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If given a wave function which is an element of a Hilbert space $\mathcal{H}$ which is a tensor product of two other Hilbert spaces $\mathcal{H}_1\otimes\mathcal{H}_2$, how do you formulate measurement of the function in one of the sub-spaces, i.e. $\mathcal{H}_1$?

For example, say you have a system $\left|\psi\right>$ which consists of two qubits $\left|\psi_1\right>\otimes\left|\psi_2\right>$. The system undergoes a unitary operation, giving $\left|\psi'\right>=U\left|\psi\right>$. It should still be possible to measure just one of the two qubits, causing a partial collapse of the system.

If the two qubits are not entangled after $U$, then they can be separated as $\left|\psi'\right>=\left|\psi_1'\right>\otimes\left|\psi_2'\right>$. It seems clear that measuring i.e. $\left|\psi_1'\right>$ should not cause $\left|\psi_2'\right>$ to collapse.

If the qubits are partially entangled, it may still be possible for one of the qubits to be in some sort of superposition. For example, if $\left|\psi'\right>=\frac{\left|00\right>+\left|01\right>+\left|11\right>}{\sqrt{3}}$, then $\left|\psi_2'\right>$'s state depends on what $\left|\psi_1'\right>$ collapses to. If $\left|\psi_1'\right>\rightarrow\left|0\right>$, then could $\left|\psi_2'\right>$ still be in a superposition, namely $\frac{\left|0\right>+\left|1\right>}{\sqrt{2}}$?

To be more specific about the "formulation" part of my question, you can use e.g. $\left|\left<0\mid\psi_1'\right>\right|^2$ to find the probability that $\left|\psi_1'\right>$ collapses to $\left|0\right>$, assuming a normalized wavefunction. What would be the format for e.g. measuring a single qubit of $\left|\psi'\right>$? Would it be something like $\left(\left<0\right|\otimes\left<\psi_2'\right|\right)\left|\psi'\right>$?


Some insight I'm looking into: the expectation value of an operator $O$ is given by $\left<\psi\mid O\mid\psi\right>$. When $O$ is a projection operator, such as $\left|0\right>\left<0\right|$ (which projects to a $\left|0\right>$-basis), the expectation value is the probability that measurement yields that particular outcome. So perhaps partial measurement of a system involves projection to some subspace spanned by multiple basis states?

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  • $\begingroup$ Yes your intuition is correct. You can use these 'partial' scalar product. Since the notation can be confusing I suggest you add lowerscript as in e.g., $_{A}\langle \chi | \psi \rangle_{AB}$ if you measure system $A$. $\endgroup$ – lcv Jan 23 at 1:08
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In general, such "partial collapse" of a wavefunction cannot be expressed as the usual inner product (i.e. using $\left<0\mid\psi\right>$). Instead, you must use the expectation value of the corresponding projection operator. This works for "full collapse" to a single basis state, as the corresponding projection operator is just the outer product of the basis with itself ($\left|0\right>\left<0\right|$).

Proof:

Using the example I gave above, let $P$ be the operator that projects the first qubit to $\left|0\right>$. This can be represented by the matrix $\begin{bmatrix}1&0&0&0\\0&1&0&0\\0&0&0&0\\0&0&0&0\end{bmatrix}$. Suppose there exists some basis state $\left|B\right>$ that represents $\left|\psi_1'\right>=\left|0\right>$ while $\left|\psi_2'\right>$ is unaffected. $P$ can then be written as $\left|B\right>\left<B\right|$. If $\left|B\right>$ is given by the matrix $\begin{bmatrix}x\\y\\z\\w\end{bmatrix}$, then $\left|B\right>\left<B\right|=\begin{bmatrix}xx^*&xy^*&xz^*&xw^*\\yx^*&yy^*&yz^*&yw^*\\zx^*&zy^*&zz^*&zw^*\\wx^*&wy^*&wz^*&ww^*\end{bmatrix}$. Equating with the earlier matrix gives the following equations:

  • $z=w=0$
  • $\left|x\right|^2=\left|y\right|^2=1$
  • $xy^*=x^*y=0$

Which has no solution. Therefore, $\left|B\right>$ does not exist.

This means that the probability of measuring part of a system must be expressed using operator expectation, $\left<\psi\mid P\mid\psi\right>$.


It is possible, however, for $P$ to be expressed as a sum of outer products. In this case, $P=\left|00\right>\left<00\right|+\left|01\right>\left<01\right|$. This could be generalized to be a sum of projections to each basis of the subspace.

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  • $\begingroup$ Of course, you can expand it as a sum of overlap-squares. $\endgroup$ – Norbert Schuch Nov 8 '16 at 20:40

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