Measurement Devices are not Reversible

In the Quantum Measurements: What Are They? lesson, we introduced measurement devices and noted that we will avoid calling measurement devices gates because they are not (in almost all circumstances) represented by matrices that satisfy the conditions of a quantum gate. In particular, they are not reversible. For example, after a photon interacts with a polarization state analyzer, its state might be $|\text{vlp}⟩$. From that information, we know very little about the state describing the photon before it arrived at the device. So, there is no way to apply the measurement operation to get back to the original state.

State analyzers can, however, be represented by “projection operators” constructed as products of right vectors and left vectors. For example, the operation of a state analyzer that puts the qubit in the device’s $\ket0$ state is represented by:

$$M_0 = \ket{0} \bra{0}.$$

If a state $|S⟩ = a_0 \ket0 + a_1 \ket1$ interacts with (is acted upon by) the device, we write the following:

$$M_0 \underbrace{\ket{S}}_\text{{input state}} = \overbrace{\underbrace{\ket{0}}_{\text{ output state}} \bra{0}}^\text{projection operator} {S\rangle}.$$

The output state (consistent with the measurement postulate) is the basis state $\ket0$. The probability of getting that result is the square of $\braket{0|S}$, which we recognize as the amplitude $a_0$ of $\ket S$ associated with the basis state $\ket0$. If the qubit is subsequently detected, we have completed a quantum observation.

As we discussed in the State Amplitudes and Probabilities lesson, the probability is found experimentally by repeating the state preparation–measurement cycle many times. But from a single observation, which is the important point here, we don’t know anything about the amplitude except that it is not $0$.

What happens if we apply $M_0$ to the state on the right side of the above equation?

$$M_0 \underbrace{\ket{0}}_{\text{ state after 1st M}_0} \braket{0|S} = \ket{0} \underbrace{\braket{0|0}}_{=1}\braket{0|S} = \ket{0}\braket{0|S}.$$

For the last equality in the above equation, we used $\braket{0|0}=1$ since the basis state is a unit vector. We see that we don’t get back to the original state $\ket S$. Some authors like to say the measurement “collapses the state” and that collapse is irreversible. I prefer to say that the state analyzer device projects the state of the incoming qubit along one of the device’s basis states. Since there are many (in fact infinitely many) different incoming states that get projected to the same output state, the measurement device projection is not reversible.