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According to Wikipedia, if a system has $50\%$ chance to be in state $\left|\psi_1\right>$ and $50\%$ to be in state $\left|\psi_2\right>$, then this is a mixed state.

Now, consider the state $$\left|\Psi\right>=\frac{\left|\psi_1\right>+\left|\psi_2\right>}{\sqrt{2}},$$ which is a superposition of the states $\left|\psi_1\right>$ and $\left|\psi_2\right>$. Let $\left|\psi_i\right>$ be eigenstates of the Hamiltonian operator. Then measurements of energy will give $50\%$ chance of it being $E_1$ and $50\%$ of being $E_2$. But this then corresponds to the definition above of mixed state! However, superposition is defined to be a pure state.

So, what is the mistake here? What is the real difference between mixed state and superposition of pure states?

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    $\begingroup$ The mixed state is a statistical mixture, while superposition refers to a state carrying some other states simultaneously. $\endgroup$ – Hasan Oct 12 '13 at 9:33
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    $\begingroup$ Possible duplicate: physics.stackexchange.com/q/70436/2451 $\endgroup$ – Qmechanic Oct 12 '13 at 19:16
  • $\begingroup$ As stated, your question is fully addressed in the accepted answer. However, the underlying story is much more subtle than I used to think: physics.stackexchange.com/questions/98703/… In particular, while there is a clear distinction between a pure and a mixed state, there isn't an as clear distinction between the classical and quantum "parts" of the total probabilities in a mixed state. $\endgroup$ – Dvij Mankad May 24 at 14:26
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The state

\begin{equation} |\Psi \rangle = \frac{1}{\sqrt{2}}\left(|\psi_1\rangle +|\psi_2\rangle \right) \end{equation}

is a pure state. Meaning, there's not a 50% chance the system is in the state $|\psi_1\rangle$ and a 50% it is in the state $|\psi_2\rangle$. There is a 0% chance that the system is in either of those states, and a 100% chance the system is in the state $|\Psi\rangle$.

The point is that these statements are all made before I make any measurements.

It is true that if I measure the observable corresponding to $\psi$ ($\psi$-gular momentum :)), then there is a 50% chance after collapse the system will end up in the state $|\psi_1\rangle$.

However, let's say I choose to measure a different observable. Let's say the observable is called $\phi$, and let's say that $\phi$ and $\psi$ are incompatible observables in the sense that as operators $[\hat{\psi},\hat{\phi}]\neq0$. (I realize I'm using $\psi$ in a sense you didn't originally intend but hopefully you know what I mean). The incompatibliity means that $|\psi_1 \rangle$ is not just proportional to $|\phi_1\rangle$, it is a superposition of $|\phi_1\rangle$ and $|\phi_2\rangle$ (the two operators are not simulatenously diagonalized).

Then we want to re-express $|\Psi\rangle$ in the $\phi$ basis. Let's say that we find \begin{equation} |\Psi\rangle = |\phi_1\rangle \end{equation}

For example, this would happen if \begin{equation} |\psi_1\rangle = \frac{1}{\sqrt{2}}(|\phi_1\rangle+|\phi_2\rangle) \end{equation} \begin{equation} |\psi_2\rangle = \frac{1}{\sqrt{2}}(|\phi_1\rangle-|\phi_2\rangle) \end{equation} Then I can ask for the probability of measuring $\phi$ and having the system collapse to the state $|\phi_1\rangle$, given that the state is $|\Psi\rangle$, it's 100%. So I have predictions for the two experiments, one measuring $\psi$ and the other $\phi$, given knowledge that the state is $\Psi$.

