3
$\begingroup$

There is something i don't totally get about the density operators when we are in the mixed case.

Let's say I have this operator :

$$ \rho=\sum_i q_i | \psi_i \rangle \langle \psi_i | $$

I am doing a measurement of an observable $A$ and I find $a_i$ as result of measurement.

I know that my state will be projected in the eigenspace associated to $a_i$.

Thus my density matrix at the end should be :

$$ \rho'= \sum_i q_i \frac{P_j |\psi_i \rangle \langle \psi_i | P_j}{q_{j|i}}$$

Where $q_{j|i}=Trace(P_j |\psi_i \rangle \langle \psi_i |)$.

But the "problem" with this is that I can't consider a global operator applied on my density matrix during this process. Indeed I have to do a normalisation operation corresponding to the $q_{j|i}$ in the denominator.

In a pure case however I can say that $\rho'=\frac{P_j \rho P_j}{q_j}$ : I have a "global" operation applied on my density matrix without using any decomposition.

However, in this link : http://www-bcf.usc.edu/~tbrun/Course/lecture17.pdf on page 10, the seem to say that it is possible to apply a global operator on the density matrix in a mixed state, they write :

$$ \rho \rightarrow \sum_i \frac{p_{i|j}}{p_{j|i}}P_j |\psi_i \rangle \langle \psi_i |P_j = \frac{P_j \rho P_j}{p_j} $$

Same result on the wikipedia page : https://en.wikipedia.org/wiki/Density_matrix

I don't understand how we end up with such result. What is the mathematical trick to get the result ? I would like a deconnected explanation from http://www-bcf.usc.edu/~tbrun/Course/lecture17.pdf because I am really confused by what they do. The best would be a short proof in an answer.

Remark : I posted this message months ago but I edited my question right now because I still don't understand. However the comment written are probably not consistent anymore.

$\endgroup$
  • $\begingroup$ they are averaging out the expectation value for each pure state giving each a weight $$\frac{p_{i|j}}{p_{j|i}}$$ $\endgroup$ – Bardeen Feb 28 '18 at 10:58
  • $\begingroup$ @001 I don't understand what you mean precisely. For example for me the term should be $\frac{q_i}{p_{j|i}}$, where $q_i$ is the statistical probability (not quantum) to have the state $|\psi_i\rangle$. And $p_{j|i}$ is the quantum probability to be in the state $j$ after measurement when we prepared the state $|\psi_i\rangle$ at the beginning. $\endgroup$ – StarBucK Feb 28 '18 at 11:07
  • $\begingroup$ the two psis are going to become pho multiplied by $P_j $ in two sides and the sum of $\frac{1}{p_{j|i}}$ is $p_j$ $\endgroup$ – Bardeen Feb 28 '18 at 11:32
  • $\begingroup$ you are correct qi here is taken to be $$\ p_{i|j}$$ $\endgroup$ – Bardeen Feb 28 '18 at 11:38
  • $\begingroup$ Sum of qi/pji is pj $\endgroup$ – Bardeen Feb 28 '18 at 11:49
1
+50
$\begingroup$

So, using the measurement postulate for pure quantum states, you derived the formula:

$$\sum_i q_i \frac{P_j |\psi_i\rangle\langle\psi_i|P_j}{p_{j|i}} \tag{1}$$

where $p_{j|i} = \|P_j |\psi_i\rangle \|^2$ is the probability of measuring result $j$ starting from the state $|\psi_i\rangle$. And from this you would like to deduce the known formula:

$$\frac{P_j \rho P_j}{p_j} \tag{2}$$

where $p_j = \sum q_i p_{j|i}$ is the overall probability of measuring result $j$. But that doesn't work! If you check some examples, you will see that in general expressions $(1)$ and $(2)$ are not equal.

So $(1)$ is not correct, but why? Apparently you did everything right. You started from a statistical superposition of quantum states $|\psi_i\rangle$ with probabilities $q_i$. After the measurement the state $|\psi_i\rangle$ becomes $|\psi'_i\rangle = \frac{P_j |\psi_i\rangle}{\sqrt{p_{j|i}}}$. So we simply end up with the statistical superposition of the $|\psi'_i\rangle$, right?

Well, yes, but for the subtlety that the classical statistical probabilities have changed due to the measurement! By reading out the result $j$, we have gained information about our initial mix of quantum states and we need to update our weights $q_i \rightarrow q'_i$ to reflect this. For example, in the extreme case where $P_j |\psi_i\rangle = 0$ for some of the $i$, we know the system cannot have been in one of those quantum states to begin with, or we wouldn't have measured $j$ (and the fact that $(1)$ fails to be normalized in this case is an indication that something is not right indeed).

So, to understand better what is going on, and to find out what the correct $q'_i$ are, let us imagine that we repeat the experiment a large number $N$ of times, always preparing our initial mixed state in the same way: this is, after all, what probabilities are about. In $N_i = q_i N$ of the experiments, the initial quantum state is $|\psi_i\rangle$. Among those, the result $j$ will be measured in $N_{i,j} = q_i p_{j|i} N$ experiments after which the state will be in $|\psi'_i\rangle$. Now, we discard all experiments in which a result different from $j$ has been obtained. We are left with only $N'=\sum_i N_{i,j} = p_j N$ experiments, among which the portion of experiments that started in the state $|\psi_i\rangle$ and are now in the state $|\psi'_i\rangle$ is $\frac{N_{i,j}}{N'} = \frac{q_i p_{j|i}}{p_j}$. That's our $q'_i$!

Now, if in $(1)$ we replace $q_i$ by this $q'_i$ we do arrive at $(2)$.

$\endgroup$
1
$\begingroup$

Note that the result must be independent of any decomposition of $\rho$ into a mixture of pure states since the latter is in principle unobservable.

The correct formula for the most general filtering operation is $\rho\to P_j\rho P_j^*/p_j$ with probability $p_j=Tr(P_j\rho P_j^*)$, where the $P_j$ form a system of operators satisfying $\sum_j P_j^*P_j=1$.

In the case of a von Neumann measurement one has orthogonal projections $P_j$ to eigenspaces, where $P_j^*=P_j=P_j^2$; this is just a special case.

In the pure case where in addition $\rho=\psi\psi^*$ with $\psi$ of norm 1, this reduces to the projection of the wave function to the eigenspaces.

$\endgroup$
  • $\begingroup$ Thanks but I would like to have a proof of the formula in the case of a mixt state. I don't see how we can end up with this formula in the mixt state $\endgroup$ – StarBucK Oct 9 '18 at 18:33
  • $\begingroup$ @StarBucK: Well, I take it as a definition of what it means to perform a general physical operation (filtering or measuring) on a general state, and then get the pure case as a corollary. Note that von Neumann measurements are only a tiny class of possible measurements. If you accept the projection postulate for pure states as an axiom, it tells you very little about more general situations, whereas accepting my simple formula as an axiom tells the whole story for each case. Then no proof is needed, for the same reason why you accept the projection postulate without proof. $\endgroup$ – Arnold Neumaier Oct 10 '18 at 13:55
  • $\begingroup$ Note also that my approach has much less technical baggage - you don't even need to know the spectral theorem! $\endgroup$ – Arnold Neumaier Oct 10 '18 at 13:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.