# Understanding of measurement in quantum mechanics?

I have a computer science background with basically zero physics background. I am trying to gain a 'high-level' understanding of quantum mechanics to aid me in some computer science work.

Is my understanding correct here regarding a quantum register in a quantum computer -

When a quantum register is read, the superpositioned states 'collapse' into a single state and we only know the probability that a certain outcome has occurred.

I.e. if we have a state $|\psi \rangle= a|0\rangle + b|1\rangle$, then the process of reading a quantum register returns 2 values: The state $|0\rangle$ or $|1\rangle$, and the probability that this is the 'correct' state, which is calculated by taking $\sqrt{|a|^2 + |b|^2}$.

I would like to be precise in my understanding, I'm not looking for 'thats kind of right', if it is incorrect I would appreciate it if you could tell me exactly what I have wrong.

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No, you can still only read 1 value out - $a$ and $b$ tell you the probability of either being read; unless either $a$ or $b$ are zero, the readout is probabilistic.

[Caveat: I assume by 'read' we are talking about non-quantum results, i.e. not entangling with the qubit(s)]

We have a single qubit in the state:

$a|0\rangle + b|1\rangle$

The probabilities are then:

$P(0) = |a|^2 \\ P(1) = |b|^2$

Since these are the only 2 possible states for a qubit, we know that $|a|^2+|b|^2 = 1$

The process of reading memory in such a manner extracts the information from it, collapsing the superposition. After a read, then, the memory is back to either $a=0, b=1$ or $a=1, b=0$.

Note that $a$ and $b$ are complex numbers, so there is an entire Bloch sphere continuum of state space for each qubit.

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So if you read the register and you get a $|0>$ where do you get the probability information from? I would have imagined that the register would have to supply you with the values of $a$ and $b$? –  sonicboom Apr 7 '14 at 11:32
@sonicboom the probability information comes from actually konwing the state that your qubit or whatever was in before measuring it. In practice, how one does this is by preparing many identical systems and performing identical measurements, which will allow you to statistically infer what state you are probing. –  Danu Apr 7 '14 at 12:17
...alternatively, one can try to force your system into a certain state via clever operations on e.g. the spin of a particle or some other property that can be macroscopically controlled. –  Danu Apr 7 '14 at 12:19
@Danu So as PhotonicBoom says, we just keep running measurements until we are satisfied we have reached accurate values for $a$ and $b$? –  sonicboom Apr 7 '14 at 12:35
Yes, or the alternative method I just outlined. –  Danu Apr 7 '14 at 12:39

If you have these two states, when you take a measurement then the probability to obtain state $|0\rangle$ is $|a|^2$ and the probability of observing state $|1\rangle$ is $|b|^2$, thus its is required that $|a|^2 + |b|^2 = 1$. That means the system is definitely in one of these 2 states.

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So what is the return value of the probability register? Is it the values $a$ and $b$? –  sonicboom Apr 7 '14 at 11:31
The register will return the state, not the probability. You get the probability by doing many measurements. So if repeat the measurement 10 times and you get 4/10 times $|0>$ then $|a|^2 = 2/5$. –  Constandinos Damalas Apr 7 '14 at 11:35
That seems a bit hackish does it not? How many times would we typically need to measure to be extremely confident of the probability? –  sonicboom Apr 7 '14 at 12:13
A lot! Unfortunately it is the only way I know of to measure probabilities. I'm sure there must be other ways out there but I'm not aware of them. –  Constandinos Damalas Apr 7 '14 at 12:17