Representing a Bell measurement on non adjacent qubits

I have a state $${|\psi\rangle} = s {\Bigl(|1\rangle_1|1\rangle_2-|0\rangle_1|0\rangle_2\Bigr)}\otimes{\Bigl(|0\rangle_3|1\rangle_4-|1\rangle_3|0\rangle_4\Bigr)}\otimes{\Bigl(|0\rangle_5|1\rangle_6-|1\rangle_5|0\rangle_6\Bigr)},$$ with the subindexes meaing the particle number (for some normalising factor $$s$$). How do we represent a measurement of qubit 1 and 5 in the Bell basis, given that these two qubits are not adjacent?

• Do you mean to ask how we would represent that measurement mathematically? Aug 23 '19 at 8:10
• yes sir do you group them together and then what next? Aug 23 '19 at 8:26

The most direct way to do this using normal notation is to simply write the Bell projections using the same convention for subscripts:$$\def\ket#1{\lvert#1\rangle}\def\bra#1{\langle#1\rvert}\def\idop{\mathbf 1}$$ \begin{aligned} \bra{\Phi^+}_{1,5} \;&=\; \tfrac{1}{\sqrt 2}\Bigl(\,\bra{0}_1\bra{0}_5 \,+\, \bra{1}_1 \bra{1}_5\,\Bigr), \\ \bra{\Phi^-}_{1,5} \;&=\; \tfrac{1}{\sqrt 2}\Bigl(\,\bra{0}_1\bra{0}_5 \,-\, \bra{1}_1 \bra{1}_5\,\Bigr), \\ \bra{\Psi^+}_{1,5} \;&=\; \tfrac{1}{\sqrt 2}\Bigl(\,\bra{0}_1\bra{1}_5 \,+\, \bra{1}_1 \bra{0}_5\,\Bigr), \\ \bra{\Psi^-}_{1,5} \;&=\; \tfrac{1}{\sqrt 2}\Bigl(\,\bra{0}_1\bra{1}_5 \,-\, \bra{1}_1 \bra{0}_5\,\Bigr). \end{aligned} It might not be immediately obvious why it makes sense to just 'decorate a bra with subscripts'. The idea is that for any qubit which is not indicated in the subscripts, what we have in mind is that we take the tensor product with the identity. So, for example, we have \begin{alignat}{2} \bra{0}_1\bra{0}_5 \;&:=\; &&\bra{0} \otimes \idop \otimes \idop \otimes \idop \otimes \bra{0} \otimes \idop, \\[1ex] \bra{1}_1\bra{1}_5 \;&:=\; &&\bra{1} \otimes \idop \otimes \idop \otimes \idop \otimes \bra{1} \otimes \idop, \\[1ex] \bra{\Phi^+}_{1,5} \;&:=\; \tfrac{1}{\sqrt 2}\Bigl(\,&&\bra{0} \otimes \idop \otimes \idop \otimes \idop \otimes \bra{0} \otimes \idop \\[-1.5ex]&&&+ \bra{1} \otimes \idop \otimes \idop \otimes \idop \otimes \bra{1} \otimes \idop\,\Bigr), \end{alignat} and so on. The advantage of writing things this way with the subscripts is that if you decompose an operator such as $$\bra{\Phi^+}$$, $$\bra{\Psi^-}$$, and so forth as a sum of tensor products, you can then compose it with other operators without having to write all of the identity tensor factors, so long as you only compose together operators acting on the same tensor factor. For instance: \begin{aligned} \bra{\Phi^+}_{1,5} \ket{\psi}& \\[1ex] = \tfrac{1}{\sqrt2}\Bigl[& \bra{0}_1\bra{0}_5\ket{\psi} + \bra{1}_1 \bra{1}_5 \ket{\psi} \Bigr] \\[2ex] = \tfrac{\textstyle s}{\sqrt2}\Bigl[& \bra{0}_1\bra{0}_5 {\Bigl(|1\rangle_1|1\rangle_2-|0\rangle_1|0\rangle_2\Bigr)\Bigl(|0\rangle_3|1\rangle_4-|1\rangle_3|0\rangle_4\Bigr)\Bigl(|0\rangle_5|1\rangle_6-|1\rangle_5|0\rangle_6\Bigr)} \\&+\bra{1}_1\bra{1}_5 {\Bigl(|1\rangle_1|1\rangle_2-|0\rangle_1|0\rangle_2\Bigr)\Bigl(|0\rangle_3|1\rangle_4-|1\rangle_3|0\rangle_4\Bigr)\Bigl(|0\rangle_5|1\rangle_6-|1\rangle_5|0\rangle_6\Bigr)}\Bigr] \\[1ex] = \tfrac{\textstyle s}{\sqrt2}\Bigl[& {\Bigl(\langle0|1\rangle_1|1\rangle_2-\langle0|0\rangle_1|0\rangle_2\Bigr)\Bigl(|0\rangle_3|1\rangle_4-|1\rangle_3|0\rangle_4\Bigr)\Bigl(\langle0|0\rangle_5|1\rangle_6-\langle0|1\rangle_5|0\rangle_6\Bigr)} \\&+ {\Bigl(\langle1|1\rangle_1|1\rangle_2-\langle1|0\rangle_1|0\rangle_2\Bigr)\Bigl(|0\rangle_3|1\rangle_4-|1\rangle_3|0\rangle_4\Bigr)\Bigl(\langle1|0\rangle_5|1\rangle_6-\langle1|1\rangle_5|0\rangle_6\Bigr)}\Bigr] \\[1ex] = \tfrac{\textstyle s}{\sqrt2}\Bigl[& {-|0\rangle_2\Bigl(|0\rangle_3|1\rangle_4-|1\rangle_3|0\rangle_4\Bigr)|1\rangle_6} - {|1\rangle_2\Bigl(|0\rangle_3|1\rangle_4-|1\rangle_3|0\rangle_4\Bigr)|0\rangle_6}\Bigr] \\[2ex] = -\tfrac{\textstyle s}{\sqrt2}\Bigl(& {|0\rangle_2|1\rangle_6 + |1\rangle_2|0\rangle_6\Bigr) \Bigl(|0\rangle_3|1\rangle_4-|1\rangle_3|0\rangle_4}\Bigr). \end{aligned} Three things about this answer:

