6
$\begingroup$

Does there exist a circuit that allows you to transform a quantum state into a superposition of its inverse; i.e., transform a state into an equal superposition of all basis states that are orthogonal to the initial state?

For example transform: $$\frac{1}{\sqrt{3}}(|00\rangle + |01\rangle + |10\rangle) \text{ to } |11\rangle$$

$$\frac{1}{\sqrt{2}}(|00\rangle + |01\rangle) \text{ to } \frac{1}{\sqrt{2}}(|10\rangle + |11\rangle)$$

$$|00\rangle \text{ to } \frac{1}{\sqrt{3}}(|01\rangle + |10\rangle + |11\rangle)$$

Etc.

$\endgroup$
2
  • 4
    $\begingroup$ What exactly is an inverse of a quantum state? $\endgroup$
    – Guangliang
    Commented Mar 11, 2022 at 15:50
  • 1
    $\begingroup$ Hi, and welcome to QCSE. Is your question distinguished from this one? $\endgroup$ Commented Mar 11, 2022 at 16:01

3 Answers 3

7
$\begingroup$

It's not possible to build such a quantum gate. First of all, what would be the inverse of: $$|+\rangle=\frac12\left(|00\rangle+|01\rangle+|10\rangle+|11\rangle\right)$$ ? You can convince yourself that you can't find an inverse to this state, since every possible candidate $|\psi\rangle$ is already the inverse of another state. Thus, this gate would not be unitary, since it would not be invertible. Also, this gate would not be linear. Let us call $U$ this gate. It is easy to express $U$ in the computational basis: $$\begin{align}U|00\rangle&=\frac{1}{\sqrt{3}}\left(|01\rangle+|10\rangle+|11\rangle\right)\\U|01\rangle&=\frac{1}{\sqrt{3}}\left(|00\rangle+|10\rangle+|11\rangle\right)\\U|10\rangle&=\frac{1}{\sqrt{3}}\left(|00\rangle+|01\rangle+|11\rangle\right)\\U|11\rangle&=\frac{1}{\sqrt{3}}\left(|00\rangle+|01\rangle+|10\rangle\right)\\\end{align}$$

Thus, by linearity: $$\begin{align}U\left[\frac{1}{\sqrt{2}}\left(|00\rangle+|01\rangle\right)\right]&=\frac{1}{\sqrt{2}}\left(U|00\rangle+U|01\rangle\right)\\&=\frac{1}{\sqrt{6}}\left(|01\rangle+|10\rangle+|11\rangle+|00\rangle+|10\rangle+|11\rangle\right)\\&=\frac{1}{\sqrt{6}}\left(|00\rangle+|01\rangle\right)+\sqrt{\frac23}\left(|10\rangle+|11\rangle\right)\end{align}$$ But we also know that $U\left[\frac{1}{\sqrt{2}}\left(|00\rangle+|01\rangle\right)\right]=\frac{1}{\sqrt{2}}\left(|10\rangle+|11\rangle\right)$, hence the contradiction.

$\endgroup$
1
  • $\begingroup$ Thank you, that makes perfect sense! $\endgroup$
    – John Burke
    Commented Mar 13, 2022 at 20:34
3
$\begingroup$

If we take "inverse of a quantum state" to be inverse of elements of some encoded binary string with respect addition-mod-2, then you can get close to the operation you're looking for, but the operation will only work for moving between representations of vectors with complementary Hamming weight. As mentioned by other answers, taking the inverse of a state isn't really well defined but here I'll describe one possibly interesting interpretation.


Say you have a binary string $\mathbf{v} \in \{0, 1\}^4$. We will work with a family of states $|\psi(\mathbf{v})\rangle$ satisfying \begin{equation} |\psi(\mathbf{v})_i|^2 = \lVert \mathbf{v}\rVert_1^{-1} v_i \end{equation}

where $ \lVert \mathbf{v}\rVert_1 = \sum_i v_i$. For instance, an example of this kind of representation would be \begin{equation} \mathbf{v} = (0, 1, 1, 1) \Rightarrow |\psi(\mathbf{v})\rangle = \frac{1}{\sqrt{3}} \left(c_{01}|01\rangle + c_{10}|10\rangle + c_{11}|11\rangle\right) \end{equation}

where $c_{ij} \in \{1, -1\}$ are required if there's any hope for the desired operation to be unitary. Then define the "inverse" of $\mathbf{v}$ with respect to $\oplus_{\text{mod 2}}$ to be $$ \bar{\mathbf{v}} = \mathbf{v} \oplus (1, 1, 1, 1) = (1, 0, 0, 0) $$

This is an element-wise inverse of elements in $\mathbb{Z}_2$ with respect to $\oplus$ rather than an inverse of a vector, which wouldn't make much sense. The goal then is to find some $U$ such that \begin{align} U|\psi(\mathbf{v})\rangle &= |\psi(\bar{\mathbf{v}})\rangle \\ U^\dagger|\psi(\bar{\mathbf{v}})\rangle &= |\psi(\mathbf{v})\rangle \end{align}

However as an earlier answer demonstrated, we can prove by counterexample that no such $U$ can work for all choices of $\mathbf{v}$. What I will show then is a family of operations $U_k$ that work restricted to the set

$$ W_k = \{\mathbf{v}: \lVert \mathbf{v}_1 \rVert \in \{k, 4-k\}\} $$

that is, the set of all bistrings with weight $k$ or $4-k$. Intuitively this restriction makes sense: the "inversion" operation transforms vectors into vectors of complimentary weight. By demonstration, here are the unitaries: \begin{align} U_1 = \frac{1}{\sqrt{3}}\begin{pmatrix} 0 & 1 & -1 & 1 \\ 1 & 0 & 1 & 1 \\ 1 & 1 & 0 & -1 \\ -1 & 1 & 1 & 0 \end{pmatrix} \qquad \qquad U_2 = \frac{1}{2}\begin{pmatrix} -1 & 1 & 1 & 1 \\ 1 & -1 & 1 & 1 \\ 1 & 1 & -1 & 1 \\ 1 & 1 & 1 & -1 \end{pmatrix} \end{align}

