# Mixed-State Density Matrices Versus Physical Ensembles

I am trying to follow a discussion (distinguishing the two sorts of systems mentioned in the title of this question) in Schlossahuer's book, Decoherence: And the Quantum-To-Classical Transition. Schlosshauer mentions that it is important to distinguish between two sorts of systems which are often (erroneously) considered to be equivalent. First there is a single system where we have an ignorance-interpretable mixture of pure states (that is, we are ignorant about the true state of the system) and so we represent this with a standard (mixed state) density operator. On the other hand, there is a system made up of $$N$$ identical subsystems where there are some well-defined fraction of the $$N$$ (sub)systems, $$p_iN$$, in a given state $$|\psi_i\rangle$$ so that the overall system is in the state $$|\psi\rangle = \Pi_{j=1}^{Np_1}|\psi_j\rangle \dots \Pi_{j=1}^{Np_M}|\psi_j\rangle$$.

My question is: how do we compute the ensemble average of a given single-system observable in this latter case? Probably some of my confusion lies in my not understanding what it means to measure a single system observable. Does the operator representing the observable have the form $$O \otimes O \otimes \dots \otimes O$$? If so then, using the trace/Born rule and evaluating the trace in the product basis I would obtain:

$$\langle O \otimes O \otimes \dots \otimes O \rangle = Tr(|\psi\rangle \langle \psi |O \otimes O \otimes \dots \otimes O) = \sum_{j_1=1}^M \dots \sum_{j_N=1}^M (\langle \psi_{j_1}|\otimes\dots \otimes\langle \psi_{j_N}|)|\psi\rangle \langle \psi|(O |\psi_{j_1}\rangle \otimes \dots \otimes O |\psi_{j_N}\rangle)$$ $$\stackrel{(1)}{=} (\Pi_{j=1}^{Np_1} \langle \psi_1|O |\psi_{1}\rangle)\dots \Pi_{j=1}^{Np_M} \langle \psi_M|O |\psi_{M}\rangle$$ where in (1) I've used that only one term of the nested sums (one permutation) leads to a nonzero inner product $$\langle \psi_{j_1}|\otimes\dots \otimes\langle \psi_{j_N}|)|\psi\rangle$$.

But this sort of calculation leads me nowhere near what Schlosshauer gets (below in equation 2.39, where Schlosshauer also divides by $$N$$ to additionally compute an average I think?). Does this mean I've used the wrong operator to represent the measurement? Should it be $$O \otimes I \otimes \dots \otimes I + I \otimes O \otimes \dots \otimes I + \dots + I \otimes I \otimes \dots \otimes O$$ instead?

I've attached a full picture of the section in case my description of the problem is not clear:

• In the ensemble viewpoint, single-system observables are observables of the form $\hat{I}\otimes\cdots\otimes\hat{I}\otimes\hat{O}\otimes\hat{I}\otimes\cdots\otimes\hat{I}$. He explicitly says that you must limit yourself to single-system measurements, which outlaws operators of the form you wrote down in your second paragraph. Is this the crux of your question? Commented Mar 10, 2023 at 21:05
• I think so @March. If possible, can you explain exactly how some sun of these single-system observables leads to an ensemble average as given in 2.39?
– EE18
Commented Mar 10, 2023 at 21:08

The author explicitly says that we must limit yourselves to single-system measurements on the ensemble in order to draw an equivalence between the statistical mixture picture and the ensemble picture.

