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My question is really straightforward. Can we define a mixed state $\varphi$ to be a linear combination of pure states of the form: $$\varphi = \sum_{n=1}^{\infty}p_{n}\psi_{n}$$ where each $\psi_{n}$ is pure and $\sum_{n=1}^{\infty}p_{n} = 1$?

Usually mixed states are defined as positive, trace-class operators of the form: $$\varphi = \sum_{n=1}^{\infty}p_{n}|\psi_{n}\rangle\langle\psi_{n}|$$ but I think we can always map the projection $|\psi_{n}\rangle \langle \psi_{n}|$ to the state $\psi_{n}$ itself, right? But then, what is the interpretation of these $p_{n}$'s? Is it the probability of the system to be in the state $\psi_{n}$, since we don't have full information about its precise state? But then, since the $\psi_{n}$'s are not necessarily orthogonal, we don't necessarily have $|\langle \varphi, \psi_{n}\rangle| = p_{n}$, so I am a bit confused.

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    $\begingroup$ Your fist equation does not correspond to a mixed state. In the second equation, you need that $p_n\geq 0$, too. Regarding the other point: It depends a bit on the context how to interpret the $p_n$. In general, the convex decomposition of density operators into pure states is not unique. And as you point out, it is generally not the case that these are probabilities to find, upon a suitable measurement, the system in the state $\psi_n$. $\endgroup$ Oct 18, 2023 at 15:52
  • $\begingroup$ One operational definition is that a density operator encapsulates a physical state preparation; for example, you might imagine a "machine", preparing the system with probability $p_n$ in the state $|\psi_n\rangle\langle \psi_n|$; then if you do (repeated) experiments using this device, the correct state you have to use for the calculations is $\rho=\sum\limits_n p_n |\psi_n\rangle\langle \psi_n|$. But then again, this decomposition is in general far from unique, and a density operator can correspond to many different ensembles. $\endgroup$ Oct 18, 2023 at 15:56

2 Answers 2

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  1. The first equation is a pure state, not a mixed one (and also the constraint is $\sum_n |p_n|^2 = 1$).

  2. The second state is generically mixed (with $\sum_n p_n =1$).

  3. If we are being strict with our language, your first $\varphi$ is a pure state whereas your second $\varphi$ is a mixed density matrix.

  4. All states can be mapped to density matrices via $$ |\psi \rangle \mapsto |\psi \rangle \langle \psi | $$ Only pure density matrices can be mapped to states (by definition). Mixed density matrices cannot be mapped to states.

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    $\begingroup$ Thanks! So, basically a mixed state cannot be represented as a vector of the underlying Hilbert space? Only as an operator? $\endgroup$
    – MathMath
    Oct 18, 2023 at 19:49
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    $\begingroup$ yes. that's the crucial point. $\endgroup$
    – Prahar
    Oct 18, 2023 at 20:07
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In general you seem to be confused about probability amplitudes versus probabilities. See below for further discussion.

My question is really straightforward. Can we define a mixed state $\varphi$ to be a linear combination of pure states of the form: $$\varphi = \sum_{n=1}^{\infty}p_{n}\psi_{n}$$ where each $\psi_{n}$ is pure and $\sum_{n=1}^{\infty}p_{n} = 1$?

Given that each of the $\psi_n$ is a state vector then your $\varphi$ above is a (potentially unnormalized) pure state. (It is only a properly normalized state if only one of the $p_n=1$ and all others are zero.)

Normally one would write down a normalized pure state as $$ \varphi = \sum_{n=1}^{\infty}a_{n}\psi_{n}\;, $$ where $$ \sum_n|a_n|^2 = 1 $$

Usually mixed states are defined as positive, trace-class operators of the form: $$\varphi = \sum_{n=1}^{\infty}p_{n}|\psi_{n}\rangle\langle\psi_{n}|$$

Here, with the condition $\sum_n p_n=1$ in combination with the condition $p_n\ge 0$, your above equation for $\varphi$ would define a density operator.

Whether the state is "pure" or "mixed" depends on the values of the $p_n$. If only one $p_n=1$ and all others are zero then the state is pure, otherwise it is mixed.

but I think we can always map the projection $|\psi_{n}\rangle \langle \psi_{n}|$ to the state $\psi_{n}$ itself, right?

Yes. Sure.

But then, what is the interpretation of these $p_{n}$'s?

You seem to be confusing the probabilities $p_n$ with the probability amplitude $a_n$. If you literally make the linear combination: $$ \sum_n p_n \psi_n\;, $$ then the $p_n$ are just real coefficients of some improperly normalized pure state (unless only one of the $p_n$ is one, in which it is normalized).

Is it the probability of the system to be in the state $\psi_{n}$,

No, $p_n$ is generally not the probability to be in $\psi_n$ when $\varphi = \sum_n p_n \psi_n$.

Again, you are probably getting confused between probability and probability amplitude. It would only make sense to interpret the $p_n$ as an amplitude if only one is non-zero (and equal to one), in which case the sum is trivial.

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