What is a density matrix?

Discussion in 'Physics & Math' started by arfa brane, Jul 25, 2015.

  1. arfa brane call me arf Valued Senior Member

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    These are used in QM experiments when the state of some system of particles (like a beam of them) isn't "known in advance". You might know though that for instance a beam of unpolarised light has photons in it which are polarised horizontally or vertically, and that means the probability of horizontal polarisation is 1/2 and vertical polarisation is 1/2--the per-particle statistics.

    So the density of probabilities always sums to 1, the unpolarised beam looks like: \( \begin{pmatrix} \frac {1} {2} & 0 \\ 0 & \frac {1} {2} \end{pmatrix} \) a matrix with trace 1.

    If the beam passes through a vertically aligned polarising filter, this 'projects' all the vertically polarised photons through the filter and absorbs the other half of the beam with horizontal polarisation. Everything has to be linear, so the filter effectively adds a matrix like this:

    \( \begin{pmatrix} 0 & 0 \\ 0 & -\frac {1} {2} \end{pmatrix} \) to the density matrix. The emerging beam has 1/2 the number of photons as the incident beam (ideally), but its density matrix is again trace 1: \( \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} \).
     
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  3. arfa brane call me arf Valued Senior Member

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    . What's wrong with this statement?

    What's wrong is, it isn't something you know, or can ever know.
    In fact the incoming beam is a collection of photons (maybe mixed frequencies) in which each photon is in a superposition of the allowed polarisation states, which are pairwise orthogonal: horizontal is orthogonal to vertical, right-circular is orthogonal to left-circular; the superposition of the first pair (of states) is equivalent to the superposition of the second pair (up to measurement, by anything). You don't have a combined beam of photons, half of which are polarised one way, the other half the other way.
    Each particle is capable of being transmitted or absorbed with equal probability (hence the matrix above describing the density for the incoming beam). The filter is what makes the photons "choose" one or the other, so to get actual 50/50 rates (so you can measure the change in intensity), use lots of photons.

    A closer description of what's happening is each photon is like a quantum coin that can land heads or tails, the filter is "tossing" a lot of coins to get them to land heads or tails and it "keeps" the ones that aren't heads.
     
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  5. arfa brane call me arf Valued Senior Member

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    And here's a pic:

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    Pretty ordinary looking; low-cost etc.

    But the filter is "tuned" to the source (it's visible and infrared light, mostly). If the source was emitting x-rays it would have smaller spacing and be made of something that doesn't degrade or get hot say.

    The filter is a waveguide, but the notion of "waves" is easily substituted with the notion of probabilities, and each photon has a probability amplitude of being left or right circularly polarised, so there isn't a beam of photons with half "turning" to the left, the other half "turning" to the right, it's a beam where each photon is doing both: the filter uses a photon, in some sense, to polarise another. It can be said (it is) that the emerging beam is in a prepared state, a pure state.
     
    Last edited: Jul 25, 2015
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  7. arfa brane call me arf Valued Senior Member

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    This isn't accurate either. It suggests that the incoming beam has a higher entropy (random polarisations) than the polarised, emergent beam.
     
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  8. arfa brane call me arf Valued Senior Member

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    So here's a thought experiment: use three filters.

    With one filter, the angle of polarisation depends on the alignment; with two filters in line the angle (of the plane of polarisation) depends on the angle the last filter is set to. Most people know that two filters aligned at right angles will block all the light (but that depends on how good the filters are, in practise some light will "get through").

    With a third filter between the first two, set at 45° light passes through again. What's happening?
    If the first filter is say, oriented vertically, the middle filter will repolarise this light (absorbing some of it), at 45°, then it can pass through the third horizontal filter.
    The part that's hard to explain is, why is the emerging light 1/8 of the initial intensity? Does this change if you change the angle of the intervening filter to say 30° ?

    You could, along with the filters, use some kind of object and project its shadow on a screen (lets make it a chess piece). Where can you position this object so it casts a shadow? Before, after, in between two of the filters? Does it matter where the object is?
     
  9. arfa brane call me arf Valued Senior Member

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    Suppose instead of visible light and an optical filter, you use microwave radiation and a wire grill with appropriate dimensions.

    The same thing happens (depending on how ideal the wire grill is as a polarising waveguide), and for the same reasons. Long-wavelength photons have a larger squared amplitude than visible photons--the waveguide has larger dimensions. A classical Maxwellian explanation for what happens is that plane waves interact with the wires in the grill, and those waves which have the smallest cross section (are perpendicular to the wires) are absorbed the least, those parallel to the wires are absorbed the most.

    Again, each photon is instead like a quantum coin, in the other view. The wire grill flips each coin (qubit) to see if its heads or tails. Sounds a bit weird, maybe.
     
