# How to perform a state density modification for a single targeted state only?

I have a question about single target state modification... Suppose we have a 3 qubit state density distribution as follows (prenormalized):

$$\begin{bmatrix} |000\rangle & 3 \\ |001\rangle & 4 \\ |010\rangle & 5 \\ |011\rangle & 6 \\ |100\rangle & 7 \\ |101\rangle & 8 \\ |110\rangle & 9 \\ |111\rangle & 10 \\ \end{bmatrix}$$

and we would like to do a "whack-a-mole" and whack $$|010\rangle$$ down to $$0$$ while keeping all the other proportions unchanged, i.e.

$$\begin{bmatrix} |000\rangle & 3 \\ |001\rangle & 4 \\ |010\rangle & \color{red}0 \\ |011\rangle & 6 \\ |100\rangle & 7 \\ |101\rangle & 8 \\ |110\rangle & 9 \\ |111\rangle & 10 \\ \end{bmatrix}$$

How might one do this using physically allowed quantum state manipulations?

You're using some unusual notation and terminology that don't entirely fit. I'm assuming you question is that you are starting from a state $$|\psi\rangle=3|000\rangle+4|001\rangle+5|010\rangle+6|011\rangle+7|100\rangle+8|101\rangle+9|110\rangle+10|111\rangle)/\sqrt{380}$$ and you want to convert it into $$|\phi\rangle=3|000\rangle+4|001\rangle+6|011\rangle+7|100\rangle+8|101\rangle+9|110\rangle+10|111\rangle)/\sqrt{355}.$$ You have a vast range of options for how you might do this. By far the simplest is essentially the suggestion of TristanNemoz where you run a probabilistic protocol. In terms of a quantum circuit, this looks like

and works provided the measurement gives the answer 0, which happens with probability $$\frac{355}{380}=\frac{71}{76}$$. The circuit is simple, and has the further advantage that you don't actually need to know what state $$|\psi\rangle$$ is, only the term that you want to knock out. The disadvantage, of course, is that it is probabilistic.

