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

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.

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.

• It is also not solvable with non-unitary operations such as lcu? It seems like it should be doable with ancilla at the cost of a success probability Sep 22 at 6:26
• In fact, finding the real/imaginary parts of an amplitude was exactly the motivation for this post! The amplitude estimation procedure seems the only game in town, but leaves a lot to be desired Sep 22 at 6:29

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.

• I don't agree with your solution. In the cases you consider, there is no relative phase, which can be captured. For example, the case where $a=1/\sqrt{2}, d = 1/\sqrt{2}$. Is there not a transformation which can preserve this relative phase? Sep 22 at 1:33
• Multiplying through by a global phase, I can always cancel one of the complex components, so $|\phi\rangle$ is always representable with 3 non-zero components Sep 22 at 1:37

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.

$$^1$$ 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.

• Thanks Adam, much more clear how the problem arises. If I encoded all of my unitary operations and states in this way, and then mapped it to a quantum computer, it seems like one would have the ability to implement not just linear, but also anti-linear operations, which seems quite appealing in a lot of cases. Is there anything preventing me from working in this representation and using only real-valued matrices and states? Sep 22 at 19:57
• As far as we know Nature does not provide anti-linear operations. Suppose you do have an anti-linear operation, e.g. complex conjugation $K$ and act with it on the second qubit in $|00\rangle+|11\rangle$. One one hand, we have $|0\rangle\otimes K|0\rangle+|1\rangle\otimes K|1\rangle=|00\rangle+|11\rangle=[1,0,0,1]^T$. On the other, $|00\rangle+|11\rangle=|{+i}\rangle|{-i}\rangle+|{-i}\rangle|{+i}\rangle$, so we get $|{+i}\rangle|{+i}\rangle+|{-i}\rangle|{-i}\rangle=[1,0,0,-1]^T$. Again, operation is ill-defined. The issue is that tensor product allows scalars to move freely between subsystems. Sep 23 at 14:53
• There is also another complication: any dynamical evolution allowed in quantum physics is continuously connected to identity as I explain in this answer. Anti-linear operations aren't, so they are not valid evolutions under the postulates of quantum mechanics. Sep 23 at 14:57
• That said, you did make a valid observation that anti-linear operations on pure states translate into linear (and hence seemingly valid) operations on density matrices (without considering the environment). Nevertheless, the resulting operations on density matrices are not quantum channels. For example, $\rho^\dagger=\rho$ for every density matrix, so complex conjugation is transposition $\overline\rho=\rho^T$, but transposition is not completely positive. Sep 23 at 15:01