## Friday, October 26, 2018

### Interfere and it didn't happen

I am a bit late for the party, but also wanted to share my two cents on the paper "Quantum theory cannot consistently describe the use of itself" by Frauchiger and Renner. After sitting down and working out the math for myself, I found that the analysis in this paper and the blogpost by Scot (including many of the the 160+ comments, some by Renner) share a lot with what I am about to say but maybe I can still contribute a slight twist.

#### Coleman on GHZS

My background is the talk "Quantum Mechanics In Your Face" by Sidney Coleman which I consider as the best argument why quantum mechanics cannot be described by a local and realistic theory (from which I would conclude it is not realistic). In a nutshell, the argument goes like this: Consider the three qubit state state

$$\Psi=\frac 1{\sqrt 2}(\uparrow\uparrow\uparrow-\downarrow\downarrow\downarrow)$$

which is both an eigenstate of eigenvalue -1 for $\sigma_z\otimes\sigma_z\otimes\sigma_z$ and an eigenstate of eigenvalue +1 for $\sigma_x\otimes\sigma_x\otimes\sigma_z$ or any permutation. This means that, given that the individual outcomes of measuring a $\sigma$-matrix on a qubit is $\pm 1$, when measuring all in the z-direction there will be an odd number of -1 results but if two spins are measured in x-direction and one in z-direction there is an even number of -1's.

The latter tells us that the outcome of one z-measurement is the product of the two x-measurements on the other two spins. But multiplying this for all three spins we get that in shorthand $ZZZ=(XXX)^2=+1$ in contradiction to the -1 eigenvalue for all z-measurments.

The conclusion is (unless you assume some non-local conspiracy between the spins) that one has to take serious the fact that on a given spin I cannot measure both $\sigma_x$ and $\sigma_z$ and thus when actually measuring the latter I must not even assume that $X$ has some (although unknown) value $\pm 1$ as it leads to the contradiction. Stuff that I cannot measure does not have a value (that is also my understanding of what "not realistic" means).

#### Fruchtiger and Renner

Now to the recent Nature paper. In short, they are dealing with two qubits (by which I only mean two state systems). The first is in a box L' (I will try to use the somewhat unfortunate nomenclature from the paper) and the second in in a box L (L stands for lab). For L, we use the usual z-basis of $\uparrow$ and $\downarrow$ as well as the x-basis $\leftarrow = \frac 1{\sqrt 2}(\downarrow - \uparrow)$  and $\rightarrow = \frac 1{\sqrt 2}(\downarrow + \uparrow)$ . Similarly, for L' we use the basis $h$ and $t$ (heads and tails as it refers to a coin) as well as $o = \frac 1{\sqrt 2}(h - t)$ and $f = \frac 1{\sqrt 2}(h+f)$.  The two qubits are prepared in the state

$$\Phi = \frac{h\otimes\downarrow + \sqrt 2 t\otimes \rightarrow}{\sqrt 3}$$.

Clearly, a measurement of $t$ in box L' implies that box L has to contain the state $\rightarrow$. Call this observation A.

Let's re-express $\rightarrow$ in the x-basis:

$$\Phi =\frac {h\otimes \downarrow + t\otimes \downarrow + t\otimes\uparrow}{\sqrt 3}$$

From which one concludes that an observer inside box L that measures $\uparrow$ concludes that the qubit in box L' is in state $t$. Call this observation B.

Similarly, we can express the same state in the x-basis for L':

$$\Phi = \frac{4 f\otimes \downarrow+ f\otimes \uparrow - o\otimes \uparrow}{\sqrt 3}$$

From this once can conclude that measuring $o$ for the state of L' one can conclude that L is in the state $\uparrow$. Call this observation C.