Importantly, $U_1^\dagger U_1 = U_2^\dagger U_2 = I$, a requirement of unitarity. Then you have the following: \begin{align} U_1 |11\rangle &= \frac{1}{\sqrt{3}}\left(|00\rangle + |01\rangle -|10\rangle\right) \\ |00\rangle &= U_1^\dagger \frac{1}{\sqrt{3}}\left(|01\rangle + |10\rangle - |11\rangle\right) \\ \end{align}

and so on, with spurious minus signs appearing because of the need for unitarity. Meanwhile the weight-2 case gives \begin{align} U_2 \frac{1}{\sqrt{2}}\left( |00\rangle + |01\rangle \right) &= \frac{1}{\sqrt{2}}\left( |10\rangle + |11\rangle\right)\\ U_2 \frac{1}{\sqrt{2}}\left( |00\rangle + |11\rangle \right) &= \frac{1}{\sqrt{2}}\left( |10\rangle + |01\rangle\right)\\ U_2 \frac{1}{\sqrt{2}}\left( |00\rangle + |10\rangle \right) &= \frac{1}{\sqrt{2}}\left( |01\rangle + |11\rangle\right)\\ \end{align} in which case no minus signs appear for these choices of $|\psi\rangle$. But in both situations, the unnormalized, squared-amplitudes of $|\psi \rangle$ are satisfying a relation of "inverse" with respect to elementwise addition-mod-2.

$\endgroup$
1
  • $\begingroup$ Thanks for the help $\endgroup$
    – John Burke
    Commented Mar 13, 2022 at 20:36
0
$\begingroup$

The main reason why this is impossible is that there is no unique "inverse" for a state unless the orthogonal complement of that state is one dimensional, because there is more than one way of finding basis states.

To wit, if the initial state is orthogonal to some orthonormal set of states $$\{|\psi_1\rangle,\cdots,|\psi_N\rangle\}$$ then it will also be orthogonal to the orthonormal set of states $$\{e^{i\phi_1}|\psi_1\rangle,\cdots,e^{i\phi_N}|\psi_N\rangle\}$$ for any set of angles $\{\phi_1,\cdots,\phi_N\}$ (in fact, arbitrary unitary operations on the space spanned by the $|\psi_i\rangle$ will also suffice). This means that any of the states $$|\Psi_\perp (\phi_1,\cdots,\phi_N)\rangle=\frac{e^{i\phi_1}|\psi_1\rangle+\cdots+ e^{i\phi_N}|\psi_N\rangle}{\sqrt{N}}$$ could be used as a candidate for the inverse of the original state. But, since these states are all different in general ($|\psi_1\rangle+|\psi_2\rangle\neq |\psi_1\rangle-|\psi_2\rangle$), we now realize that we can't specify one particular inverse to our initial state unless $N=1$.

You might say that we choose some "canonical" unitary that always sets the phases to be $e^{i\phi_j}=1$. But that cannot be, because nothing makes some basis state $|\psi_j\rangle$ more standard than some other basis state $|\Phi_j\rangle=-|\psi_j\rangle$. This argument can be generalized to saying that there is no preferred set of basis states for any subspace with more than one dimension and so we can never make a unique operation for this inverse.

(You may argue that we can choose the computational basis to do everything, but that is just for our notational convenience, while unitary operators must perform the same way regardless of the basis in which you express them.)


We can also think about this in terms of density matrices $\rho_\perp=|\Psi_\perp\rangle \Psi_\perp|$ for the desired inverse state $|\Psi_\perp\rangle$. In some basis, the original density matrix will look like $$\rho=|\Psi\rangle\langle \Psi|=\begin{pmatrix}\psi_{0}\psi_0^\ast&\psi_{1}\psi_0^\ast&\cdots&\psi_{0}\psi_n^\ast&0\cdots&0\\ \psi_{1}\psi_0^\ast&\psi_{1}\psi_1^\ast&\cdots&\psi_{1}\psi_n^\ast&0\cdots&0\\ \vdots&\vdots&\ddots&\vdots&\vdots\ddots&\vdots\\ \psi_{n}\psi_0^\ast&\psi_{n}\psi_1^\ast&\cdots&\psi_{n}\psi_n^\ast&0\cdots&0\\ 0&0&\cdots&0&0\cdots&0\\ \vdots&\vdots&\ddots&\vdots&\vdots\ddots&\vdots\\ 0&0&\cdots&0&0\cdots&0 \end{pmatrix}.$$ The stated goal is to have $\rho_\perp$ being an equal superposition in the subspace orthogonal to $\rho$, so it must look like $$\rho_\perp\propto\begin{pmatrix}0&0&\cdots&0&0\cdots&0\\ 0&0&\cdots&0&0\cdots&0\\ \vdots&\vdots&\ddots&\vdots&\vdots\ddots&\vdots\\ 0&0&\cdots&0&0\cdots&0\\ 0&0&\cdots&0&1\cdots&1\\ \vdots&\vdots&\ddots&\vdots&\vdots\ddots&\vdots\\ 0&0&\cdots&0&1\cdots&1 \end{pmatrix}.$$ But this property of having a matrix of ones is basis-dependent, because if you change the basis for that subspace you will no longer have a matrix of ones (in fact, this is a pure state so in some basis it will look like a single 1 and the rest all 0s), and a unitary can't know anything about the basis in which you express it, so there cannot be a unitary performing this transformation.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.