Controlled-measurement of a quantum register

Question

Given a state vector $\left[\alpha,\beta,\gamma,\delta\right]$ which is not known a priori, does there exist an operation, which I will call "controlled-measurement", which results in the ensemble

$\left[\alpha,\sqrt{|\beta|^2+|\delta|^2},\gamma,0\right]$

with probability ${|\beta|^2}\over{|\beta|^2+|\delta|^2}$ and

$\left[\alpha,0,\gamma,\sqrt{|\beta|^2+|\delta|^2}\right]$

with probability ${|\delta|^2}\over{|\beta|^2+|\delta|^2}$, and informs me in which of these two states is the system now? (Assume that the phase of the roots is inconsequential.)

Failing this, does there exist an operation, which I will call "controlled-initialization", which operates on the same unknown state vector but deterministically results in the first case of the above ensemble? If not, what law of quantum mechanics is violated by this operation?

If either operation is impossible, what law of quantum mechanics precludes it?

Answer 1

Neither your notion of controlled measurement, nor your notion of controlled initialisation, are valid quantum operations. In the two cases, I don't think that there is a simple, single axiom of quantum mechanics that it violates; but they do violate known constraints of quantum mechanical transformations. The impossibility of "controlled measurement" is best understood in terms of density operators as a description of quantum mechanical states (including probabilistic mixtures of two or more state-vectors); and the impossibility of "controlled initialisation" has to do with what realisable transformations map every state to some pure state (a density operator representing the situation of definitely having some single wave-vector as the state vector).

#1. Regarding "controlled measurements"

The first operation ("controlled measurement") is not allowed by quantum mechanics. The reason ultimately is that it is not a linear transformation of the state. Obviously, because there is a probability involved, I'm not talking about unitary transformations; but all operations in quantum mechanics — including Schrödinger evolution and projective measurement — can be realised as completely positive, trace-preserving linear transformations of density operators. As soon as we talk about performing different transformations with different probabilities, it makes the most sense to talk about density operators, because they can incorporate probabilistic mixtures of different pure states. The thing is that all transformations which can be realized quantum-mechanically are those which are linear, and which furthermore preserve the set of density operators (it maps positive semidefinite matrices to positive semidefinite matrices, and maps matrices with trace 1 to matrices with trace 1).