  10. arfa brane call me arf Valued Senior Member

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    'heh'

     
    Last edited: Jul 27, 2015
  11. arfa brane call me arf Valued Senior Member

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    So, because of the mathematics involved, a density matrix has more than one way it can be generated, so I'll show you what the notation I've been using on partial matrices was about in the context of QIS:

    You have, like the number of possible outcomes of the toss of a single coin, a matrix that describes this "heads or tails" probability:

    \( \begin{pmatrix} \frac{1} {2} & 0 \\ 0 & \frac{1} {2} \end{pmatrix} \), the QM context says this is two equal squared probability amplitudes, but there is more than one way to get the density matrix in that case, because:

    \( \begin{pmatrix} \frac{1} {\sqrt {2}} & 0 \\ 0 & \frac{1} {\sqrt{2}} \end{pmatrix}\begin{pmatrix} \frac{1} {\sqrt {2}} & 0 \\ 0 & \frac{1} {\sqrt{2}} \end{pmatrix} = \begin{pmatrix} 0 & \frac{1} {\sqrt {2}} \\ \frac{1} {\sqrt {2}} & 0 \end{pmatrix}\begin{pmatrix} 0 & \frac{1} {\sqrt {2}} \\ \frac{1} {\sqrt {2}} & 0 \end{pmatrix} \). And that's only the ones with all positive entries.

    In "my" notation: \( [\frac{1} {\sqrt {2}} (\rho_1 + \rho_2)]^2 = [\frac{1} {\sqrt {2}} (\overline{\rho_1} + \overline{\rho_2)} ]^2 \) (don't know why the editor is extending the overline like that).
     
    Last edited: Jul 29, 2015
  12. krash661 [MK6] transitioning scifi to reality Valued Senior Member

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    arfa, examine ramanujan theorems, and then elaborate. i'm interested in your words.
     
  13. arfa brane call me arf Valued Senior Member

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    Sorry, I'm not taking requests at the moment. Contact my PA and see if they can pencil it in somewhere,

    Anyways, back to the theme of quantum information. An unpolarised beam of light has an unknown state of polarisation; you can "guess" that each photon has a definite but randomly oriented polarisation, but this "information" is destroyed when the beam passes through a polarising filter--ergo there is no input polarisation information because information is a conserved quantity. If it's physical, it's also informational--no physics, no information.

    The only solution is to treat each photon as having (being) a probability it will emerge from, or be absorbed by, the filter. A polarising filter has the same kind of density matrix as a half-silvered mirror, or beam-splitter where each photon has an equal probability of being reflected or transmitted.
    The reason the emerging beam in the polarisation example (experiment) has a density matrix equal to \( \rho_1 \) is the same reason that the 50% of photons transmitted through a 50:50 beam-splitter (counted by a detector) have wavefunctions that, when propagated backwards in time have a probability of 1 that they emerged from the source and not elsewhere, although forwards in time there is a probability of 1/2 that any photon will reach the detector and 1/2 they won't.

    Which explains in a not very definitive way, why quantum 'operations' are irreversible.
     
  14. arfa brane call me arf Valued Senior Member

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    ... So there's an algebra over the space of density matrices, when we have equiprobable states, like the faces of a coin, we have to think of the two outcomes of tossing a coin in terms of squared probability "amplitudes" of entirely abstract quantum objects--as Dirac says, trying to define what these are isn't profitable, there is only their behaviour which is described by equally abstract mathematical objects.

    If we think of a density matrix as an object which has four "slots" or places to put vectors. We can also consider "motion" of these objects--an electron in motion has a spin vector which is aligned (or projected onto) its direction of motion so it points in the same or the opposite direction. Electrons have mass and charge, you can change their velocity or reverse it, reversing their direction of motion also reverses the spin if it's not disturbed.

    Because of the semigroup action (right multiplication), the term \( (\rho_1 - {\rho_2})^2 \) is also positive, and in fact \( (\rho_2 - \rho_1)^2 = (-\rho_1 - \rho_2)^2 \) are too. This is not the case for the other pair:

    \( (\overline{\rho_1} - \overline{\rho_2})^2 = (\overline{\rho_2} - \overline{\rho_1})^2 = -I_2 \), but

    \( (\overline{\rho_1} + \overline{\rho_2})^2 = ( -\overline{\rho_1} - \overline{\rho_2})^2 = I_2 \).

    The mathematics isn't "mysterious", but squares of probability amplitudes is just a little. If a probability amplitude is a continuous waveform with positive and negative amplitudes, squaring it produces a waveform with everything positive.

