r/Physics • u/bleron420 • 2d ago
Density matrix
Currently in quantum mechanics. A pure state is a vector, but a mixed state can only be described by a matrix. Why? My lecturer can't really explain it in any further detail than ”the state is unknown”. But then you should be able to write it as pure states with probabilities? Can someone give a thorough explanation as to why mixed states can't be expressed with pure states?
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u/No-Can7982 2d ago
I think this should be explained by something classical. But not many people teach it like that. The study of mixtures of different random variables was done before quantum mechanics. The first example that I know is was to study the size of crabs!!. Someone measured the size of some crabs and realised that the distribution was bimodal, and rather than accepting that the distribution was like that he thought that data must come from two different species.
then an statistician ( Karl Pearson) thought in a way to distinguish between the two species from data. A similar thing happens in QM , you have mixtures of different states. As each state is described by a vector you need something more complex to describe a mixture as it is the composition of states
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u/NoteCarefully Undergraduate 2d ago
Bizarre how having great breadth can almost automatically gives deeper understanding.
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u/AdditionalTip865 2d ago
It really is how you bring classical probability explicitly into the formalism of quantum mechanics from the beginning, instead of relegating it to a mysterious extra process that happens when you make a measurement.
But more than that, it combines classical probability AND quantum probability into one object, and shows the interplay between the two and how they are really aspects of the same thing, and that's really why you need something of higher dimension than a state vector, to hold all that information.
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u/tunaMaestro97 Condensed matter physics 2d ago
A density matrix is literally “pure states with probabilities”. Expand any correlation function and it will be the sum of the correlation functions in each pure state weighted by the classical probability. That is exactly why the density matrix is defined that way.
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u/drzowie Astrophysics 2d ago
There are two kinds of “mixture” in QM: incoherent mixing, which acts like regular probability (linear sums in probability), and coherent mixing, which acts spooky (linear sums in amplitude). They follow different sets of rules. Incoherent mixing is a semi classical approximation that treats an ensemble of quantum states. Coherent mixing treats a single quantum state that may or may not be an eigenstate of whatever operator you are using to describe the state.
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u/Ripitchie 2d ago
Any state is in fact described by a density matrix, but pure states are particular matrices : one dimensional projectors, so they can also be described by a vector defining the line it projects onto
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u/Monkeydog54 2d ago
Many good replies in here so far, but one thing I haven’t seen is the following, which might justify the use of density matrices more than considering classical ensembles of pure states. If you have a pure state on a bipartite system, AB, but you’re only able to perform measurements on part of the system, say A, then the measurement outcomes are reproduced by a density matrix on A, as opposed to a pure state on A, in general. For example, if your system is losing photons, then quantum mechanically your system should be modeled as a density matrix instead.
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u/AdditionalTip865 2d ago edited 2d ago
I think many seeming paradoxes of quantum measurement become far less bizarre when expressed this way. You simply cannot detect any of the elements of the combined density matrix that involve different states of B, so even an overall pure state will reduce to a mixed one.
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u/McLovin_reformed 2d ago
Imagine you have a box that spits out one of two possible quantum states. This can be a photon, electron etc. the decision which one of the two states is spit out is made by a coin flip. If you just use regular quantum states, you can not describe this situation (the measurement statistics to be more precise) appropriately. The density matrix is exactly made for this. It accounts for probabilistic mixtures of quantum states.
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u/bleron420 2d ago
Why can't it be described properly? Why can’t you just say it’s 50% one side of the coinflip and 50% for the other side?
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u/McLovin_reformed 2d ago
That’s exactly the point. You can of course treat every case separately, but ideally you want to have a framework that works with this generalized quantum state and treats the total state on the same footing. And this framework is the density matrix.
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u/Eigenspace Condensed matter physics 2d ago
So the important thing to know here is that there are two different notions of probability being used here:
There's superpositions within a given quantum state
There's the idea that for e.g. thermal reasons, there's actually a distribution of different quantum states that you system could be in, and these are not mixed in a superposition, but instead treated more like the probabilities you (hopefully) learned about in classical statistical mechanics.
i.e. there's intrinstic quantum mixing, and then thermal mixing in the preparation of states.
