, simple enough, tractable enough, to have enough of a multitude of the other merits that a theory can have, and in fine good enough to be used routinely as an Engineering tool. Let us not argue whether Physics wags the dog or Engineering wags the dog. –  Peter Morgan Apr 5 '11 at 11:56

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