    This diagram is of an experiment with photons that has negative terms in it:

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    The solid black vertical bars represent 50:50 beam splitters; there is a pair of identical photons propagated at 45° from left to right in each case, and there are four possibilities:

    (1) is the upper photon reflected, the lower photon transmitted, (2) is both photons transmitted, (3) is both photons reflected, (4) is the converse of (1).

    The idea is that 1 and 4 get subtracted from each other, and 2 and 3 are also subtracted from each other. This is because 2 and 3 are indistinguishable if the photons are--the beam splitter can't "record" this information, so the terms have to cancel. This canceling is equivalent to destructive interference. 1 and 4 also "interfere destructively" because they can't both occur.

    So the first two outcomes are positive, the last two are negative, in the sum over states. Another reason, apart from considering a destructive interference, that (4) is the negative of (1) is because the reflection is negative relative to (1), there is a phase difference between the reflections in (1) and (4).
     
    Last edited: Jul 29, 2015
  15. arfa brane call me arf Valued Senior Member

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    A way to pick it all apart in terms of density matrices goes something like this:

    Consider one photon initially: take the upper one in (1) and (2). This has a density matrix: \( \begin{pmatrix} \frac{1} {2} & 0 \\ 0 & \frac{1} {2} \end{pmatrix}\); reflection and transmission are equiprobable. Similarly the density matrix for the lower photon from (1) and (3) is the same so the sign isn't negative.
    The negative signs appear when you consider both photons, now the probabilities for each of the four states is \( \frac {1} {4} = \frac {1}{2} \frac {1} {\sqrt{2}}\frac {1} {\sqrt{2}} \). So with all four states there are products of probability amplitudes.

    Of course, (1) and (4) are distinguishable from each other, and from (2) or (3), but there are only three detectable outcomes, unless the photons aren't identical.
     
    Last edited: Jul 29, 2015
  16. arfa brane call me arf Valued Senior Member

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    And again with the coin analogy. Now we have a pair of identical coins and we have to map the two states: "reflected", "transmitted", to them.
    I'll go with heads = transmitted.

    Again, the beam-splitter is just tossing this pair of coins, two detectors in the respective outgoing beams will both register a photon half the time. That is, they see both coins are heads or both tails. The other half the time only one detector registers anything, corresponding to one coin heads, the other tails. That seems like a simplification but seriously, it's as good as anything I've heard, and it's also an idea you see in online lectures by the likes of MIT staff, so ...

    Ok so to keep it real, consider again each state (what you can expect to measure) in the diagram: (1) and (2) have the lower photon being transmitted (heads), the upper one is reflected or transmitted (uhm, heads or tails), so the first two have a positive sign. (3) and (4) have the lower photon reflected (tails) so they have a negative sign because of the phase angle between {1,2}, and {3,4}--the optical quirk that means the latter pair has negative probability. But reflections just invert things--a peak becomes a trough, if you square the amplitude it's still positive. The sign appears because there are two particles and each has independent polarisation statistics.

    Now you might ask, if both detectors register, is there an alternative universe where only one of them registers? Well, before you send the photons to the beam-splitter, the system: particles, beam-splitter and detectors, is supposedly in a superposition of all the possible outcomes--it's what the universe looks like every instant, not just the experiment(s).

    What an experiment does do is tie down the concept of locality: concretely two photons are interacting simultaneously and locally with a matter-field. The states of the detectors are "entangled with the measurement": we have conservation of probabilities as a kind of flow, a probability current which is unitary (hence the zero sum-over-states, which just says information is conserved over measurements).
    Half the measurements consist of (2) and (3), a sum of nonzero eigenstates, you subtract these states (not the measurements) from each other to keep everything unitary in the system (in fact, the beam-splitter is doing the measurement, the detectors could be removed and the probabilities won't change.
     
    Last edited: Jul 29, 2015
  17. arfa brane call me arf Valued Senior Member

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    Another look at the diagram:

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    It's from the wiki page on the Hong-Ou-Mandel effect, or what happens when the interacting photons aren't identical.

    The diagram is a state diagram, with a zero sum over states. The coin analogy is useful because it's as good a way as any to describe the behaviour of photons and a 50:50 beam-splitter. But it doesn't explain the diagram, or why it has negative terms in it.

    The "information is conserved over measurements" notion is very close to saying: "total angular momentum is conserved in the system". In each state, the initial condition is invariant--two identical photons are emitted so they each meet the same chunk of matter. There are identical angles of incidence and reflection/transmission. A reflection at the same angle introduces a change in momentum (the spin vector), these changes are opposite in sign so they sum to zero.

    That's why the first two states have a positive sign--the lower photon doesn't change its spin. When they both interact with the beam-splitter their spins can end up pointing in the same direction or 'opposite' directions in Hilbert space.