We can see that your mapping isn't linear, as follows. Consider the state $\def\conj{^{\ast}} |\psi\rangle = \bigl[ \alpha \;\; \beta \;\; 0 \;\; \delta \bigr]^\dagger$. The initial density matrix corresponding to this state is then $$ \rho \;=\; |\psi\rangle\langle\psi| \;=\; \begin{bmatrix} \alpha\alpha\conj & \alpha\beta\conj & 0 & \alpha\delta\conj \\ \beta\alpha\conj & \beta\beta\conj & 0 & \beta\delta\conj \\ 0 & 0 & \;\;0\;\; & 0 \\ \delta\alpha\conj & \delta\beta\conj & 0 & \delta\delta\conj \end{bmatrix}.$$ Note that all of the non-zero entries of the matrix are linearly independent parameters from each other. The evolution you describe would yield each of the following states with probability $\beta\beta\conj/(\beta\beta\conj + \delta\delta\conj)$ and $\delta\delta\conj/(\beta\beta\conj + \delta\delta\conj)$ respectively, $$\begin{align*} |\phi_1\rangle \;&=\; \begin{bmatrix} \alpha \\ \sqrt{\beta\beta\conj + \delta\delta\conj} \\ 0 \\ 0 \end{bmatrix}, &\qquad |\phi_2\rangle \;&=\; \begin{bmatrix} \alpha \\ 0 \\ 0 \\ \sqrt{\beta\beta\conj + \delta\delta\conj} \end{bmatrix}, \end{align*}$$ which have the corresponding density matrices $$\begin{align*} |\phi_1\rangle\langle\phi_1| \;&=\; \begin{bmatrix} \alpha\alpha\conj & \alpha\sqrt{\beta\beta\conj + \delta\delta\conj} & 0 & 0 \\ \alpha\conj\sqrt{\beta\beta\conj + \delta\delta\conj} & \beta\beta\conj + \delta\delta\conj & 0 & 0 \\ 0 & 0 & \;\;0\;\; & 0 \\ 0 & 0 & 0 & \;\;0\;\; \end{bmatrix}, \\[3ex] |\phi_2\rangle\langle\phi_2| \;&=\; \begin{bmatrix} \alpha\alpha\conj & 0 & 0 & \alpha\conj\sqrt{\beta\beta\conj + \delta\delta\conj} \\ 0 & \;\;0\;\; & 0 & \;\;0\;\; \\0 & 0 & \;\;0\;\; & 0 \\ \alpha\conj\sqrt{\beta\beta\conj + \delta\delta\conj} & 0 & 0 & \beta\beta\conj + \delta\delta\conj \end{bmatrix}. \end{align*}$$ The resulting transformation is then the mapping $$\begin{align*} M(\rho) \;&=\; M\left( \begin{bmatrix} \alpha\alpha\conj & \alpha\beta\conj & 0 & \alpha\delta\conj \\ \beta\alpha\conj & \beta\beta\conj & 0 & \beta\delta\conj \\ 0 & 0 & \;\;0\;\; & 0 \\ \delta\alpha\conj & \delta\beta\conj & 0 & \delta\delta\conj \end{bmatrix} \right) \\[2ex]&=\; \frac{\beta\beta\conj}{\beta\beta\conj + \delta\delta\conj} |\phi_1\rangle\langle\phi_1| \;+\; \frac{\delta\delta\conj}{\beta\beta\conj + \delta\delta\conj} |\phi_2\rangle\langle\phi_2| \\[2ex]&=\; \begin{bmatrix} \alpha\alpha\conj & \alpha\tfrac{\beta\beta\conj}{\sqrt{\beta\beta\conj + \delta\delta\conj}} & 0 & \alpha\tfrac{\delta\delta\conj}{\sqrt{\beta\beta\conj + \delta\delta\conj}} \\ \alpha\conj\tfrac{\beta\beta\conj}{\sqrt{\beta\beta\conj + \delta\delta\conj}} & \beta\beta\conj & 0 & 0 \\ 0 & 0 & \;\;0\;\; & 0 \\ \alpha\conj\tfrac{\delta\delta\conj}{\sqrt{\beta\beta\conj + \delta\delta\conj}} & 0 & 0 & \delta\delta\conj \end{bmatrix}.\end{align*}$$ This is not a linear function of the non-zero parameters $\alpha\alpha\conj$, $\alpha\beta\conj$, $\alpha\delta\conj$, etc. So the function $M$ on density operators is not a linear one, and thus not a physically valid transformation of density operators.

#2. Regarding "controlled initialisations"

The second operation ("controlled initialisation") is also not possible, quantum-mechanically, for an arbitrary initial state. The only deterministic operations (ones which yield a pure state with certainty) are the ones which erase information and prepare a new constant state, and unitary operations. Your map is obviously not constant. It isn't unitary either, because it doesn't preserve inner products: we can see this by considering the two vectors $\bigl[ \alpha, \beta, \gamma, \beta \bigr]$ and $\bigl[ \alpha, \beta, \gamma, -\beta \bigr]$ for $\beta \ne 0$, and any values of $\alpha, \gamma \in \mathbb C$, which may even be orthogonal but are ultimately mapped to the same vector. So the notion you present of "controlled initialization" is not valid either.

