I have a question about entangled state. Suppose I consider the entangled state $\frac{1}{\sqrt{2}}(|00\rangle + |11\rangle)$. I saw an argument for how measurement of the first bit is affected by whether or not the second bit is measured because this is an entangled state in an article I am reading. The argument is clear to me and I describe it below, so please keep reading:

If I measure the first bit without first measuring the 2nd bit, then the probability that I get $|0\rangle$ is 1/2. This is clear to me. If I measure the 1st bit after measuring the 2nd bit, the probability that it is $|0\rangle$ is 0 or 1 depending on whether the measured value of the 2nd bit was $|0\rangle$ or $|1\rangle$. This is clear to me too. But what I am confused about is this:

If I measure the first bit after the second bit, this is what I would calculate the probability of measuring the first bit being $|0\rangle$ to be: $\frac{1}{2}\times 1 + \frac{1}{2}\times 0 = \frac{1}{2}$. The first term is the prob. of measuring the second bit as $|0\rangle$ and then the 1st bit as 0 and the 2nd term is the prob of measuring the 2nd bit as 1 and then measuring the 1st bit as zero. If the probability of measuring the first bit as 0 is to same regardless of whether the 2nd bit was measured or not is unchanged, what is the effect of the entangled state? Or am I missing something?

  • $\begingroup$ HINT: There are other local things you can do with a qubit, other than just measuring its state. $\endgroup$ – Ali Jun 17 '13 at 9:26

First of all, the correctly normalized entangled state is $$ \frac{1}{\sqrt{2}}(|00\rangle + |11\rangle)$$ i.e. the factor $1/2$ instead of $1/\sqrt{2}$ in the original wording of the question gives a non-normalized state.

Second of all, when we're only interested in measurements in which nothing else than the measurement of the first bit itself; and the second bit itself is ever being measured, then the interpretation of the state above is completely simple. It's a superposition that assigns $p=1/2$ to the possibility that both bits are $0$, $p=1/2$ that both bits are measured as one, and $p=0$ to the possibilities that the bits are different ($10$ or $01$).

It doesn't matter at all which of the bits is measured first. In fact, in typical realizations of the two qubits, the two qubits don't interact at all once they're separated in the aforementioned entangled state. In many cases, we can make them spacelike-separated which means that, because of the limitations of special relativity, they can't interact even in principle. So of course that one of the two measurements won't influence the other.

An effect of the entangled state is that the measured (and predicted) values of the spin are correlated. In this case, the values of the spin are the same even though this shared value of the qubits isn't determined prior to the measurement (it can be both $0$ or $1$). So they can't be treated as independent bits whose values objectively exist prior to the measurement.

Already this simple entanglement can't be reproduced by any classical theory. This is not clear if we measure just the value of the bits themselves because we could assume that the state of the two qubits was objectively $00$ or was objectively $11$ even though we may have been ignorant about which one it was. However, this assumption about the objective nature of the state as $00$ or $11$ prior to the measurement may be disproved if we measure some different quantities. For example, if the qubits above are interpreted as $J_z$ components of the two spin-1/2 particles' spin, the measurements of $J_x$ done on both particles will also be perfectly correlated (the probabilities of $00,01,10,11$ would still be $1/2,0,0,1/2$) while the classical theory assuming an objective "up-up" state (or "down-down" state, but only one of them) would predict that the probabilities of measuring $00,01,10,11$ for the signs of $J_x,J'_x$ would be $1/4,1/4,1/4,1/4$. No classical theory can guarantee these relatively high correlations that hold regardless of the type of measurements we choose. These limitations of any (local) classical theories are known as Bell's inequalities. They may be proven to hold in all candidate local (because relativistic) classical theories of the correlations; but they may be shown to be violated in Nature (and shown to be violated in quantum mechanics which agrees with Nature).

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    $\begingroup$ thanks! I get it! also, thanks for pointing out that i forgot the square root sign for the normalization factor. i've edited the question. $\endgroup$ – user34801 Jun 17 '13 at 9:49

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