Map $n$ qubit state with complex amplitudes to $n+1$ qubit state with real amplitudes

Question

For the simplest case, consider a single qubit state $|\psi\rangle$, and assume access to a state preparation unitary $V$ satisfying $$ V|0\rangle = |\psi\rangle $$ and $$ V|1\rangle = |\perp\rangle. $$

In the computational basis, we will write $|\psi\rangle$ as $$ |\psi\rangle = \begin{pmatrix} a + i b,\newline c + id \end{pmatrix}. $$

Now, consider a new state $|\phi\rangle$ where $\phi$ has as amplitudes the real and imaginary components of of $|\psi\rangle$ as $$ |\phi\rangle = \begin{pmatrix} a\newline b\newline c\newline d \end{pmatrix} $$

Clearly $|\phi\rangle$ is still a normalized state, and there has been no "change" in information content to prepare, just a slightly different representation.

Does there exist a unitary matrix $U$ satisfying \begin{equation} U:|\psi\rangle|0\rangle \rightarrow |\phi\rangle? \end{equation}

If so, is there a construction of it with elementary quantum gates?

I've tried a few calculations, but I'm getting stuck on the fact that using say controlled applications of $V$ depends only on the norm of amplitude, not it's real and imaginary components, so it's not clear how to separate them out this way in a coherent fashion. I also tried implementing something like an LCU of $V$ with $V^\dagger$ to see if that gave me what I wanted, but not quite.

Does anyone have suggestions on where to look? Is this possible? Is there a violation of no-cloning possibly happening somewhere preventing this? Much appreciated.

Answer 1

Call the operation you want to construct $D$ and call the qubit that ends up storing the real/imaginary distinction $q$.

If I gave you $D$, you could apply $D$ then $Z_q$ then $D^{-1}$. The overall effect of that sequence would be to negate the imaginary component of all amplitudes. In other words, you are able to conjugate the state vector of the system by using $D$.

Conjugation is non-unitary, so giving you $D$ violated the unitarity postulate of quantum mechanics. Therefore $D$ is not constructible from elementary operations, since they all follow the unitarity postulate.

Answer 2

This will violate unitarity of the transformation $U$. Consider states $$\begin{bmatrix}\frac1{\sqrt2} \\ \frac1{\sqrt2}\end{bmatrix} (b=d=0)$$ and $$\begin{bmatrix}i\frac1{\sqrt2} \\ i\frac1{\sqrt2}\end{bmatrix} (a=c=0)$$

The states before applying $U$ to them are the same up to a global phase, the states after applying $U$ are orthogonal, so applying $U$ does not preserve the inner product of these states.

Answer 3

TL;DR: The proposed map fails to be insensitive to the global phase. For example, it can tell apart $|0\rangle\equiv[1,0]^T$ from $|0\rangle\equiv[i,0]^T$ even though these two different mathematical objects represent the same physical state. This failure finds its most dramatic expression in the density matrix formulation.

Global phase

The answers by Craig and Mariia are excellent, but - judging by the comments - not entirely convincing. My guess is that this is due to the fact that they both rely on subtle intuitive grasp of the role that the global phase plays in the usual formalism of quantum mechanics. This formalism is not fully rigorous in the sense that the mathematical objects it employs, most notably state vectors, are not in a one-to-one correspondence with the elements of the physical theory$^{1,2}$. To work around this, people typically rely on intuitive understanding that global phase is unobservable.

Fortunately, there is a rigorous formalism for quantum mechanics that is free of global phase. Let's see what it says about the proposed map.

The "map" isn't even a map

The map $D$ sends $|\psi\rangle=[a+ib,c+id]^T$ to $|\phi\rangle=[a,b,c,d]^T$. Let's write down the density matrices corresponding to the input state $$ \begin{align} \psi&=|\psi\rangle\langle\psi|=\begin{bmatrix} a^2+b^2&ac+bd-i(ad-bc)\\ ac+bd+i(ad-bc)&c^2+d^2\\ \end{bmatrix}\tag1 \end{align} $$ and the output state $$ \begin{align} \phi&=|\phi\rangle\langle\phi|=\begin{bmatrix} a^2&ab&ac&ad\\ ab&b^2&bc&bd\\ ac&bc&c^2&cd\\ ad&bd&cd&d^2\\ \end{bmatrix}.\tag2 \end{align} $$ All we want is a map $D$ that sends $\psi\mapsto\phi$... But, wait! This is not well defined as a mathematical function! Let alone one that satisfies all the constraints imposed by quantum mechanics.

To see this, consider the action of $D$ on a state like $|0\rangle\langle 0|=\begin{bmatrix}1&0\\0&0\end{bmatrix}$. A function is supposed to assign a single output to each input, but our definition of $D$ associates an infinite number of possible outputs to $|0\rangle\langle 0|$ including $$ \begin{bmatrix}1&0&0&0\\0&0&0&0\\0&0&0&0\\ 0&0&0&0\end{bmatrix},\quad \begin{bmatrix}\frac12&\frac12&0&0\\\frac12&\frac12&0&0\\0&0&0&0\\0&0&0&0\end{bmatrix},\quad \begin{bmatrix}0&0&0&0\\0&1&0&0\\0&0&0&0\\ 0&0&0&0\end{bmatrix}.\tag3 $$ This is of course directly related to what the other answers explain. The proposed map fails to be insensitive to the global phase. For example, it can tell apart $|0\rangle\equiv[1,0]^T$ from $|0\rangle\equiv[i,0]^T$ even though these two different mathematical objects represent the same physical state.

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^{$^1$ This is due to the fact that changing the global phase of a ket does not change the physical state it represents.
$^2$ Nevertheless, one can actually employ this formalism rigorously by keeping in mind that quantum amplitudes are homogeneous coordinates in a projective space. Clearly, this is not being done in the description of the map in question.}