# Do unitary matrices acting on entangled states always give a quantum state?

I'm trying to understand what happens when Alice(Bob) apply a unitary to her(his) part of an entangled state. Let us consider the following unitary transformations: $$U_1 = \frac{1}{\sqrt{2}} \begin{bmatrix} 0 & 0 & 1 & -1\\ 0 & 0 & 1 & 1\\ -1&-1&0&0\\ 1&-1&0&0 \end{bmatrix} \text{ and }~ U_2 = \frac{1}{\sqrt{3}} \begin{bmatrix} 0 & 1 & 1 & 1\\ -1 & 0 & -1 & 1\\ -1& 1 & 0&-1\\ -1&-1&1&0 \end{bmatrix}.$$

Suppose Alice and Bob share the state $$|\psi\rangle_{AB}= \frac{1}{\sqrt{2}}(|00\rangle _A|00\rangle_B+|01 \rangle_A|01\rangle_B)$$. To improve readability I will say that $$|\psi\rangle_{AB}=\frac{1}{\sqrt{2}}(|0\rangle_A|0\rangle_B+|1\rangle_A|1\rangle_B).$$

My problem is when I try to calculate the resulting state after Alice applies $$U_1$$ and Bob applies $$U_2$$ to their part of $$|\psi\rangle_{AB}$$:

\begin{align} (U_1 \otimes U_2) &\frac{1}{\sqrt{2}}(|0\rangle_A|0\rangle_B+|1\rangle_A|1\rangle_B)\\ &= \frac{1}{\sqrt{2}} (U_1|0\rangle_AU_2|0\rangle_B + U_1|1\rangle_AU_2|1\rangle_B)\\ &= \frac{1}{\sqrt{2}} \Big(\frac{1}{\sqrt{2}}(-|2\rangle+|3\rangle) \frac{1}{\sqrt{3}}(-|1\rangle-|2\rangle-|3\rangle) + \\ &~~~~~~\frac{1}{\sqrt{2}}(-|2\rangle-|3\rangle) \frac{1}{\sqrt{3}}(|0\rangle+|2\rangle-|3\rangle)\Big)\\ &= \frac{1}{\sqrt{2}}\Big[ \frac{1}{\sqrt{6}} \big(|2\rangle|1\rangle+|2\rangle|2\rangle+|2\rangle|3\rangle -|3\rangle|1\rangle-|3\rangle|2\rangle-|3\rangle|3\rangle\big) \\ &\phantom{= \frac{1}{\sqrt{2}}~~} +\frac{1}{\sqrt{6}} \big(-|2\rangle|0\rangle-|2\rangle|2\rangle+|2\rangle|3\rangle -|3\rangle|0\rangle-|3\rangle|2\rangle+|3\rangle|3\rangle\big)\Big], \end{align}

here is where I'm having trouble, I'm not sure if I can just simplify this expression. In other words, to allow interference between the states or not.

Simplifying the terms would give: \begin{align} &= \frac{1}{2\sqrt{6}} \big(|2\rangle|1\rangle+|2\rangle|3\rangle -|3\rangle|1\rangle-|3\rangle|2\rangle -|2\rangle|0\rangle+|2\rangle|3\rangle -|3\rangle|0\rangle-|3\rangle|2\rangle)\\ &= \frac{1}{2\sqrt{6}} \big(|2\rangle|1\rangle+2|2\rangle|3\rangle -|3\rangle|1\rangle-2|3\rangle|2\rangle -|2\rangle|0\rangle-|3\rangle|0\rangle) \end{align} that is not a valid quantum state because $$4(\frac{1}{2\sqrt{6}})^2 + 2(\frac{1}{\sqrt{6}})^2 = \frac{1}{2} \neq 1$$.

tl;dr: What's the resulting state after $$(U_1 \otimes U_2) |\psi\rangle_{AB}$$?

