# Many-Worlds Interpretation and GHZ States

I'm working through a problem set, and I've come across the following problem:

In this problem, you'll explore something that we said in class about the Many-Worlds Interpretation of quantum mechanics: namely, that "two branches interfere with each other if and only if they produce an outcome that's identical in all respects." Consider the n-qubit "Schrodinger cat states" $$\frac{|0\cdots 0 \rangle + |1 \cdots 1 \rangle}{\sqrt2}$$ a) What probability distribution over n-bit strings do we observe if we Hadamard the first $$n-1$$ qubits, then measure all n qubits in the $$\{ |0 \rangle , |1 \rangle \}$$ basis?

b) Is this the same distribution or a different one than if we had applied the same measurement to the state $$\frac{|0 \cdots 0 \rangle \langle 0 \cdots 0 | + |1 \cdots 1 \rangle \langle 1 \cdots 1 |}{2}$$

c) What probability distribution over n-bit strings do we observe if we Hadamard all $$n$$ qubits, then measure all n qubits in the $$\{ |0 \rangle , |1 \rangle \}$$ basis?

d) Is this the same distribution or a different one than if we had applied the same measurement to the state $$\frac{|0 \cdots 0 \rangle \langle 0 \cdots 0 | + |1 \cdots 1 \rangle \langle 1 \cdots 1 |}{2}$$

I have solved the problem as follows:

a) Equal probability of seeing any n-bit string of qubits.

b) Different: that mixed state has a 50/50 shot of seeing $$|0 \cdots 0 \rangle$$ or $$|1 \cdots 1 \rangle$$

c) Equal probability of seeing any n-bit string of qubits that have an even number of $$|1\rangle$$s

d) Different. Same state as b.

What I don't understand is what this has to do with the Many-Worlds Interpretation! Could someone explain the significance of this exercise? Thanks!

• Why don't you ask your teacher and post the answer? (Or is this not from your class?) – Norbert Schuch Oct 11 '18 at 22:04
• One more comment: It seems you are mis-interpreting b (and d): The question asks what happens if you take that state, Hadamard n-1 bits, and measure -- "same measurement" here refers to the WHOLE scheme described in a, I'd say. (Otherwise, b and d would be exactly the same, which wouldn't make much sense.) Then you will NOT get only 0000 and 1111! – Norbert Schuch Oct 11 '18 at 22:08
• @NorbertSchuch I'm self teaching from lecture notes / problem sets I found online, so I can't ask the professor, sadly. Also OH that makes more sense for b and d – Joe Oct 11 '18 at 22:10
• In any case, if you want to learn quantum info, just ignore Many Worlds (at least for now). – Norbert Schuch Oct 11 '18 at 22:11
• They're from Scott Aaronson at UT Austin. Specifically from problem set 6. Link: scottaaronson.com/blog/?p=3943 – Joe Oct 11 '18 at 22:13

I should probably start by describing my philosophical standpoint: I would never talk about "many worlds" or some such. However, I certainly believe that it is possible that everything, including measurement, is unitary. That apparently makes me a many-worldian. It's not necessary to buy wholesale into a picture of diverging worlds. And I think this question demonstrates my (possible) view point quite well.

So, I think of measurement as being like a controlled-not, but because the measurement device is a large device, it's like targetting many qubits. So, if you've got a qubit in $$|0\rangle+|1\rangle$$ that you might be measuring, the post-measurement state would be $$\frac{1}{\sqrt{2}}\left(|000\ldots 0\rangle+|111\ldots 1\rangle\right)$$ (Aside: you might even consider yourself part of the measuring device. At that point, if you're in the 0 branch, you see results 0, and the you're in the 1 branch, you see 1 results). The issue is, if this is true, why do we end up describing something post-measurement using classical probabilities, i.e. $$\frac12\left(|000\ldots 0\rangle\langle000\ldots 0|+|111\ldots 1\rangle\langle 111\ldots 1|\right)?$$ The answer is basically, that, if you made sure that you did something with every quantum system that has become entangled as a result of the measurement, you would be able to see a difference between the quantum and classical descriptions (answer c$$\neq$$d). Miss just one qubit (which may be part of the measurement device, or somewhere else in the environment. there's so many that you'll always miss some), and all subsequent results are as if you had the classical probability distribution (answer a=b).

