# About the definition of mixed states

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.

• 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 $\psi_n$. Oct 18, 2023 at 15:52
• 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. Oct 18, 2023 at 15:56

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.

• Thanks! So, basically a mixed state cannot be represented as a vector of the underlying Hilbert space? Only as an operator? Oct 18, 2023 at 19:49
• yes. that's the crucial point. Oct 18, 2023 at 20:07

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.