Using now C, B and A one is tempted to conclude that observing L' to be in state $o$ implies that L is in state $\rightarrow$. When we express the state in the $ht\leftarrow\rightarrow$-basis, however, we get

$$\Phi = \frac{f\otimes\leftarrow+ 3f\otimes \rightarrow + o\otimes\leftarrow - o\otimes \rightarrow}{\sqrt{12}}.$$

so with probability 1/12 we find both $o$  and $\leftarrow$. Again, we hit a contradiction.

One is tempted to use the same way out as above in the three qubit case and say one should not argue about contrafactual measurements that are incompatible with measurements that were actually performed. But Frauchiger and Renner found a set-up which seems to avoid that.

They have observers F and F' ("friends") inside the boxes that do the measurements in the $ht$ and $\uparrow\downarrow$ basis whereas later observers W and W' measure the state of the boxes including the observer F and F' in the $of$ and $\leftarrow\rightarrow$ basis.  So, at each stage of A,B,C the corresponding measurement has actually taken place and is not contrafactual!

#### Interference and it did not happen

I believe the way out is to realise that at least from a retrospective perspective, this analysis stretches the language and in particular the word "measurement" to the extreme. In order for W' to measure the state of L' in the $of$-basis, he has to interfere the contents including F' coherently such that there is no leftover of information from F''s measurement of $ht$ remaining. Thus, when W''s measurement is performed one should not really say that F''s measurement has in any real sense happened as no possible information is left over. So it is in any practical sense contrafactual.

To see the alternative, consider a variant of the experiment where a tiny bit of information (maybe the position of one air molecule or the excitation of one of F''s neutrons) escapes the interference. Let's call the two possible states of that qubit of information $H$ and $T$ (not necessarily orthogonal) and consider instead the state where that neutron is also entangled with the first qubit:

$$\tilde \Phi = \frac{h\otimes\downarrow\otimes H + \sqrt 2 t\otimes \rightarrow\otimes T}{\sqrt 3}$$.

Then, the result of step C becomes

$$\tilde\Phi = \frac{f\otimes \downarrow\otimes H+ o\otimes \downarrow\otimes H+f\otimes \downarrow\otimes T-o\otimes\downarrow\otimes T + f\otimes \uparrow\otimes T-o \otimes\uparrow\times T}{\sqrt 6}.$$

We see that now there is a term containing $o\otimes\downarrow\otimes(H-T)$. Thus, as long as the two possible states of the air molecule/neuron are actually different, observation C is no longer valid and the whole contradiction goes away.

This makes it clear that the whole argument relies of the fact that when W' is doing his measurement any remnant of the measurement by his friend F' is eliminated and thus one should view the measurement of F' as if it never happened. Measuring L' in the $of$-basis really erases the measurement of F' in the complementary $ht$-basis.

## Wednesday, October 17, 2018

### Bavarian electoral system

Last Sunday, we had the election for the federal state of Bavaria. Since the electoral system is kind of odd (but not as odd as first past the post), I would like to analyse how some variations (assuming the actual distribution of votes) in the rule would have worked out. So, first, here is how actually, the seats are distributed: Each voter gets two ballots: On the first ballot, each party lists one candidate from the local constituency and you can select one. On the second ballot, you can vote for a party list (it's even more complicated because also there, you can select individual candidates to determine the position on the list but let's ignore that for today).

Then in each constituency, the votes on ballot one are counted. The candidate with the most votes (like in first past the pole) gets elected for parliament directly (and is called a "direct candidate"). Then over all, the votes for each party on both ballots (this is where the system differs from the federal elections) are summed up. All votes for parties with less then 5% of the grand total of all votes are discarded (actually including their direct candidates but this is not of a partial concern). Let's call the rest the "reduced total". According to the fraction of each party in this reduced total the seats are distributed.

Of course the first problem is that you can only distribute seats in integer multiples of 1. This is solved using the Hare-Niemeyer-method: You first distribute the integer parts. This clearly leaves fewer seats open than the number of parties. Those you then give to the parties where the rounding error to the integer below was greatest. Check out the wikipedia page explaining how this can lead to a party losing seats when the total number of seats available is increased.