There certainly exist deterministic options, assuming that you know both $$|\psi\rangle$$ and $$|\phi\rangle$$. You just need to choose any unitary that satisfies $$U|\psi\rangle=|\phi\rangle.$$ You have to complete its action on an orthonormal basis that includes $$|\psi\rangle$$, with the choice of what the corresponding output states are giving you a lot of freedom to potentially alter how easy/hard the unitary is to construct. Here is one such example (which I have no desire to decompose into gates!): $$\left( \begin{array}{cccccccc} \frac{110}{113}+\frac{12 \sqrt{\frac{3}{13871767}}}{5}+\frac{9}{10 \sqrt{1349}} & \frac{2 \left(339-2 \sqrt{30849}\right)}{565 \sqrt{1349}} & 4 \sqrt{\frac{3}{13871767}}+\frac{3}{2 \sqrt{1349}} & -\frac{6}{113}+\frac{24 \sqrt{\frac{3}{13871767}}}{5}+\frac{9}{5 \sqrt{1349}} & -\frac{7}{113}+\frac{4 \sqrt{\frac{21}{1981681}}}{5}+\frac{21}{10 \sqrt{1349}} & -\frac{8}{113}+\frac{32 \sqrt{\frac{3}{13871767}}}{5}+\frac{12}{5 \sqrt{1349}} & -\frac{9}{113}+\frac{36 \sqrt{\frac{3}{13871767}}}{5}+\frac{27}{10 \sqrt{1349}} & -\frac{10}{113}+8 \sqrt{\frac{3}{13871767}}+\frac{3}{\sqrt{1349}} \\ \frac{15}{\sqrt{122759 \left(703+4 \sqrt{30849}\right)}} & \frac{8+\sqrt{30849}}{5 \sqrt{1349}} & -\frac{\sqrt{30849}-182}{91 \sqrt{1349}} & \frac{30}{\sqrt{122759 \left(703+4 \sqrt{30849}\right)}} & -\frac{\sqrt{30849}-182}{65 \sqrt{1349}} & \frac{40}{\sqrt{122759 \left(703+4 \sqrt{30849}\right)}} & \frac{45}{\sqrt{122759 \left(703+4 \sqrt{30849}\right)}} & -\frac{2 \left(\sqrt{30849}-182\right)}{91 \sqrt{1349}} \\ -\frac{5 \sqrt{\frac{3}{10283}}}{2} & 0 & \frac{\sqrt{\frac{339}{91}}}{2} & -5 \sqrt{\frac{3}{10283}} & -\frac{5 \sqrt{\frac{7}{4407}}}{2} & -\frac{20}{\sqrt{30849}} & -\frac{15 \sqrt{\frac{3}{10283}}}{2} & -\frac{25}{\sqrt{30849}} \\ -\frac{6}{113}+\frac{24 \sqrt{\frac{3}{13871767}}}{5}+\frac{9}{5 \sqrt{1349}} & \frac{4 \left(339-2 \sqrt{30849}\right)}{565 \sqrt{1349}} & 8 \sqrt{\frac{3}{13871767}}+\frac{3}{\sqrt{1349}} & \frac{101}{113}+\frac{48 \sqrt{\frac{3}{13871767}}}{5}+\frac{18}{5 \sqrt{1349}} & -\frac{14}{113}+\frac{8 \sqrt{\frac{21}{1981681}}}{5}+\frac{21}{5 \sqrt{1349}} & -\frac{16}{113}+\frac{64 \sqrt{\frac{3}{13871767}}}{5}+\frac{24}{5 \sqrt{1349}} & -\frac{18}{113}+\frac{72 \sqrt{\frac{3}{13871767}}}{5}+\frac{27}{5 \sqrt{1349}} & -\frac{20}{113}+16 \sqrt{\frac{3}{13871767}}+\frac{6}{\sqrt{1349}} \\ -\frac{7}{113}+\frac{4 \sqrt{\frac{21}{1981681}}}{5}+\frac{21}{10 \sqrt{1349}} & -\frac{70}{\sqrt{457311 \left(703+4 \sqrt{30849}\right)}} & 4 \sqrt{\frac{7}{5945043}}+\frac{7}{2 \sqrt{1349}} & -\frac{14}{113}+\frac{8 \sqrt{\frac{21}{1981681}}}{5}+\frac{21}{5 \sqrt{1349}} & \frac{290}{339}+\frac{28 \sqrt{\frac{7}{5945043}}}{5}+\frac{49}{10 \sqrt{1349}} & -\frac{56}{339}+\frac{32 \sqrt{\frac{7}{5945043}}}{5}+\frac{28}{5 \sqrt{1349}} & -\frac{21}{113}+\frac{12 \sqrt{\frac{21}{1981681}}}{5}+\frac{63}{10 \sqrt{1349}} & -\frac{70}{339}+8 \sqrt{\frac{7}{5945043}}+\frac{7}{\sqrt{1349}} \\ -\frac{8}{113}+\frac{32 \sqrt{\frac{3}{13871767}}}{5}+\frac{12}{5 \sqrt{1349}} & -\frac{80}{\sqrt{457311 \left(703+4 \sqrt{30849}\right)}} & \frac{4}{\sqrt{1349}}+\frac{32}{\sqrt{41615301}} & -\frac{16}{113}+\frac{64 \sqrt{\frac{3}{13871767}}}{5}+\frac{24}{5 \sqrt{1349}} & -\frac{56}{339}+\frac{32 \sqrt{\frac{7}{5945043}}}{5}+\frac{28}{5 \sqrt{1349}} & \frac{275}{339}+\frac{32}{5 \sqrt{1349}}+\frac{256}{5 \sqrt{41615301}} & -\frac{24}{113}+\frac{96 \sqrt{\frac{3}{13871767}}}{5}+\frac{36}{5 \sqrt{1349}} & -\frac{80}{339}+\frac{8}{\sqrt{1349}}+\frac{64}{\sqrt{41615301}} \\ -\frac{9}{113}+\frac{36 \sqrt{\frac{3}{13871767}}}{5}+\frac{27}{10 \sqrt{1349}} & \frac{6 \left(339-2 \sqrt{30849}\right)}{565 \sqrt{1349}} & 12 \sqrt{\frac{3}{13871767}}+\frac{9}{2 \sqrt{1349}} & -\frac{18}{113}+\frac{72 \sqrt{\frac{3}{13871767}}}{5}+\frac{27}{5 \sqrt{1349}} & -\frac{21}{113}+\frac{12 \sqrt{\frac{21}{1981681}}}{5}+\frac{63}{10 \sqrt{1349}} & -\frac{24}{113}+\frac{96 \sqrt{\frac{3}{13871767}}}{5}+\frac{36}{5 \sqrt{1349}} & \frac{86}{113}+\frac{108 \sqrt{\frac{3}{13871767}}}{5}+\frac{81}{10 \sqrt{1349}} & -\frac{30}{113}+24 \sqrt{\frac{3}{13871767}}+\frac{9}{\sqrt{1349}} \\ -\frac{10}{113}+8 \sqrt{\frac{3}{13871767}}+\frac{3}{\sqrt{1349}} & \frac{4}{\sqrt{1349}}-8 \sqrt{\frac{91}{457311}} & \frac{5}{\sqrt{1349}}+\frac{40}{\sqrt{41615301}} & -\frac{20}{113}+16 \sqrt{\frac{3}{13871767}}+\frac{6}{\sqrt{1349}} & -\frac{70}{339}+8 \sqrt{\frac{7}{5945043}}+\frac{7}{\sqrt{1349}} & -\frac{80}{339}+\frac{8}{\sqrt{1349}}+\frac{64}{\sqrt{41615301}} & -\frac{30}{113}+24 \sqrt{\frac{3}{13871767}}+\frac{9}{\sqrt{1349}} & \frac{239}{339}+\frac{10}{\sqrt{1349}}+\frac{80}{\sqrt{41615301}} \\ \end{array} \right)$$