• generally speaking, a unitary matrix always sends pure states to pure states, and a matrix such as $U\otimes I$ with $U$ unitary is also unitary. Hence yes, $(U\otimes I)$ will always send entangled states to valid (and entangled) states
– glS
Commented Jun 23, 2022 at 6:28
• Before coming up with this problem I thought that any unitary applied to a entangled state will send to another valid state (entangle or not). I agree with you. But, If I set $U=U_1 \otimes U_2$ and apply $(U \otimes I)$ to $|\psi\rangle_{AB}|0\rangle$ I still get the same problem don't I? Commented Jun 23, 2022 at 12:58
• you don't get any problem. You'll get $U|\psi\rangle_{AB}$, which is a normalised vector (hence a state) if $U$ is unitary
– glS
Commented Jun 23, 2022 at 13:02
• I get that the theory says I will get to a normalised vector. But on this case $U |\psi\rangle_{AB} =$ the last state of my question. That is not normalised Mithrandir24601 suggested this normalization factor. But I haven't found any formalization of this. I thought any unitary applied to an entangled state would give a valid state without needing to do anything else. Commented Jun 23, 2022 at 13:10
• either accept one of the answers if you think it suitably addressed your question, or just post an answer yourself and accept that one. There's no real reason to delete the post even if it was due to a calculation mistake.
– glS
Commented Jun 23, 2022 at 17:59

Yes, you can absolutely simplify down. The problem you're having isn't one of simplification, rather one of counting all the terms in the normalisation, although you appear to have picked up a stray $$\sqrt{2}$$ at the end of the question - the final line should read $$$$\left(U_1\otimes U_2\right)\left|\psi\right\rangle_{AB} = \frac{1}{2\sqrt{3}} \big(|2\rangle|1\rangle+|2\rangle|3\rangle -|3\rangle|1\rangle-|3\rangle|2\rangle -|2\rangle|0\rangle+|2\rangle|3\rangle -|3\rangle|0\rangle-|3\rangle|2\rangle).$$$$
Now, if we normalise, while there are indeed 8 terms in the RHS, that doesn't necessarily mean that the normalisation is automatically $$1/\sqrt{8}$$ - in cases such as this, you'll find that you end up double-counting the terms $$\left|2\rangle|3\right\rangle$$ and $$\left|3\rangle|2\right\rangle$$ (just as in 1), so it's best to make this explicit by writing $$$$\left(U_1\otimes U_2\right)\left|\psi\right\rangle_{AB} = \frac{1}{2\sqrt{3}} \big(|2\rangle|1\rangle+2|2\rangle|3\rangle -|3\rangle|1\rangle-2|3\rangle|2\rangle -|2\rangle|0\rangle -|3\rangle|0\rangle).$$$$
Now, when you go to normalise, you'll find that \begin{align}\left\langle\psi_{AB}|\psi_{AB}\right\rangle &= \left\langle\psi_{AB}|\left(U_1\otimes U_2\right)^\dagger\left(U_1\otimes U_2\right)|\psi_{AB}\right\rangle \\ &= \frac{1}{12}\left(1 + 4 + 1 + 4 + 1 + 1\right) = 1, \end{align} as expected.
1 For a simpler example, this is perhaps made more explicit by considering e.g. \begin{align}\left(\left\langle 0\right| + \left\langle 0\right|\right)\left(\left| 0\right\rangle + \left| 0\right\rangle\right) = \left\langle 0\right|\left(\left| 0\right\rangle + \left| 0\right\rangle\right) + \left\langle 0\right|\left(\left| 0\right\rangle + \left| 0\right\rangle\right) = 4.\end{align}
• Many thanks for your explanation. But, how can I formalize that $\frac{1}{2\sqrt{6}} \big(|2\rangle|1\rangle+2|2\rangle|3\rangle -|3\rangle|1\rangle-2|3\rangle|2\rangle -|2\rangle|0\rangle-|3\rangle|0\rangle) \implies \frac{1}{2\sqrt{3}} ( |2\rangle|1\rangle+2|2\rangle|3\rangle +\dotsm )$ As they are not the same. They don't represent the same vector. But some how I have to consider the $\frac{1}{2\sqrt{3}}$ after applying the unitary matrices. Commented Jun 23, 2022 at 12:53
• Maybe another way of saying is: If I have a $n$-qubit state $n>1$ and measure one of the qubits, to calculate the resulting state I have to normalize it accordingly to the measured value. The concept and math of this case is clear to me. But is there any formalization of unitary applied to an entangled state to get to this normalization factor? Or this is kind of an implicit thing? Thanks once again. Commented Jun 23, 2022 at 13:03