Single-system observables are observables of the form $$\hat{I}\otimes\cdots\otimes\hat{I}\otimes\hat{O}\otimes\hat{I}\otimes\cdots\otimes\hat{I}$$, where $$\hat{I}$$'s are all identity operators on the individual systems. We want to measure the physical variable corresponding to the operator $$\hat{O}$$ on each subsystem and average the results. The corresponding operator for this measurement is $$\hat{M} = \frac{1}{N}\sum_{j=1}^N\hat{I}\otimes\cdots\otimes\hat{I}\otimes\underbrace{\hat{O}}_{j\textrm{'th spot}}\otimes\hat{I}\otimes\cdots\otimes\hat{I}\,.$$ Then, consider the state $$\lvert \Psi \rangle = \prod_{j=1}^N\lvert j:\psi_{n_j} \rangle\,,$$ where $$\lvert j:\psi_{n_j} \rangle$$ means that subsystem $$j$$ is in state $$\psi_{n_j}$$. The expectation value of the operator $$\hat{M}$$ in the state is \begin{align} \langle\hat{M}\rangle &= \langle\Psi\rvert\hat{M}\lvert \Psi\rangle \\&= \left(\prod_{l=1}^N\langle l:\psi_{n_l} \lvert\right) \left( \frac{1}{N}\sum_{j=1}^N\hat{I}\otimes\cdots\otimes\hat{I}\otimes\underbrace{\hat{O}}_{j\textrm{'th spot}}\otimes\hat{I}\otimes\cdots\otimes\hat{I} \right) \left(\prod_{k=1}^N\lvert k:\psi_{n_k} \rangle\right) \\&= \frac{1}{N}\sum_{j=1}^N \left(\prod_{l=1}^N\langle l:\psi_{n_l} \lvert\right) \left( \hat{I}\otimes\cdots\otimes\hat{I}\otimes\underbrace{\hat{O}}_{j\textrm{'th spot}}\otimes\hat{I}\otimes\cdots\otimes\hat{I} \right) \left(\prod_{k=1}^N\lvert k:\psi_{n_k} \rangle\right) \\&= \frac{1}{N}\sum_{j=1}^N \left(\langle j:\psi_{n_j} \lvert\prod_{l\neq j}\langle l:\psi_{n_l} \lvert\right) \left(\hat{O}\lvert j:\psi_{n_j} \rangle\prod_{k\neq j}\lvert k:\psi_{n_k} \rangle\right) \\&= \frac{1}{N}\sum_{j=1}^N \left(\prod_{l\neq j}\langle l:\psi_{n_l} \lvert\right) \left(\prod_{k\neq j}\lvert k:\psi_{n_k} \rangle\right) \langle j:\psi_{n_j} \lvert\hat{O}\lvert j:\psi_{n_j} \rangle \\&= \frac{1}{N}\sum_{j=1}^N \langle j:\psi_{n_j} \lvert\hat{O}\lvert j:\psi_{n_j} \rangle\,. \end{align} This expression is equal to the author's if you rearrange the sum to be over repeated single-particle states.

• @EE18 I can add more detail and explanation later if you request it. Commented Mar 10, 2023 at 21:24
• It's perfect, thank you.
– EE18
Commented Mar 10, 2023 at 23:14

If you write out the state in question explicitly, it would look like a direct product of N states as follows: $$|\Psi\rangle = \underbrace{|\psi_1\rangle|\psi_1\rangle\ldots|\psi_1\rangle}_{Np_1} \underbrace{|\psi_2\rangle|\psi_2\rangle\ldots|\psi_2\rangle\ldots}_{Np_2} \ldots \underbrace{|\psi_m\rangle|\psi_m\rangle\ldots|\psi_m\rangle}_{Np_m}$$

If you evaluate the operator $$\hat O$$ on the very first direct product state you get: $$\langle\psi_1|\hat O|\psi_1\rangle\;.$$ Note that this operator, as it acts on the entire space, looks like: $$\hat O\otimes \underbrace{1\otimes 1 \ldots \otimes 1}_{N-1}$$.

If you evaluate the operator $$\hat O$$ on the second direct product state you get: $$\langle\psi_1|\hat O|\psi_1\rangle\;,$$ where there is not a typo above, the index is still "1". Note that this operator, as it acts on the entire space, looks like: $$\hat 1\otimes \hat O \otimes \underbrace{1\otimes 1 \ldots \otimes 1}_{N-2}$$.

If you evaluate the operator $$\hat O$$ on the $$Np_1$$-th direct product state you get: $$\langle\psi_1|\hat O|\psi_1\rangle\;.$$ I.e., you get a total of $$Np_1$$ contributions of the form $$\langle\psi_1|\hat O|\psi_1\rangle$$ from the first $$Np_1$$ direct product states.

If you evaluate the operator $$\hat O$$ on the $$(Np_1+1)$$-th direct product state you get: $$\langle\psi_2|\hat O|\psi_2\rangle\;,$$ and you get a total $$Np_2$$ contributions of this form.

And so on.

Therefore when you do a standard average over all $$N$$ contributions, you find: $$\frac{1}{N}\left( \underbrace{\langle\psi_1|\hat O|\psi_1\rangle+\ldots+\langle\psi_1|\hat O|\psi_1\rangle}_{Np_1} + \underbrace{\langle\psi_2|\hat O|\psi_2\rangle+\ldots+\langle\psi_2|\hat O|\psi_2\rangle}_{Np_2} +\ldots + \underbrace{\langle\psi_m|\hat O|\psi_m\rangle+\ldots+\langle\psi_m|\hat O|\psi_m\rangle}_{Np_m} \right)$$ $$=\frac{1}{N}\left(Np_1\langle\psi_1|\hat O|\psi_1\rangle +Np_2\langle\psi_2|\hat O|\psi_2\rangle +\ldots +Np_m\langle\psi_m|\hat O|\psi_m\rangle \right)$$ $$=\frac{1}{N}\sum_{i=1}^m Np_i\langle\psi_i|\hat O|\psi_i\rangle$$ $$=\sum_{i=1}^m p_i\langle\psi_i|\hat O|\psi_i\rangle$$

• I've accepted the other answer but want to thank you as well for your answer. It helped me greatly.
– EE18
Commented Mar 10, 2023 at 23:14
• You're welcome.
– hft
Commented Mar 10, 2023 at 23:18