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$\newcommand{\ket}[1]{\left| {#1} \right> }$ I have no academic background in physics, but I'm attempting to study quantum computation.

I have read that a quantum system of two qubits is represented by normalized combinations of the basis $\left\{ \ket{00},\ket{01},\ket{10},\ket{11} \right\} $. A measurement was defined only for a single qubit of the two - that is, a state represented by $\ket{0}\otimes(a\ket{0}+b\ket{1})+\ket{1}\otimes(c\ket{0}+d\ket{1})$ can have its first qubit measured and collapse to either $\ket{0}\otimes(a\ket{0}+b\ket{1})$ or $\ket{1}\otimes(c\ket{0}+d\ket{1})$ with the appropriate probabilities, and a similar thing can be done to the second qubit.

The paper did say that the are other kinds of measurement, such as comparing the two qubits, but they are all equivalent to performing this type of measurement after a unitary transformation. From this I infer that the general representation of measurement is taken by splitting a basis $\mathcal{B}$ into two new bases $\mathcal{B}_1, \mathcal{B}_2$ of equal sizes, and then the state collapses to its projection to either $\mathcal{B}_1$ or $\mathcal{B}_2$ with appropriate probabilities.

What I don't understand is why do $\mathcal{B}_1$ and $\mathcal{B}_2$ have to be of equal sizes - for example, would it be possible for me to measure a system $a\ket{00}+b\ket{01}+c\ket{10}+d\ket{11}$ with the bases $\left\{\ket{00}\right\},\left\{\ket{01},\ket{10},\ket{11}\right\}$, so that it will collapse to $\ket{00}$ with probability $|a|^2$ and to $b\ket{01}+c\ket{10}+d\ket{11}$ with probability $|b|^2+|c|^2+|d|^2$?

If not, is there an explanation (for a non-physicist) as to why this is?

Thanks in advance.

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Care to reference the paper? I'm no expert on quantum information (though I do do quantum field theory), but I can't see any reason why the spaces have to be equal sizes. I don't think they do. It sounds like you are describing a PVM for which, I think, there are no restrictions on the dimension. –  Michael Brown Feb 23 '13 at 14:14
    
If a took a spin-1 particle (spin states +1,0,-1) and put it through a Stern-Gerlach apparatus and blocked the 0 and -1 components then that would be a counter example, wouldn't it? –  Michael Brown Feb 23 '13 at 14:15
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@MichaelBrown I am reading this paper, page 12, the second full paragraph. –  Alfonso Fernandez Feb 23 '13 at 14:24
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up vote 3 down vote accepted

Yes, in general physical systems, one can make measurements that split the basis unequally. Using normal physical language, we may say the same thing as follows: It is possible to measure an observable $J$ such that the spaces corresponding to the different possible eigenvalues of $J$ have different dimensions.

But they're not measurements of a "qubit". If the information is given in "qubits", that assumes that the size of the remaining information is independent on the result of the measurement of the "qubit", so the two sub-bases are equally large. In other words, we assume that the measured qubit is independent from the remaining ones. If it is independent, the Hilbert space is isomorphic to $H_{\rm measured}\otimes H_{\rm remains}$ and such a tensor product explicitly has the same dimension of the vectors with $b_{\rm measured}=0$ and $b_{\rm measured}=1$.

To give you an example of a measurement that collapses into one of the spaces of unequal dimensions, consider the measurement of the total angular momentum $J$ of a two-electron system where the angular momentum only arises from the spin. There are various bases on this 4-dimensional space such as $$|\rm up,up\rangle, |\rm up,down\rangle, |\rm down,up\rangle, |\rm down,down\rangle $$ but there is also another natural basis classifying the state according to the total $J^2$ i.e. $J$ and the projection $J_z$. These are the states written as $|J,J_z\rangle$: $$|0,0\rangle, |1,-1\rangle, |1,0\rangle, |1,1\rangle$$ The second and fourth state in the last displayed equation are "down down" and "up up" states, respectively, but the first and third state in the last displayed equation are the difference and the sum of "up down" and "down up", respectively.

Now, looking at the basis in the last displayed equation, we see that the measurement of $J$ either returns $0$ and "collapses" the state vector into a state in a one-dimensional Hilbert space, or it returns $J=1$ and "collapses" it into a state in a three-dimensional Hilbert space. Any other splitting of a sufficiently large Hilbert space dimension is possible, too: One can define observables (matrices) whose eigenstate subspaces split the original Hilbert space in any imaginable way. However, these measurements are usually not considered in quantum computation where we want to keep the independence of qubits.

This is analogous to the statement about classical computers where we could also find out whether a byte is a prime by "one measurement". However, classical computers usually doesn't have circuits for such direct "unequal" operations to start with. Instead, we do a sequence of classical computer operations (analogy of unitary transformations) and reduce the problem to a particular bit's being 0 or 1 at the end.

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+1 Confirmed my intuition and elaborated constructively –  Michael Brown Feb 23 '13 at 14:19
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