But now let's say that there's a 50% chance that the system is in the pure state $|\psi_1\rangle$, and a 50% chance the system is in the pure state $|\psi_2\rangle$. Not a superposition, a genuine uncertainty as to what the state of the system is. If the state is $|\psi_1 \rangle$, then there is a 50% chance that measuring $\phi$ will collapse the system into the state $|\phi_1\rangle$. Meanwhile, if the state is $|\psi_2\rangle$, I get a 50% chance of finding the system in $|\phi_1\rangle$ after measuring. So the probability of measuring the system in the state $|\phi_1\rangle$ after measuring $\phi$, is (50% being in $\psi_1$)(50% measuring $\phi_1$) + (50% being in $\psi_2$)(50% measuring $\phi_1$)=50%. This is different than the pure state case.

So the difference between a 'density matrix' type uncertainty and a 'quantum superposition' of a pure state lies in the ability of quantum amplitudes to interfere, which you can measure by preparing many copies of the same state and then measuring incompatible observables.

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  • $\begingroup$ So basically, in measurement of pure state I'd get probability density for sum of eigenstates, while for mixed state I'd just get sum of their probability densities, right? $\endgroup$ – Ruslan Oct 12 '13 at 8:02
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    $\begingroup$ I'm not entirely sure what you mean. Given a state, mixed or pure, you can compute the probability distribution $P(\lambda_n)$ for measuring eigenvalues $\lambda_n$, for any observable you want. The difference is the way you combine probabilities, in a quantum superposition you have complex numbers that can interfere. In a classical probability distribution things only add positively. $\endgroup$ – Andrew Oct 12 '13 at 14:41
  • $\begingroup$ Wonderful answer, thanks. 2 subquestions if I may: i)when you said that the non-commutativity of the two operators leads to one being in a superposition when written in the basis of the other and vice versa, is it because otherwise both would be in eigenstates and thus contradict the compatibility? ii) why was it necessary to choose incompatible operators for the example you wanted to show? By the way why do we still call it a collapse when a mixed state is measured? It is weird because we already know it is in an eigenstate but we just dont know which one, right? $\endgroup$ – user929304 Nov 19 '15 at 8:46
  • $\begingroup$ @Andrew How would you present you mixed state and density matrix in your example? $\endgroup$ – Alex Oct 3 '16 at 19:33
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    $\begingroup$ J.J.Sakurai gave a good explanation of the mixed ensemble and pure ensemble. He said, " A pure ensemble is by definition a collection of physical systems such that every member is characterized by the smae ket $|\alpha \rangle$. In contrast, in a mixed ensemble, a fraction of the members with relative population $\omega_1$ are characterized by $|\alpha^{(1)}\rangle$; some other fraction with relative population $\omega_2$, by $|\alpha^{(2)}\rangle$; and so on." $\endgroup$ – Wang Yun Jan 8 at 11:55
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The sentence of Wikipedia :

"For example, there may be a 50% probability that the state vector is $| \psi_1 \rangle$ and a 50% chance that the state vector is $| \psi_2 \rangle$ . This system would be in a mixed state."

is false.

The difference between pure states and partially or completely mixed states, is only a difference of structure of the density matrix.

For a pure (supposed normed) state $\psi$, the density matrix is $\rho =|\psi\rangle \langle \psi|$, and this matrix has rank one, so in some basis, $\rho$ may be written $\rho = \text{Diag}(1,0,0.......0)$

Density matrix with rank different of one correspond to partially or completely mixed states.

Compare a pure and a mixed density matrix (in a basis $\psi_1 , \psi_2$):

$$\rho_\text{pure} =\frac{1}{2} \begin{pmatrix} 1&1\\1&1 \end{pmatrix}, \quad \quad \rho_\text{mixed } =\frac{1}{2} \begin{pmatrix} 1&0\\0&1 \end{pmatrix}$$ where the pure density matrix is build from a pure state $\psi = \frac{1}{\sqrt{2}}(\psi_1 + \psi_2)$, with $\langle \psi_1| \psi_2 \rangle = 0$, and where the mixed density matrix is a classical statistical matrix.