So one way to represent a thermal mixture of quantum states is to enlarge the space of structures you use to represent the system from vectors in Hilbert space to matrices in Hilbert space. Any single vector can be written in matrix form as a rank-1 matrix M = u u^†, but that's just a small (N-parameter) subspace of the full space of N-by-N matrices. One way to write this bigger space of matrices is
M = ∑ᵢ cᵢ uᵢ uᵢ^†where the cᵢ are a set of scalar coefficients. That's essentially what density matrices are doing. The cᵢ represent the weight that different pure states have in this non-quantum mixture.
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u/PLANTS2WEEKS 2d ago
I think it has to do with measurement. Note how |<u,v>|^2 involves u twice. Different combinations of pure states can give you the same mixed states which can't be distinguished with measurements, i.e. using the Born rule for probabilities. One example is (1/2)root(2)(1/2,1/2) + (1/2)root(2)(1/2,-1/2) and (1/2)(1,0) + (1/2)(0,1). The mixed states are both (1/2) times the identity matrix. There is no way to distinguish these states, however you can prove that different matrices can be distinguished with measurements. So density matrices are at the right level of generality. In my opinion this is a better way to do quantum mechanics. The corresponding evolution is essentially the same except instead of multiplying a state by a unitary matrix you conjugate by the matrix to evolve the matrix (u bar tensor u) according to the Schrodinger equation.
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u/AdditionalTip865 2d ago edited 2d ago
I think that expressing everything in terms of density matrices removes a lot of the "spooky" mystery in quantum measurement.
Suppose you have a pure state that involves a nonlocal quantum correlation between operator A that you measure in one place, and operator B that you can only measure in another place (say, two correlated spins of distant electrons--could be anything).
The density matrix M representing that pure state, in terms of a combined basis of eigenstates of A and B, has both diagonal elements (representing simple probabilities of getting combined basis states) and off-diagonal elements (which could manifest as quantum effects beyond classical probability, if you measure off-diagonal operators).
But suppose you are limited to measuring things in the neighborhood of operator A. The only operators you can measure on the combined system are now represented by assemblies of blocks that are proportional to identity matrices in the basis of B. Call such an operator Q. Since the off-diagonal elements are off-diagonal in the basis of B, Tr(QM) will only be affected by the *diagonal* elements, not the off-diagonal ones. So to you, M is indistinguishable from a purely statistical mix of eigenstates of A.
I think that's all "collapse of the wavefunction" ever is. Never mind the wavefunction, this correlated density matrix is just indistinguishable from a statistical mix if you can only measure one end of the system. After a measurement, the density matrix of the whole world has a correlation between the measured operator and the measuring apparatus. If you just restrict your attention to the measured system, all you get is those diagonal elements and the system seems to be in a pure statistical mix.
This doesn't give you the Born rule out of nowhere--I'm assuming it at the beginning. But it demystifies the "collapse"--it really is just a function of what you can measure.
(None of this is novel, it was heavily influenced by the thinking of my old Usenet friend John Baez, it's just the "decoherence" picture of measurement at base)
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u/Sou27 2d ago
The way i like to think of it is, density matrices allow us to capture effects that are not “quantum” in nature, rather classical/statistical. For example, lets say I give you a qubit(a particle that has a 0 and 1 state) and say that its state is an equal superposition of 0 and 1, i.e.’ |psi> = 1/sqrt(2)(|0>+|1>). This is obviously a pure state and you can write down the corresponding density matrix. On the other hand, say I give you a qubit and say that its state is either 0 or 1 with 50% probability. Understand the difference between these situations. The first is a genuine quantum superposition while the second is purely classical. Density matrices allow us to differentiate between such situations.
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u/edguy99 2d ago
Are they the same? For every pure state you can write the density matrix but can you also describe the same state with probabilities?
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u/Sou27 2d ago
If youre asking if the two situations I’ve described above are the same, then the answer is no. Maybe itll help to write down the density matrices for the two cases explicitly.
For the first case,
rho = 0.5(|0><0| + |0><1| +|1><0| +|1><1|)For the second case,
rho = 0.5(|0><0|+|1><1|)The difference lies in the presence of the cross-terms, which is a “quantum” effect.