Answer 2

What you're looking for (I assume, based on your example) is a quantum operation which takes

$$ (\alpha, \beta, \gamma, \delta)$$

to

$$ (\alpha, u, \gamma, 0)$$ with probability $\frac{|\beta^2|}{| \beta^2| + |\delta^2|}$

and

$$ (\alpha, 0, \gamma, u')$$
with probability $\frac{|\delta^2|}{|\beta^2| + |\delta^2|}.$

If you insist on getting the result of the measurement somehow, then this is impossible, because if you start with a few copies of a state $$(\alpha, \epsilon_1, \gamma, \epsilon_2),$$ then assuming you can do what you want, you will be able to approximate the ratio between $\epsilon_1$ and $\epsilon_2$ regardless of how small $\epsilon_1$ and $\epsilon_2$ are, something which violates the principles of quantum mechanics.

ANSWER TO REVISED QUESTION:

You want to start with $(\alpha, \beta, \gamma, \delta)$, and get $$(\alpha, e^{i\theta}\sqrt{|\beta^2|+|\delta^2|}, \gamma, 0).$$

This is impossible. Suppose first that the phase is always 1. This means that you can start with $(\frac{\alpha}{2}, \pm\frac{\beta}{\sqrt{2}}, \frac{\alpha}{2},0)$, and get $(\frac{\alpha}{2}, \frac{\beta}{\sqrt{2}}, \frac{\alpha}{2},0)$. However, if you started with a mixture of the two vectors $(\alpha\frac{1}{2}, \pm\beta\frac{1}{\sqrt{2}}, \alpha\frac{1}{2},0)$, this is the same as a mixture of the vectors $(\frac{1}{\sqrt{2}}, 0,\frac{1}{\sqrt{2}},0)$ and $(0,1,0,0)$. Your operation takes a mixture of these two vectors and turns it into a superposition of these two vectors. But turning a mixture into a superposition is impossible by the laws of quantum mechanics.

This means that the operation must preserve phases. But now, we somehow need a quantum operation that maps points on the Bloch sphere composed by the second and fourth coordinates of your vector into a single dimension. This is the same as putting a complex phase on each point of the unit sphere so that opposite points have the opposite phase. That is, you want a continuous map of the unit sphere onto the circle with antipodes mapped to antipodes. I am fairly sure this is topologically impossible (although I'd appreciate anybody who can cite a theorem proving this).

Answer 3

Such an operation is indeed physically realizable. Suppose you wish to measure qubit $b$ if qubit $a$ is in the 1 state. Then just measure $a$, and if you get 1 then measure $b$ (e.g. the lab assistant puts $b$ into the measurement apparatus or not depending on the outcome of the $a$ measurement). But maybe you wanted to accomplish this without measuring $a$. In that case, prepare an ancillary state $c$, do a controlled-controlled-not operation between $a,b,c$, then measure $c$. The controlled-controlled-not operation is $\textrm{CCNOT}_{abc} = \left|0\right>\left<0\right|_a \otimes I_b \otimes I_c + \left|1\right>\left<1\right|_a \otimes \textrm{CNOT}_{bc}$ where $\textrm{CNOT}_{bc} = \left|0\right>\left<0\right|_b \otimes I_c + \left|1\right>\left<1\right|_b X_c$.

If $a$ was 1 then you have done a controlled-not between $b$ and $c$ which in effect copies the state of $b$ to $c$. By then measuring $c$ you get the same result as if you had measured $b$. On the other hand, if $a$ was 0 then the controlled-not doesn't happen and $c$ is left in its initial state. Measuring $c$ then gives 0 without having any effect on $b$.

Just like with any controlled-operation, there will be an impact on the qubit $a$. Specifically, if $b$ is in the 1 state then by doing this controlled measurement you end up gaining information about $a$, causing it to decohere or collapse as if it had been measured. If $b$ was in the zero state then there will be no effect on $a$ (notice that this circuit is symmetric between the $a$ and $b$ qubits).

There is no possible way of doing a measurement of $b$ controlled by $a$ without having an effect on $a$. The reason is that the measurement outcome (or non-outcome) will always reveal information about whether the measurement had been done, which in turn reveals information about the state of $a$.