# von Neumann measurement model of a qubit with continuous detector

I have a two state qubit system with initial state $$|\psi_s\rangle_i = a|0\rangle+b|1\rangle$$ and a detector with initial state $$|\psi_d\rangle_i = \int_{-\infty}^{\infty}\left(N \exp[-\frac{q^2}{2\sigma^2}+ik q]\right)^{\frac{1}{2}}|q\rangle dq$$ where $$N$$ is the constant such that $$|\psi_d\rangle_i$$ is a normalized state. The combined initial state can be written as $$|\psi\rangle_i = |\psi_s\rangle_i |\psi_d\rangle_i \, .$$ The interaction Hamiltonian is $$H/\hbar = -g (\sigma_z \otimes P)$$, where $$P$$ is the momentum operator. Therefore the unitary operator acting on $$|\psi\rangle_i$$ is $$U=\exp[i g T (\sigma_z \otimes P)]$$. After the interaction, the system state becomes $$|\psi \rangle_T = U|\psi\rangle_i \, .$$ The von Neumann measurement model says that we should perform a projective measurement on the detector and then perform partial trace over detector in order to get the system state after measurement.

What would be the appropriate measurement operator for the projective measurement over detector?

By linearity, $$$$U\vert\psi_d\rangle_i\Big(a\vert 0\rangle+b\vert 1\rangle\Big)=a\vert\psi, 0\rangle_T+b\vert\psi, 1\rangle_T$$$$ where, $$$$\vert\psi, \sigma\rangle=U\vert\psi_d\rangle_i\vert\sigma\rangle$$$$ So to get the final state we need to consider how $$U$$ acts when the qubit is in the state with definite spin.

When the spin is definite you may substitute $$\sigma_z$$ with $$\pm 1$$ depending on thebqubit spin. But then the evolution operator is simply a shift operator for $$q$$ in one direction or another.

The initial state of the detector is a gaussian wavepacket concentrated near zero. We can denote it as, $$$$\vert\psi_d\rangle_i=\vert q\approx 0\rangle$$$$ Then the resulting state of the combined system may be written as, $$$$a\vert q\approx -gT\rangle\vert 0\rangle + b\vert q\approx +gT\rangle\vert 1\rangle$$$$ Imagine that $$q$$ denotes the position of the arrow on the dial of your detector. Now this position is entangled i.e. correlated with the spin of the qubit. I.e. all you need is to measure $$q$$. The simplest observable that will give you only two possible answers is the sign of $$q$$ - negative for $$0$$ and positive for $$1$$ that can be written as, $$$$\widehat{\mathrm{sgn}(q)}=-\int\limits_{-\infty}^0dq\vert q\rangle\langle q\vert+\int\limits_0^{+\infty}dq\vert q\rangle\langle q\vert$$$$

However note that, $$$$\langle q\approx -gT\vert q\approx+gT\rangle \neq 0$$$$ This implies that no matter what observable you measure for your detector therebalways be an error in your measurement of the qubit spin. After tracing out the detector the resulting density matrix of the qubit will not be, say $$\vert 0\rangle\langle 0\vert$$, but this will be just an approximation. The larger $$gT$$ is compared to $$\sigma$$ the better. When you take this nonzero overlap into acvount in terms of the qubit your measurement will not be ideally projective but will be described with POVM formalism.

• Thanks for your reply. This type of measurement is known as weak measurement instead of projective measurement. Now the question is what is your measurement operator which tells about the sign of the detector state. How would you write this kind of measurement operator? – Parveen Oct 5 '18 at 7:36
• @Parveen I thought that should be rather obvious but ok. I've added it. Hopefully you'll make the calculations of effects of its measurement on the density matrix of the qubit yourself. – OON Oct 5 '18 at 8:43
• @Parveen and please note that I never talked about the "sign of the detector state" but about the sign of $q$ that allowsbus to differentiate the states of the detector associated with the states of qubit – OON Oct 5 '18 at 8:51
• @ONN: Thanks for your reply. I am getting it now what you are trying to say. After the interaction, the combined state of the system will be $$|\psi\rangle_{int} = \int_{-\infty}^{\infty} a|0\rangle \left(N \exp[\frac{-(q-gT)^2}{2\sigma^2}-i k(q-gT)]\right)^{1/2}|q\rangle dq + \int_{-\infty}^{\infty} b|1\rangle \left(N \exp[\frac{-(q+gT)^2}{2\sigma^2}-i k(q+gT)]\right)^{1/2}|q\rangle dq$$ Now I am still confused about how would the operator $sgn(q)$ will act on the above detector state. Will you please specify this as well. I know this must be very simple but I am a beginner in this. – Parveen Oct 7 '18 at 7:32
• @Parveen It changes the sign of the wavefunction for the negative $q$ and leaves it the same for positive $q$. E.g. the function $\psi(q)=q$ will become $\psi(q)=|q|$. Note that discontinuities are allowed – OON Oct 7 '18 at 20:23