It is easy to see that the probability density to find the system in state $1$, is the same for the two density matrices :

$$p_1 = Tr(\rho P_1) = Tr (\rho |\psi_1\rangle \langle \psi_1|) = \rho_{11}=\frac{1}{2}$$

In the same way, one finds , for the two matrices, : $p_2 = \rho_{22}=\frac{1}{2}$

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Apart from the already mathematically detailed answers given above, perhaps it would be useful to have a physical picture in mind -- the double slit experiment.

The classical 50:50 picture corresponds to the case where you send, at random i.e. 50% chance, through either one of the slits. This will result in no interference pattern on the receiving screen. This is a maximally mixed state, and has no information content.

A quantum superposition sends the particle through both slits at once, and this will produce an interference at the screen. I'm using the language "both slits at once" because we physicists are raised on these sort of language, and there's really no way of getting around it. Bohr himself likes to say that we're suspended by words. This state can be used to transmit information; say one guy modulates the positions of the slit, and so the resulting fringes seen by another guy at the screen modulates as well, and the information is contained in the modulation. Of course, this modulation will ultimately be limited by the speed of the particles, which is limited by the speed of light. Being a pure state, this means that the fringe contrast is perfect, so the information is transmitted optimally.

This suggests a fundamental difference between classical probabilities and quantum probabilities; the latter has phase, can interfere, and produce deterministic outcomes.

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  • $\begingroup$ This is an interesting answer! Can you provide some references? $\endgroup$ – gen Apr 20 at 16:47
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There's an equivalence between the two cases, where they both can be studied and represented using Pauli-matrices, which are the generators of the SU(2) group (which is a mathematical equivalence).

However, physically, every case represents a different system. The first system could be a multi-body system with a many electrons that are 50/50 polarized up and down, while the second could be a single electron, whose quantization axis isn't along its polarization axis, and let's say it's perpendicular to it, and that's how you get the superposition that gives you also a 50/50 result, where the electron can show up as being oriented to up and down in a superposition of the two states.

So notice that in the first system you had a mixture of particles/states in a single container. So BOTH states exist. While in the second case you had a single object being measured, and due to the probabilistic nature of Quantum Mechanics, you're getting that 50/50.

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There are ways to distinguish these two states.

For example, suppose we apply some kind of potential to these systems so that over a period of time they go through the unitary transformation

$|\psi_1\rangle \rightarrow (|\psi_1\rangle+|\psi_2\rangle)/\sqrt{2}$

$|\psi_2\rangle \rightarrow (|\psi_1\rangle-|\psi_2\rangle)/\sqrt{2}$

(Eg. you could implement this by applying a an RF field to a spin-1/2 particle in a magnetic field as in an NMR device.)

If you now measure the energy for the first system you have a 50/50 chance of getting $E_1$ or $E_2$. But the second system will give energy $E_1$.

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Quantum mechanics has a strict mathematical formulation in eigenstates of certain mathematical equations expressed in complex numbers. This means that there exist phases between the different solutions, and these phases are constant in time. A superposition of these eigenstates to form a new eigenstate retains the phases between the two psis.

Edit, as my answer was confusing.

Often the superposition of many states where the complete quantum mechanical solution is known is called a mixed state. In this mixed state the phases (angular information of the wavefunctions) are known and a density matrix connecting the different solutions has off diagonal elements which keep the phases between the entangled wave functions.

Mixed and superposition are two ways of describing the same physical situation.
The mixed/superposed_states density matrix is describing a coherent state. If all the off diagonal elements of the matrix are zero within measurement accuracies the many particle state is incoherent and the wave functions are not entangled.