    With the coin analogy, it might be an idea to make the states (2) and (3) the same as one coin heads, the other tails (you can't tell which is which), and (1) and (4) the same as both heads or both tails, a more intuitive idea of a distinguishable state. That's what changing the basis is, doing a thing like that.

    But notice how, in the context of distinguishability, the information is exactly half of what you can tell with a pair of actual coins (since even if they're identical they have distinguishable positions). Quantum information is always like that, you get half of what you expect classically.
     
    Last edited: Jul 31, 2015
  18. arfa brane call me arf Valued Senior Member

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    Maybe a better way of explaining the negative sign for (3) is that there are two reflections, they can't both have a positive sign otherwise the law of conservation of energy is violated. This explains why (4) is also negative, the lower reflection is negative if the upper one is positive.
     
  19. arfa brane call me arf Valued Senior Member

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    Here's a way to think about energy and momentum conservation: consider a pair of identical photons emitted towards the same side, rather than both sides of the beam-splitter.

    They can both be transmitted, both reflected, or one transmitted, the other reflected; call these \( P_{tt},\, P_{rr},\, P_{tr},\, P_{rt} \) . You have two detectors, so the first two states are distinguishable, the last two are not. This compares with the initial setup, and with its time-reversal (consider the reversal of states (1) and (4), for instance).

    If both are transmitted from the same side (compared with both transmitted from opposite sides), neither sees a change in momentum. If both are reflected, they both see the same change in momentum: two photons in, and two out with the same overall change.
    When the third possibility occurs, one changes momentum the other doesn't.

    Now consider the double reflection from both sides, in the Hong-Ou-Mandel setup (except there is no effect--the photons are indistinguishable from each other, this is the initial setup): both photons change momentum.
    If you choose the upper reflection to be a positive one (for that photon), the lower reflection cannot be positive because if it was it would be the same as the upper reflection--one of the reflections would have to somehow be a transmission instead.

    So, if we ignore the "-" signs, what we have is four transformations: \( P_{tt},\, P_{rr},\, P_{tr},\, P_{rt} \).
    We can compare the statistics (detector data) between the setup and one where two photons are 'input' to the same side of the splitter. And, we can consider the time-reversal. Everything should be conserved.

    Also notice, the four states are another way to see a Klein 4-group: the identity (two transmissions), two single, and one double reflections. It is indeed ubiquitous.
     
    Last edited: Jul 31, 2015
  20. arfa brane call me arf Valued Senior Member

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    It's ok to play around with changes to the setup, like having both emitters on the same side of the beam-splitter.

    Here's another straightforward change: with the setup as in the diagram, emit two photons at a time, so both detectors always fire and detect one, two or three photons each [damn, that's just wrong, because all four photons can end up moving in parallel]. What does this do?

    If you draw a state diagram that looks like a double transmission plus a double reflection (I displace these slightly for convenience), you can "take it apart" in eight ways that each have two photons. If you start subtracting these parts from the first "master" diagram, interesting things appear. What you're doing is adding together two different systems of coordinates--the emitters are rotated apart or together--and seeing what the difference is.

    So, today's exercise is to find the relation between permutations like that, and choosing some permutation of V (apart from its representation) so all the signs commute.
     
    Last edited: Aug 1, 2015
  21. arfa brane call me arf Valued Senior Member

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    Another, maybe more intuitive view of the eigenstate diagrams, is that the time reversal of (1) or (4) implies that the two photons were emitted from the same source (or a pair of sources) by a double transmission or a double reflection. Either state has a probability of 0 of occuring, so their sum is also 0.

    If you do sum the states together (sum over (1) and (2), sum over (3) and 4) then subtract the partial sums), you get a double transmission minus a double reflection from one side as the remainder, which must be equal to 0 because they don't occur.

    (1) + (2) is a combined state with one upper reflection, one upper-to-lower transmission, and two lower-to-upper transmissions, (3) + (4) has one upper reflection, one upper-to-lower transmission and two lower reflections, so (1) + (2) - ((3) + (4)) = (a sum over) two states which are not eigenstates of the system.
     
  22. arfa brane call me arf Valued Senior Member

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    One more detail about the Hong-Ou-Mandel experiment: the detectors measure intensity correlations.

    Although there are two photon sources, generating a pair of these simultaneously isn't at all like tossing a pair of coins. The generation of a single photon is bounded by the uncertainty principle, so that the generation of two photons is generally not likely to occur simultaneously. When they are perfectly simultaneous the detectors see no coincidence--the identical photons cancel; when the photons aren't identical or interact at different times there is an intensity correlation.
     

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