• This calculation has effectively removed qubits 1 and 5 from the state — we're basically describing a destructive measurement. If we wanted to describe a projective measurement which doesn't destroy the qubits involved, we'd have to take a tensor product with the state $$\ket{\Phi^+}_{1,5}$$. We can certainly write out what that superposition would look like, simply by expanding $$\ket{\Phi^+}_{1,5} = \tfrac{1}{\sqrt 2}\bigl[ \ket{0}_1 \ket{0}_5 + \ket{1}_1 \ket{1}_5 \bigr]$$, and including this in the description of the post-measurement state.

• I've moved the tensor factors of qubits 2 and 6 so that they are adjacent. It's not difficult to see that they are in a tensor product with qubits 3 and 4, in any case, and it doesn't matter if we change the order given that we're using the subscripts to keep track of which qubit is which — just like listing some variables in different orders doesn't change what values those variables have. This allows us to see that after the measurement, the states of qubits 2 and 6 are in the state $$\ket{\Psi^+}$$ is the measurement result on qubits 1 and 5 was $$\ket{\Phi^+}$$ (and similar things will happen for the other possible outcomes of the measurement).

• Notice that the normalisation is affected. We would normally chose $$s$$ so that the original vector had a norm of $$1$$, so for example $$s = 1/2\sqrt{2}$$. This new state has a leading factor of $$s/\sqrt{2} = 1/4$$, so it is not difficult to see that it is sub-normalised. Specifically, the vector here has norm $$1/2$$, which corresponds to the fact that a Bell measurement on qubits 1 and 5 will produce the state $$\ket{\Phi^+}$$ with probability $$1/4 = (1/2)^2$$.

• when you are describing about the Bell projections shouldn't $|\Psi^+\rangle$ be $|\Phi^+\rangle$ ?? Aug 23 '19 at 10:37
• Typo fixed, thanks. Aug 23 '19 at 10:42
• First of all,thanx for taking the pain and wiring out such a crystal clear answer. just adding to the 2nd point of your answer, that would also tell that qubits $3$ and $4$ are in state $|\Psi^-\rangle_{34}$ Aug 23 '19 at 10:52
• Sure, qubits 3 and 4 are in the state $\lvert \Psi^-\rangle$. Actually, they are to begin with as well; and as we are not doing anything to either of them, they remain in this state. Aug 23 '19 at 10:57
• I have one doubt sir, why have you taken the $-$ sign outside, because if it is inside the first bracket, then it will correspond to a different bell state? How do we decide whether to leave that sign or not? Aug 23 '19 at 12:43