Finally, another method that you could consider if you wanted something deterministic, but you didn't know $$|\psi\rangle$$ is amplitude amplification, but I suspect that's going beyond the scope of what you want.

To get the big unitary matrix, I used a piece of Mathematica code:

psi0 = {3, 4, 5, 6, 7, 8, 9, 10}/Sqrt[380];
ns1 = Orthogonalize[NullSpace[{psi0}]];
ns1 = Join[{psi0}, ns1]
phi0 = {3, 4, 0, 6, 7, 8, 9, 10}/Sqrt[355];
ns2 = Orthogonalize[NullSpace[{phi0}]];
ns2 = Join[{phi0}, ns2]
U = FullSimplify[
Plus @@ (Transpose[{#[[2]]}].{#[[1]]} & /@ Transpose[{ns1, ns2}])]


This is actually major overkill. It's good enough to find any unitaries $$U_1$$ such that $$U_1|000\rangle=|\psi_0\rangle$$ and $$U_1|000\rangle=|\phi_0\rangle$$ (which are relatively straightforward to construct) and then just run the circuit $$U_1^\dagger U_2$$. Indeed, this paper shows (and ultimately gives a construction) for how a unitary can be constructed using no more than 4 controlled-not gates!.

• Thank you very much! I get the first circuit now. Regarding the second matrix, did you use an algorithm to find the unitary matrix that maps $\Psi$ to $\phi$, or do we just keep trying till we get the desired distribution? I suppose any arbitrary prescription of state vector distribution can in principle be prepared using the correct unitary matrix? Jul 19, 2023 at 9:25

The projectors $$\mathcal{M}_0=|010\rangle\!\langle010|$$ and $$\mathcal{M}_1=I-|010\rangle\!\langle010|$$ define a measurement. Applying this measurement, you will get outcome $$0$$ with probability $$\frac{5^2}{3^2+\cdots+10^2}=\frac{5}{76}$$ and outcome $$1$$ with probability $$\frac{71}{76}\approx93.42\%$$.

If you get outcome $$1$$, then the resulting state will be the one that you want.

• thank you, i get what you mean now. Jul 19, 2023 at 9:19