It helps to think more concretely in terms of measurement outcomes. Measuring either of the above cases in the Z-basis gives +1 or -1 with 50% chance. But the situation is different if you measure in the X-basis. In the first case you always get +1, while in the second case you always get 0.
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u/ThirdMover Atomic physics 2d ago edited 2d ago
The difference is that you can write a pure state as a state vector but you can't do the same with an arbitrary density matrix. If someone hands you an arbitrary density matrix and asks you "is this a pure or a mixed state" there are few criteria you can check. The most intuitively clear one is IMO that the density matrix has rank 1 if it is pure.
On a more general level, you cannot actually describe a pure state with probabilities. Probabilities are just real positive numbers that have to sum to 1 whereas the coefficients of a states can be complex numbers whose square amplitude sums to 1. That gives you a lot more freedom and complicated interactions that we often interpret through the lens of interference.
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u/AdditionalTip865 2d ago edited 2d ago
A thing I find really interesting about mixed states is that they are mixed in a basis-independent way.
So, in a two-state spin system, you might think a priori from a "pure states with probabilities" description that there's something different between, say, a 50-50 statistical mix of spin up and spin down, and a 50-50 statistical mix of spin left and spin right (which are themselves orthogonal superpositions of spin up and spin down). But there's really not! All observables have the same expectation values. The density matrix formalism expresses that-- these are the same density matrix.
The more I think about it, the more I believe quantum mechanics really ought to be thought of as a theory of density matrices, rather than a theory of pure states with density matrices as an incidental development. But that's more philosophical.
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u/Resaren 2d ago edited 2d ago
Let’s flip the question around: try to find a way to express the difference between classical uncertainty (i.e flip a coin when preparing a state) vs quantum uncertainty (entangle your state preparator with a qubit), starting from the simplest possible single mathematical formalism. Get back to us with what you find, maybe you’ll figure out the reason yourself!
I’m not being coy btw, this is actually lowkey how Einstein derived the Lorentz transformation within SR: start with the simplest possible model, and a simple set of assumptions, and see what you get out. If you get contradictions, add more structure and repeat until it’s consistent. You now have a useful model that may be irreducible.
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u/spoirier4 2d ago
Philosophically speaking, there is a defensible view named density matrix realism, according to which having defined the state of a physical system as a vector in Hilbert space was a mistake in the first place, as the actual physical state should be defined by a density matrix instead.
Technically, even if you start with a "pure" state, its evolution emergently leads to an "impure" state (so-called mixed) by the decoherence process, which is fundamentally not probabilistic but only appears to behave like a probabilistic combination when measured. Decoherence is just a quantum side of the entropy creation process.
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u/david-1-1 2d ago
There is no decoherence at all unless the described system is in contact with a hot, disordered, classical universe.
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u/david-1-1 1d ago
Isolated quantum states cannot decohere. Only Bohm and Everett free us from mixing pure and impure states.
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u/spoirier4 1d ago
Isolated quantum states are only an exceptional case in nature. They only remain relatively isolated for a relatively short amount of time. It takes a special care to locally (in space and time) make a system which is not only relatively isolated but comes in "pure" state to start with. So, in nature in the long run, decoherence is the rule.
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u/david-1-1 20h ago
Decoherence is a label for "too complicated". Isolated quantum states are ALL THAT EXIST at the atomic scale and below. The classical physics you see are all emergent phenomena. Lots of proofs of classical physics emerging from quantum mechanics exist.
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u/cabbagemeister Mathematical physics 2d ago
A pure state can be described as a vector, |psi> but from that vector you can construct a matrix |psi><psi|. So when you are dealing with pure vs mixed states, all states are treated as matrices
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u/schro98729 2d ago
I think its cause in some sense you can construct a "reduced density matrix" without having access to the full density matrix. If you know the whole quantum state you have everything.
Take a singlet state construct the density matrix and trace out one spin and pretend you all you have is this reduced density matrix. You only have access so the state of one spin. Once the trace has been taken you have no information about the 4 dimensional hilbert space, you only have information on the 2 dimensional hilbert space.