In a sense there are two types of superpositions, one type is where a total boundary conditions obeying solution to the problem exists, and this is approximated by a density matrix where the phases are retained , and of superpositions where the density matrix is diagonal and the individual wavefunctions are within measurement errors independent of each other, i.e. measuring quantities for particle A does not affect/change the wavefunction and quantities that may be measured of particle N. Mixed is used mostly for the first meaning of superposition, for a total quantum mechanical state.

$$\rho_{mn}= \sum_i p_i\langle u_m|\psi_n\rangle\langle \psi_n|u_n\rangle= \langle u_m|\hat{\rho}|u_n\rangle$$

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  • $\begingroup$ Can you tell me the meaning of "phase", please? $\endgroup$ – user36790 Jun 28 '15 at 18:19
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    $\begingroup$ have a look en.wikipedia.org/wiki/Phase_%28waves%29 . The wave function gets its name because it has mainly sinusoidal expressions. $\endgroup$ – anna v Jun 28 '15 at 18:44
  • $\begingroup$ Ok, then can I ask why in superposition the "phases" are retained? Meant to say what do you mean by the statement above? $\endgroup$ – user36790 Jun 28 '15 at 18:54
  • $\begingroup$ @user36790 above was for your question. I edited my answer and changed it because it was confusing the issues $\endgroup$ – anna v Jun 28 '15 at 19:03
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    $\begingroup$ particle A from particle N will control whether the superposition of the individual wave functions is entangled/coherent or not. If it is coherent, it is called mixed. If it is incoherent it is just a superposition of independent wavefunctions which is the classical limit. $\endgroup$ – anna v Jun 29 '15 at 5:11
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Inspired by Wang Yun's comment, I read through of J.J. Sakurai's "Modern Quantum Mechanics", p174-176. I think that his usage of the phrases 'mixed ensemble'/'pure ensemble' are more appropriate (and less confusing) than 'mixed state'/'pure state'. Using the phrase 'pure state' leads to confusion with the phrase 'superposition of states'.

A couple paraphrased examples from his book:

Ex. 1:

A mixed state is like having a high school graduating class that is 50% male and 50% female. When we pick a student at random, the probability of being male (or female) is 0.5. A quantum superposition would be like a student who is a coherent linear superposition of being both male and female.

Ex. 2:

Consider an oven emitting silver atoms. The atoms can either be spin up or down. There is no preferred direction and hence the atoms are unpolarized, a "random ensemble". If we now send the beam of atoms through a Stern-Gerlauch experiment, we'd expect the beam to be split into its two spin states. If we select one of those beams, we would have a 'pure ensemble' (Sakurai uses 'ensemble' instead of 'state'). The beam would then be polarized.

Now if we took another Stern-Gerlauch experiment which is rotatable, and passed our polarized beam through it, the intensities of the two output beams (from the second S-G experiment), would vary as the second S-G experiment is rotated.

The difference between the 'pure ensemble' and 'random ensemble' is illustrated because rotating a Stern-Gerlauch experiment on an 'random ensemble' input yields output polarized beams of constant and equal intensity, regardless of the angle it is rotated.

The 'pure ensemble' (the polarized beam) will have an angle where the output beams from the rotatable S-G experiment will be 0.

The 'random ensemble' and the 'pure ensemble' are the two extremes of what is known as a 'mixed ensemble'. It is important to note that an ensemble is a collection of physical systems (i.e. multiple particles). A 'mixed ensemble' can be thought of a mixture of 'pure ensembles'.

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I also find this to be confusing. However, I think that the Wikipedia "Quantum State" explanation of the difference is less confusing than the Wikipedia "Density matrix" explanation.
It states that the mathematical difference between the two is that the trace of the density matrix of a pure state is 1, but the trace of the density matrix of a non-pure mixed state is less than one.

The issues of first preparing and second measuring pure versus non-pure mixed states adds further complexity.

The quantum superposition can be a pure state, but I think you can also prepare mixtures of two different quantum superpositions,

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    $\begingroup$ The trace of a density matrix is always one. Otherwise the probabilities of measuring the possible values that an observable may take won't add up to one. $\endgroup$ – Adomas Baliuka Nov 2 '16 at 3:58

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