Hopefully you now see how that should correspond to the questions you're being asked. As other commenters have indicated, however, the measurement process that you're performing on the two different systems is the full combination of Hadamards+computational basis measurement.

• @Joe: This is about as good as it's going to get. Even as someone more sympathetic to MWI than to any other well-defined interpretation, I think the link of the exercise to the MWI is tenuous at best. The idea that "If you made sure that you did something with every quantum system that has become entangled [...], you would be able to see a difference between the quantum and classical descriptions [...]. Miss just one qubit [...] and all subsequent results are as if you had the classical probability distribution" is the most salient point I can see about this exercise in relation to the MWI. – Niel de Beaudrap Oct 16 '18 at 11:24
• This is a very good answer. Thank you. My only question is how are you getting that answer a = answer b? I'm getting different answers. – Joe Oct 16 '18 at 15:18
• @Joe given that there is one qubit that is not measured, if you trace out that qubit, then the state is the same in both cases. – DaftWullie Oct 16 '18 at 18:08

You are 100% correct that this question has nothing to do with the Everett interpretation (also known as "many-worlds interpretation") of quantum mechanics, and in fact I would even agree with your professor's description of the many-worlds interpretation that "two branches interfere with each other if and only if they produce an outcome that's identical in all respects."

The "many-world's interpretation" is an interpretation of wavefunction collapse in which if the wavefunction is in the state $$\frac{1}{\sqrt{2}}\left(|0\rangle + |1\rangle\right)$$ and you get $$|0\rangle$$ when you measure the state, there is a different parallel universe somewhere in which $$|1\rangle$$ was measured.

The picture right at the top of the many-worlds Wikipedia page depicts exactly this (the cat dies in one universe and survives in another):

What does it mean to be a "different" universe, if the universe is everything? Well Hawking talked about this in the first page of his book Brief History of Time. He defines our universe to be literally everything that is physically capable of having any influence on any measurement that anything in our universe ever makes. Therefore there is not much value in talking about other universes, because by definition we will not be able to make any measurements to test any theories we have about other universes, and if we could, then by definition they would be part of our universe (so not actually a different universe).

So certainly use the many-worlds interpretation to conceptualize wavefunction collapse if it helps you, or think of the Bohr interpretation or the Einstein interpretation (whatever helps you conceptualize wavefunction collapse). In all cases, the outcome of any measurements in our universe should not change, and that's why these are called interpretations and not theories (it's called "many-worlds interpretation" not "many-worlds theory").

So it was a trick question. Probabilities do not depend on the "interpretation" they use, they depend on the theory you use: classical mechanics, quantum mechanics (Schroedinger's equation), quantum electrodynamics, quantum chromodynamics, string theory, loop quantum gravity, or whatever you wish.

You are therefore correct, and may also want to point out to your professor that s/he is missing a square root symbol in the denominator, otherwise these states are not normalized and the probabilities given as the square moduli of the coefficients, will not add up to 1, which violates the definition of probability that all probabilities should add up to 1.

*The image is courtesy of Christian Schirm who made the image himself and put it on Wikipedia with the Creative Commons license.

• Actually, the missing square root was my error! I edited it in the post – Joe Oct 11 '18 at 21:09
• I'm reasonably confident in my answers. Note that part (a) is just Hadamard-ing the first $n-1$ bits. What you said is my answer for part (c) though! – Joe Oct 11 '18 at 21:14
• I know. But you're not changing the LAST bit of $|0 \cdots 0 \rangle$ or $|1 \cdots 1 \rangle$. Thus we don't get the cancelling you describe. That will only happen if we hadamard all $n$ qubits. – Joe Oct 11 '18 at 21:16
• You are 100% correct, because |xxx0> cannot be added together with |xxx1>. You seem to know what you're doing. What are your thoughts on my answer about the rest? It is my hope that people coming here eventually "accept" some answer, otherwise I have failed in my goal of answering the question. – user1271772 Oct 11 '18 at 21:19
• I've read it and it makes sense. I'm reasonably comfortable with the ideas you have described. I'm hesitant to believe the exercise has no relation to the Many-Worlds Interpretation, since the problem was included specifically to illustrate a point about it. – Joe Oct 11 '18 at 21:22