On Cluster States: Measuring in $|+rangle$ and $|-rangle$ basis vs. some other computational basis

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Based on your question, it feels as you might have some small misunderstandings, so allow me to (hopefully) clear them up.

Quantum measurements?

In its most simple form, a quantum measurement on a qubit is a 'check': you pick a basis for the qubit (i.e. two orthogonal vectors, e.g. $|0rangle$ and $|1rangle$, or $|+rangle$ and $|-rangle$). Then, the measurements 'checks' if the qubit's state is the one or the other basis vector. The measurement result is just a label that tells you which of the two it was: usually they're labelled with a $0$ or $1$, but this is arbitrary.

So how is this check performed, what are the 'rules'? Let's say I measure in the basis ${|psi_{1}rangle, |psi_{2}rangle}$. Since these two vectors form a basis, I can write any state of the qubit as $|psirangle = alpha|psi_{1}rangle + beta |psi_{2}rangle$. The basic rule of measurements is known as the Born rule: upon measurement of the state $|psirangle$ in the aforementioned basis, the measurement returns 'it was $|psi_{1}rangle$' with probability $|alpha|^{2}$, and the measurement returns 'it was $|psi_{2}rangle$' with probability $|beta|^{2}$.

The measurement always returns a classical value as the measurement result (usually this is just one bit indicating which of the two options it was). I wouldn't say, though, that any of these measurements are classical. There is not one basis which is more or less classical than any other.

Measurements in some specific bases

Now, some bases are more special than others.

• First and foremost, we have the computational basis: ${|0rangle,|1rangle}$. There is just one computational basis, and this is it. There are no other computational bases!

• The basis ${|+rangle, |-rangle}$ is a different basis. It is often called the Hadamard basis, because if you perform a Hadamard gate on the computational basis states, you get these basis states.

• The basis $|pmbeta (alpha)rangle = e^{ialpha/2}|0rangle pm e^{-ialpha/2}|1rangle$ is yet another basis. It often turns up in measurement based quantum computation, but a measurement in this basis is not any more or less 'quantum' than a measurement in any other basis.

How exactly do you dis-entangle a qubit?

Let's say I have two qubits, entangled, for instance in the state $|0rangle|0rangle + |1rangle|1rangle$ (up to a normalization factor). If I measure the second qubit in any basis, I force it to be a specific state (e.g. $|psi_{1}rangle$) of the basis I'm performing the measurement in. This means that I can now write the state as $|lrangle |psi_{1}rangle$, for whatever state $|lrangle$ on the first qubit. The two qubits are not entangled anymore, as I can write the state of either one without taking into account the state of the other. Any measurement will always dis-entangle qubits, as I force the qubit I'm performing the measurement on to be a specific basis state - so after the measurement I can always write down the state of that one qubit without any regard for any other.

How do you perform a measurement in any basis?

This, strictly speaking, depends on the specific physical implementation that you are trying to implement the experiment. But, normally speaking, measurements are possible in only the computational basis. If I want to measure in the basis $|psi_{1}rangle, |psi_{2}rangle$, I just first have to rotate from that basis to the computational basis. That is, I perform the unitary $U_{basis}$:

$$ U_{basis} = |0ranglelangle psi_{1}| + |1rangle langle psi_{2}| $$ which maps the state $|psi_{1}rangle$ to the state $|0rangle$, and the state $|psi_{2}rangle$ to the state $|1 rangle$.

In the specific case of a measurement in the $|pmbeta(alpha)rangle$ basis, the unitary can be implemented as first a $R_{z}(-alpha)$ rotation (mapping $|pmbeta(alpha)rangle$ to ${|+rangle, |-rangle}$), followed by a Hadamard operation (mapping ${|+rangle, |-rangle}$ to ${|0rangle, |1rangle}$).