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Let's assume following two density matrices are corresponding to the A and B in the Stern-Gerlach apparatus bellow (I know Stern-Gerlach is a more of a physics experiment but I think it can equally be considered a quantum computing problem):

$\rho_A=\frac{1}{2}\begin{pmatrix}1&1\\1&1\end{pmatrix}$

$\rho_{B}=\frac{1}{2}\begin{pmatrix}1&-1\\-1&1\end{pmatrix}$

We also have the following two projection operators:

$p_{z+}=\begin{pmatrix}1&0\\0&0\end{pmatrix}$

$p_{z-}=\begin{pmatrix}0&0\\0&1\end{pmatrix}$

In such case how should we calculate and combine the density matrices to find out the very last Z device output?

It seems to me that the result has to be $\begin{pmatrix}1&0\\0&0\end{pmatrix}$, but I don't know how this calculation has to be done so that it takes into account the interference, in a coherent way!

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By definition a density matrix is given by:

$$\rho= \sum p_i |\psi_i \rangle \langle \psi_i|$$

Where there is a probability $p_i$ of finding the state to be $|\psi_i\rangle$. Now if we are "combining" two quantum states represented by $\rho_A= \sum p_{A_i} |\psi_i \rangle \langle \psi_i|$ and $\rho_B= \sum p_{B_i} |\psi_i \rangle \langle \psi_i|$ then the "combined" state will be found in state $|\psi_i\rangle$ with probability $\frac{p_{A_i}+p_{B_i}}{2}$. This follows from directly from basic probability. And so the combined density matrix will be given by:

$$\rho_{A+B}= \sum \frac{p_{A_i}+p_{B_i}}{2} |\psi_i \rangle \langle \psi_i|$$

It seems that you already somewhat realize this, but the analysis that follows from what you laid out obtains the wrong result, and this is because the projection operators you defined are not accurate for what is happening. We do not in fact have two independent quantum states $\rho_a$ and $\rho_b$ being "combined" but what is happening is a little more subtle.

From the input of the $X$ measurement to the input of $Z$ measurement we do indeed have $P_{z+}$ and $P_{z-}$, but because we never collapse our state and just recombine them after the $Z$ measurement, the net effect of that gate and the recombination is given by $P_{z+}+P_{z-}=\mathbb{1}$. And so the objects you defined in the question as $\rho_a$ and $\rho_b$ never actually exist. Infact at the input of the $Z$ measurement our state is given by:

$$\big(P_{z+}+P_{z-} \big)| z+\rangle=| z+\rangle$$

And correspondingly we measure $100\%$ of spins to be positive in the $z$ basis

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  • $\begingroup$ Thanks. A couple of questions though: 1: I might be wrong but I think you meant the X gate is neutral rather than Z, that is $P_{x+}+P_{x-}=1$, am I right? 2: Then how come if we do the calculation using state vectors instead of density matrices and projections then it works out fine without taking into consideration the point you mentioned? That is $|\uparrow\rangle=\frac{1}{\sqrt{2}}(|\rightarrow\rangle+|\leftarrow|)=\frac{1}{\sqrt{2}}[\frac{1}{\sqrt{2}}(|\uparrow\rangle+|\downarrow\rangle)+\frac{1}{\sqrt{2}}(|\uparrow\rangle-|\downarrow\rangle)]=|\uparrow\rangle$ Am I missing something? $\endgroup$
    – al pal
    Sep 9 at 17:28
  • $\begingroup$ Also is it that $(P_{x+}+P_{x-})|z+\rangle\neq(P_{z+}|z+\rangle)+(P_{z-}|z+\rangle)$? $\endgroup$
    – al pal
    Sep 9 at 18:40
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Perhaps I am misunderstanding the question, but I don't agree that you get one measurement result with 100% probability.

Let me run through the calculation first with pure states (as much as I can!).

We start on the left of the diagram, and when the particles pass through the first measurement device, we are selecting the $|z+\rangle$ state by blocking the $|z-\rangle$ output. We express $$ |z+\rangle=\frac{1}{\sqrt{2}}(|x+\rangle+|x-\rangle) $$ so that tells us that if we measure using the $x$ basis (the next step), the outcomes are 50:50. Let us now recombine the two outputs so that we cannot tell which measurement result we got. We're forced to already use the density matrix formalism at this point. The output from the $x$ measurement (and recombining of paths) is $$ \frac12|x+\rangle\langle x+|+\frac12|x-\rangle\langle x-|=I/2. $$ This is the state that is input to the final $Z$ measurement on the right-hand side. Clearly, you get 50:50 outcomes.

The density matrix analysis is exactly the same. We start with $$ \rho=P_{z+}. $$ The probabilities of measurement outcomes $x\pm$ are $\langle x\pm|\rho|x\pm\rangle=\frac12$ and the corresponding outcome states are $P_{x\pm}$, and so the net state after we've recombined the two outcomes is $$ \frac12(P_{x+}+P_{x-})=I/2 $$ and analysis of the final measurement is exactly the same as before because it's the same formalism.

If you look, for example, at Figure 6.9 in http://physics.mq.edu.au/~jcresser/Phys301/Chapters/Chapter6.pdf this agrees with my analysis.

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  • $\begingroup$ Thanks for the answer, but isn't the Equation (6.33) in the document you linked stating that if no observation is performed (that is the results are combined without measurement) then the probability of observing $|\uparrow\rangle$ after the very last gate is 1 and the probability of observing $|\downarrow\rangle$ is 0? $\endgroup$
    – al pal
    Sep 10 at 16:15
  • $\begingroup$ Oh, I agree that the answer is completely different if you don’t make the x measurement. But this is not what I understood from your question, so perhaps some clarification is required if that’s what you intended. $\endgroup$
    – DaftWullie
    Sep 11 at 17:50
  • $\begingroup$ Yes that was my question... why happens when x measurement is not performed. But anyway thanks for the link, it was very helpful. $\endgroup$
    – al pal
    Sep 13 at 18:16
  • $\begingroup$ If no $X$ measurement is performed, the calculation is very straightforward. The output from the first $Z$ measurement is the input to the second $Z$ measurement and you don't have to think about $X$ bases at all. $\endgroup$
    – DaftWullie
    Sep 14 at 7:38

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