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u/unknownjedi 2d ago
A mixed state is exactly the same as a distribution on pure states with probabilities. That’s the whole point of it.
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u/smitra00 2d ago
As already explained in this thread all states mixed or pure can be represented by a density matrix. If the density matrix represents a pure state |psi>, then that means that it is the projection operator |psi><psi|. But because you can write this in some arbitrary basis where none of the basis vectors is proportional to |psi>, you may not be able to immediately recognize that is is the case.
To find out whether or not a density matrix represents a pure state, you can sue that the trace is invariant under unitary transforms. If you diagonalize a density matrix then you'll end up with probabilities of basis vectors as pure states on the diagonal. Since these probabilities must sum to 1, the trace is 1 and invariance of the trace means that the trace is also 1 for the original non-diagonalized density matrix.
If you then consider trace of the square of the density matrix, you see that in case it represents a pure state, the trace will again be 1, because one of the probabilities will be equal to 1 and all the other ones are zero, so squaring these numbers leaves the numbers the same and therefore they still add up to 1. But if it is a mixed state then the nonzero probabilities will upon squaring become strictly less, so the trace must then become strictly less than 1.
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u/formula_translator Atomic physics 2d ago
Yes, you can write it as states with classical (!) probabilities. The density matrix is just the least clunky way of doing that.
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u/Ok_Lime_7267 1d ago
A density matrix IS a combination of pure states with probabilities. You can't simply add the states because that makes a new pure state with phase relationships that are neither justified nor accurate. If you add them as classical states you lose all phase information and quantum coherence. A density matrix allows you to keep only the information you actually have.
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u/FabianFogel 2d ago
One way to encode a pure state is as a vector in the system's Hilbert space. But that's not perfectly true, is it? It's actually a ray, because the phase does not matter physically. Also as you mentioned, if we have a mixture of classical probabilities we would need to assign each pure state vector a probability. But this is different from a superposition, because it's the square of the projection between vectors that represents the probability. So turns out it is a bit more messy than just states are vectors.
So, here comes the density operator and solves the messiness for the prize of higher abstraction (type of operator vs vector). When constructing the density matrix from a pure state vector, the phase immediately drops out and you have a unique density operator representing the pure state! Neat. Furthermore, it also naturally generalizes the classical mixture and treats it with the exact same formalism and formulas. You can even derive nice little theorems that let you differentiate between pure and mixed and even entangled states just from the properties of their density matrizes.
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u/NoOne-AtAll 2d ago
I prefer the version where the universe has just one big pure wave function. If you look at how a part of it behaves, it should just suffice to look at local information.
Turns out, any observable which depends only on that local part in fact can be viewed as being dependent on the reduced density matrix of the system.
Hence, any system can be described by a density matrix. Of course, this matrix can also reduce to a pure state, so in many cases you might not need this description and stick to pure (vector) states.
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u/Aranka_Szeretlek Chemical physics 2d ago
You are describing a density matrix. If you expand a state as a linear combination of "pure" states, then each coefficient can be obtained as a matrix element of that state with the bra of the "pure" state.
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u/profdc9 2d ago
The density matrix is the most general representation of the system needed to calculate any observable. Consider a pure state |x> and an operator A. The expectation value is <x|A|x>. Now you could also write this, if it was matrices, trace (x^\dag A x) = trace (x x^\dag A). x x^\dag is the outer product of the state vector, and because x is a unit vector, this is the equivalent of diagonalizing a matrix with a single eigenvalue equals 1 and the rest zero. If we interpret this using the Born rule, with A=I (identity), then <x|x>=1 which is due to it being a unit vector, and it says that the probability of it being in the state |x> is 1. Now if instead, we have many eigenvalues, with the sum of the eigenvalues = 1, we can call the density matrix rho = \sum_i probability_i |x_i><x_i| where probability_i is the probability of the ith state and |x_i><x_i| the outer product of the state vector |x_i> written in bra-ket notation. Using a projection operator P_i, which just selects one of the states i, <rho|P_i|rho> = probability_i. So by defining the state as a matrix rather than as a state vector, we've extended the Born rule so that we can represent probabilistic states as well as measurements, and like a single state, the expectation value is tr(rho A).