# Distinguishing two quantum states practically

Suppose we have two states

$$|x\rangle = 1 |0\rangle + 0 |1\rangle$$

and $$|y\rangle = \sqrt{1-\epsilon^2} |0> + \epsilon |1>$$ where

say $\epsilon = 10^{-20}$

Can we distinguish between $|x\rangle$ and $|y\rangle$ practically?

How many times will we have to repeat the measurement experiment in order to be sure that $|x\rangle \neq |y\rangle$?

Because $|x\rangle = |0\rangle$ and $|y\rangle$ is very very very close to $|0\rangle$.

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For the state $|x\rangle$, if you repeat the same experiment with this initial state $N$ times, you never get $|1\rangle$ as the outcome. However, if you repeat the same experiment with $|y\rangle$ as the initial state, the probability to get $|1\rangle$ in each copy of the experiment is $\epsilon^2$. The results $|1\rangle$ will appear infrequently, distributed as in a Poisson process.

To prove that the coefficient in front of $|1\rangle$ is nonzero, you need to get the outcome $|1\rangle$ at least once. It will approximately occur after $1/\epsilon^2$ repetitions of the experiment, in your case $10^{40}$ repetitions (clearly, your numbers mean that $|x\rangle$ and $|y\rangle$ are indistinguishable in practice because $10^{40}$ is very large). The probability that you will never get $|1\rangle$ after many more experiments than $10^{40}$ is dropping to zero exponentially.

If you want to measure $\epsilon$ with a relative error $\delta$, you need roughly $1/\delta^2\epsilon^2$ copies of the experiment because the relative errors in a measured quantity goes down like $1/\sqrt{N}$.

One should use all these situations for emphasizing the main difference between classical and quantum physics: in classical physics, small changes of the state imply small changes in the things we can measure. In quantum physics, small changes of the state vector may bring large changes in the outcome of a single experiment, but the probability of such a large change is very small.

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yes, but then you have the problem that there is no reliable way to make 'copies' from quantum systems, and i'm not clear if there are ways to obtain good enough replicas (what is 'good enough'?) – lurscher Nov 16 '11 at 17:24
@Luboš Motl Generally how long does it take to perform 1 such experiment? – Pratik Deoghare Nov 16 '11 at 17:31
Dear @Pratik, there's obviously no "general" prescription for how long time a measurement takes. Different experiments are different. Many of them can be fast but it is obviously nonsense to invent a general duration, like 2 nanoseconds, for "every" experiment. – Luboš Motl Nov 16 '11 at 17:47
Dear @lurscher, "yes, but then you have the problem..." - I personally don't have any problem related to these words. If you can't produce the exact same experimental situation arbitrarily many times, then you can't accurately measure any properties of the wave function describing the situation. The wave function simply isn't an observable, in the technical - and also informal - sense. This proposition isn't a "problem"; instead it is one of the "most fundamental postulates and deepest insights of [quantum] physics". At most, it may be someone's psychological problems - see Psychology SE. – Luboš Motl Nov 16 '11 at 17:49
Dear @Pratik, thanks for your interest. Quantum measurements end up being mundane things. For example $|0\rangle$ and $|1\rangle$ above refer to the electron with spin up or down. You send the electron into a magnetic field that deflects it up or down depending on the spin (the electron is a small magnet), and then you detect it BEEP up or down (after a second, or any amount of time). That's the measurement - a sort of destructive one. Most quantum measurements are "destructive"; to say the least, every measurement in quantum mechanics changes something about the state of the measured object. – Luboš Motl Nov 16 '11 at 17:57