Because this is what happens in the next step: Remember that we already allocated a number of seats to constituency winners in the first round. Those count towards the number of seats that each party is supposed to get in step two according to the fraction of votes. Now, it can happen, that a party has won more direct candidates than seats allocated in step two. If that happens, more seats are added to the total number of seats and distributed according to the rules of step two until each party has been allocated at least the number of seats as direct candidates. This happens in particular if one party is stronger than all the other ones leading to that party winning almost all direct candidates (as in Bavaria this happened to the CSU which won all direct candidates except five in Munich and one in Würzburg which were won by the Greens).

A final complication is that Bavaria is split into seven electoral districts and the above procedure is for each district separately. So there are seven times rounding and adding seats procedures.

Sunday's election resulted in the following distribution of seats:

After the whole procedure, there are 205 seats distributed as follows

• CSU 85 (41.5% of seats)
• SPD 22 (10.7% of seats)
• FW 27 (13.2% of seats)
• GREENS 38 (18.5% of seats)
• FDP 11 (5.4% of seats)
• AFD 22 (10.7% of seats)

Now, for example one can calculate the distribution without districts throwing just everything in a single super-district. Then there are 208 seats distributed as

• CSU 85 (40.8%)
• SPD 22 (10.6%)
• FW 26 (12.5%)
• GREENS 40 (19.2%)
• FDP 12 (5.8%)
• AFD 23 (11.1%)
You can see that in particular the CSU, the party with the biggest number of votes profits from doing the rounding 7 times rather than just once and the last three parties would benefit from giving up districts.

But then there is actually an issue of negative weight of votes: The greens are particularly strong in Munich where they managed to win 5 direct seats. If instead those seats would have gone to the CSU (as elsewhere), the number of seats for Oberbayern, the district Munich belongs to would have had to be increased to accommodate those addition direct candidates for the CSU increasing the weight of Oberbayern compared to the other districts which would then be beneficial for the greens as they are particularly strong in Oberbayern: So if I give all the direct candidates to the CSU (without modifying the numbers of total votes), I get the follwing distribution:
221 seats
• CSU 91 (41.2%)
• SPD 24 (10.9%)
• FW 28 (12,6%)
• GREENS 42 (19.0%)
• FDP 12 (5.4%)
• AFD 24 (10.9%)
That is, there greens would have gotten a higher fraction of seats if they had won less constituencies. Voting for green candidates in Munich actually hurt the party as a whole!

The effect is not so big that it actually changes majorities (CSU and FW are likely to form a coalition) but still, the constitutional court does not like (predictable) negative weight of votes. Let's see if somebody challenges this election and what that would lead to.

The perl script I used to do this analysis is here.

Postscript:
The above analysis in the last point is not entirely fair as not to win a constituency means getting fewer votes which then are missing from the grand total. Taking this into account makes the effect smaller. In fact, subtracting the votes from the greens that they were leading by in the constituencies they won leads to an almost zero effect:

Seats: 220
• CSU  91 41.4%
• SPD  24 10.9%
• FW  28 12.7%
• GREENS  41 18.6%
• FDP  12 5.4%
• AFD  24 10.9%
Letting the greens win München Mitte (a newly created constituency that was supposed to act like a bad bank for the CSU taking up all central Munich more left leaning voters, do I hear somebody say "Gerrymandering"?) yields

Seats: 217
• CSU  90 41.5%
• SPD  23 10.6%
• FW  28 12.9%
• GREENS  41 18.9%
• FDP  12 5.5%
• AFD  23 10.6%
Or letting them win all but Moosach and Würzbug-Stadt where the lead was the smallest:

Seats: 210

• CSU  87 41.4%
• SPD  22 10.5%
• FW  27 12.9%
• GREENS  40 19.0%
• FDP  11 5.2%
• AFD  23 11.0%