This is something standard, by I'm trying to redo this with spectral theory. Suppose we start with the usual postulates of quantum mechanics:

  1. States are unit rays on a separable Hilbert space. In particular they are described by unit vectors.

  2. Observables are hermitian operators acting on the Hilbert space.

  3. The possible values of an observable are the ones from its spectrum.

  4. If $A$ is an observable with associated projection-valued measure $\mathbb{P}_A$ given by the spectral theorem, its values on the state $\Psi$ are described by the probability measure $$\mu_A(E)=(\Psi,\mathbb{P}_A(E)\Psi),\quad \forall \text{ measurable $E\subset \sigma(A)$}$$

  5. If the measurement outcome lies in $E\subset \sigma(A)$, then the post-measurement state is the normalized projection $$\Psi'=\frac{1}{\|\mathbb{P_A(E)}\Psi\|}\mathbb{P}_A(E)\Psi$$

Now, if we assume one classical ensemble of quantum states $\{(p_i,\Psi_i)\}$ with probabilities $p_i$ we can show two things upon defining the density operator $\rho = \sum_i p_i \Psi_i(\Psi_i,\cdot)$:

  1. The probability measure, including the uncertainty in the state becomes $$\mu_{A}(E)=\operatorname{Tr}(\mathbb{P}_A(E)\rho)$$ This follows by classical probability, which states that $$\mu_A(E)=\sum_i p_i\mu_A^i(E)=\sum_i p_i (\Psi_i,\mathbb{P}_A(E)\Psi_i)$$ decomposing $\mathbb{P}_A(E)$ in one orthonormal basis the result follows.

  2. The mean values become $$\langle A\rangle = \operatorname{Tr}(A\rho)$$

    this follows directly observing that if $f : \sigma(A)\to \mathbb{C}$ is integrable against the measures $\mu_A$ and $\mathbb{P}_A$ then $$\int_{\sigma(A)} f(\lambda)d\mu_A(\lambda)=\operatorname{Tr}\int_{\sigma(A)}f(\lambda)d\mathbb{P}_A(\lambda)\rho$$

    and applying to $f(\lambda)=\lambda$ which gives the mean value.

So this line of thought is working fine, because these are the correct expressions. Now I wanted to derive similarly the post-measurement state when the result of the measurement lies in $E\subset \sigma(A)$.

My line of thought was: if the state of the system is $\Psi_i$ the post-measurement state is $\Psi_i'$ given by postulate (5). This will happen with probability $p_i$. So we get a new ensemble of states $\{(p_i,\Psi_i')\}$ with same probabilities.

If we assemble the density operator we have

$$\rho'=\sum_i p_i \Psi_i'(\Psi_i',\cdot)=\sum_i p_i \frac{1}{\|\mathbb{P}_A(E)\Psi_i\|^2}\mathbb{P}_A(E)\Psi_i(\mathbb{P}_A(E)\Psi_i,\cdot).$$

Since the projectors are hermitian we get

$$\rho'=\mathbb{P}_A(E)\sum_i p_i \frac{1}{\|\mathbb{P}_A(E)\Psi_i\|^2}\Psi_i(\Psi_i,\cdot)\mathbb{P}_A(E).$$

But now I cant proceed because of the denominator term. I mean, If I'm not mistaken, the result should be


What am I missing here? How can we arrive at the correct form for the post-selected density operator following the above reasoning?

  • 1
    $\begingroup$ I'm sorry, but isn't your notation/formalism mostly obscuring the question? As far as I can see, the question is: given I understand projective measurements on pure states, how can I derive the formula for mixed states, given the ensemble interpretation? (If not, it would be helpful if you could elucidate where this differs.) $\endgroup$ – Norbert Schuch Jan 10 '19 at 17:13

Your mistake is that you don't use that after the measurement, the ensemble for the post-measurement has reweighted probabilities. This is, if the probability for your outcome is $q_{i}$ on the state $\Psi_i$, then you have to construct the average with weights $p_i q_{i}$ times the post-measurement state for that case -- i.e., your ensemble is $(p_iq_i,\psi_i')$ times some joint normalization. This will exactly cancel the denominator you are worried about.

(P.S.: Feel free to rewrite this in your notation.)

  • $\begingroup$ thanks I got it. The probability for post-measurement state $\psi_i'$ is the probability that the prior-measurement state was $\psi_i$ and that the measurement gives the specific result. This means that the probabilities multiply $p_i q_i$. We further normalize this distribution dividing by $N = \sum p_i q_i$. The resulting $p_i' = p_iq_i/N$ are then the correct probabilities for the post-measurement ensemble right? $\endgroup$ – Gold Jan 10 '19 at 17:23
  • $\begingroup$ Exactly! _______ $\endgroup$ – Norbert Schuch Jan 10 '19 at 17:27
  • $\begingroup$ @user1620696 Why did you delete your last question after I answered it? $\endgroup$ – Norbert Schuch Jan 11 '19 at 9:16
  • $\begingroup$ because I really thought it wouldn't be useful for others. What was being claimed by my adviser was simply plain wrong and I think it was not such a usual mistake that everyone might step in sometimes. Neverthelss, if you think it's nice to keep it, I shall undelete it. $\endgroup$ – Gold Jan 11 '19 at 11:23
  • $\begingroup$ @user1620696 I think it is fair to undelete it. I think it could be helpful to others (even learning about different kind of counterexamples for entanglement with different properties), and it is fair to whoever took the effort to answer :) $\endgroup$ – Norbert Schuch Jan 11 '19 at 12:37

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