# Interpretation of qubit

1. Mar 16, 2015

This is undoubtedly one of the more basic (not to say stupid) questions ever to appear in this rubric. But here goes:
Suppose you have v= 0.6|0> + 0.8|1>. This means that there is 36% chance of measuring |0> and 64% chance of measuring |1>. Otherwise put, given a million particles in this state, around 360 thousand will be measured as |0> and around 640 thousand will be measured as |1>. So far so good.
Now, however, I am attempting to visualize the particle as containing a collection of |0>'s and |1>'s, so that upon measurement, one of these will be selected. From the above, it would then seem that, if the total number were taken to be a million, this visualization would be 360 thousand |0>'s and 640 thousand |1>'s. On the other hand, it would seem that the original statement that v= 0.6|0> + 0.8|1> would indicate that out of 1.4 million, there would be 600 thousand |0>'s and 800 |1>'s. This is obviously wrong, but why?

2. Mar 16, 2015

### atyy

The qubit is a special case, and may be able to avoid the Kochen-Specker theorem against non-contextual hidden variables (I'm not sure, have to look it up).

However, in general, if you have more than one qubit, you cannot visualize it this way unless you have hidden variables. If you don't have hidden variables, a pure state is the most complete specification of a single quantum system. If you don't have hidden variables, you cannot associate a single quantum system in a pure state with an ensemble from which the outcome is selected upon measurement.

There is of course an ensemble associated with a pure state in that complete knowledge of a state only allows you to make probabilistic predictions, so taking a frequency interpretation of probability, one needs an ensemble to check the predictions of quantum mechanics. (One might get round this for applying quantum mechanics to the universe, but that's another discussion.)

Edit: Yes, it seems the qubit can avoid the Kochen-Specker theorem, and there are non-contextual hidden variable models for the qubit. https://books.google.com/books?id=EK0WskLzuJEC&source=gbs_navlinks_s (p47)

Last edited: Mar 16, 2015
3. Mar 16, 2015

### Strilanc

First, because it's incorrect to imagine that the zeroes and ones are there ahead of time. The state $3/5 \left|0\right\rangle + 4/5 \left|1\right\rangle$ is a *complete* description of the qubit, as far as quantum mechanics goes, with nothing left out. Trying to pretend that there's more information there will lead you into paradoxes.

Second, because unitary matrices preserve the square of the amplitudes instead of the amplitudes themselves. If you try to use a non-conserved quantity as the "real" amount of 0s and 1s, then you can increase that amount with non-operations like combining systems. For example, if you pair the qubits together so you have 500 thousand $0.36 \left|00\right\rangle + 0.48 (\left|01\right\rangle + \left|10\right\rangle) + 0.64 \left|11\right\rangle$ qubit pairs instead of a million $3/5 \left|0\right\rangle + 4/5 \left|1\right\rangle$ qubits, you will find that the amount of 0s has increased to $2 \cdot (0.36 + 0.48) \cdot 500000 = 1.68 \cdot 500000 = 840000 > 600000$ despite it being the same system.

4. Mar 16, 2015

thanks, atyy and StriLanc.
First, StriLanc: your first point is well taken: I was not actually assuming that there were determinate values before measurement, but merely attempting to rephrase the probabilities in terms of ordered sets as a conceptual guide. That is, make an isomorphism that would match a probability, say of one fourth, on a single vector to a set of vectors in the ratio of one to four. (When we go towards infinity, this would have to be rephrased with limits etc, but the intuitive idea would be the same.) But that would lose its conceptual appeal very quickly, as I see from your sample calculation.
atyy: at first, when I read your response, I thought your citation of the no-go theorem was making the same point as the first paragraph from StriLanc. But then given your follow-up that the no-go theorem can be circumvented, I am no longer too sure. I would be grateful if you could elaborate on the connection here, or perhaps give me a link that does not require me to order a new book (I do not have access to a good academic library.)

5. Mar 16, 2015

### atyy

Yes, I was making the same point as Strilanc. However, for the special case of one qubit, I think the Kochen-Specker theorem against non-contextual hidden variables can be circumvented. I think it cannot be circumvented for two qubits.

I cannot find online a counter-example for one qubit except that mention in the link above. However, in articles like http://arxiv.org/abs/1203.3091 and http://arxiv.org/abs/1105.1286 they use two spins, which I think is because the one qubit has a non-contextual hidden variable model. If I find the counterexample online, I'll post it here.

Edit: Maybe http://arxiv.org/abs/0705.0181, Is there contextuality for a single qubit? by Andrzej Grudka and Pawel Kurzynski

Edit: From the references in Grudka and Kurzynski's paper, I think the non-contextual hidden variable model for a single qubit is the one in Bell's 1966 paper http://www.ffn.ub.es/luisnavarro/nuevo_maletin/Bell (1966)_Hidden variables.pdf (p449, Eq 1 - 3).

Last edited: Mar 17, 2015
6. Mar 17, 2015

### martinbn

How did you get this? 36% of 1.4 million is 504 thousand and 64% of 1.4 million is 896 tousand!

7. Mar 17, 2015