# How to create a maximally entangled state of two 4-level quantum mechanical systems?

Let's say that I have a 4-level quantum state, which is described by a linear combination of the following four eigenbases:

$$|\text{red}⟩ = \begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \end{bmatrix} , |\text{blue}⟩ = \begin{bmatrix} 0 \\ 1 \\ 0 \\ 0 \end{bmatrix} , |\text{green}⟩ = \begin{bmatrix} 0 \\ 0 \\ 1 \\ 0 \end{bmatrix} , |\text{yellow}⟩ = \begin{bmatrix} 0 \\ 0 \\ 0 \\ 1 \end{bmatrix}\,.$$

Hence; $$|\psi⟩ = \alpha|\text{red}⟩ + \beta|\text{blue}⟩+ \gamma|\text{green}⟩ + \delta|\text{yellow}⟩$$ represents an arbitrary quantum state.

What would the Dirac notation of a maximally entangled state of 2 such 4-level quantum states look like? Also, what operations would need to be performed in order to prepare such a state? Lastly, how can one check if an arbitrary state in such an eigenbasis is an entangled state or not?

• These are exactly the computational basis we generally use. $\{ |0\rangle, |1\rangle, |2\rangle, |3\rangle \} \equiv \{ |00\rangle, |01\rangle, |10\rangle, |11\rangle \} \equiv \{ |\text{red}\rangle, |\text{blue}\rangle, |\text{green}\rangle, |\text{orange}\rangle \}$. Oct 28, 2023 at 22:00
• Right, but this only describes one quantum system right? Now if I have another quantum system with the same basis, how can I create a maximally entangled state between both of them? Oct 28, 2023 at 22:06
• These basis are sufficient to describe two qubits. I am not sure exactly what your question is. One of the maximally entangled state is just $$|\psi \rangle = \frac{1}{\sqrt{2}} \bigg( | \text{red} \rangle \otimes | \text{red} \rangle + |\text{yellow} \rangle \otimes | \text{yellow} \rangle \bigg)$$ Oct 28, 2023 at 22:09
• Are you saying that you want a maximally entangled state between two systems of size $\mathbb{C}^4$ each? i.e., maximally entangled state you are looking for $\in \mathbb{C}^{16}$? Oct 28, 2023 at 22:13
• Yes that would be correct Oct 28, 2023 at 22:14

## 1 Answer

Too long for a comment so writing it here.

Say, for simplicity, \begin{align} |0\rangle &:= |\text{red}\rangle \,, \\ |1\rangle &:= |\text{blue}\rangle \,, \\ |2\rangle &:= |\text{green}\rangle \,, \\ |3\rangle &:= |\text{yellow}\rangle \,. \\ \end{align}

The Dirac notation for one of the maximally entangled state between two 4-level systems, $$A$$ and $$B$$, that you are looking for, as discussed in the comments, up to a global phase, would be:

$$|\Phi\rangle_{AB} = \frac{1}{2}\bigg(|0\rangle_A |0\rangle_B + |1\rangle_A |1\rangle_B + |2\rangle_A |2\rangle_B + |3\rangle_A |3\rangle _B \bigg)\,.$$

You can obtain the rest of the maximally entangled states between two 4-level systems by doing the unitary operation on $$|\Phi\rangle_{AB}$$ as follows:

$$|\Phi^{x,z}\rangle_{AB} = \bigg(X_A(x)Z_A(z) \otimes I_B \bigg)|\Phi\rangle_{AB}$$

where $$x,z \in \{0,1,2,3\}$$.

Here, $$X(x)$$ and $$Z(z)$$ are generalized Pauli operators for $$d=4$$, i.e., they are $$4 \times 4$$ matrices, which are also known as cyclic shift operators and phase operators, respectively.

In conclusion, there will be total $$16$$ maximally entangled states for two 4-level systems which are precisely $$\{|\Phi^{x,z}\rangle_{AB} \}$$, where $$x,z \in \{0,1,2,3\}$$.

As for checking if a given joint state of systems $$A$$ and $$B$$ is entangled or not, can be done by calculating the coherent information, $$I(A\rangle B)$$, of the given state. If it is positive, you can conclude that the given state is entangled (It is not an iff relation).

$$I(A\rangle B) = S(B) - S(AB)\,,$$

where $$S(\cdot)$$ is the Von Neumann entropy.

As for the circuit, I will need more time. If no one answers in the meantime, I will edit this answer later when I have free time.

• Kindly check the answer I have posted as a reply to this. Thank you! Oct 28, 2023 at 23:36
• There was a mistake in my previous answer. Please read it again and then repost your reply. Sorry about that. Oct 28, 2023 at 23:52
• Excuse me for my ignorance but could you give an example of what the generalised Pauli matrices ‘X(x)’ and ‘Z(z)’ for the system in question are, since I am only familiar with the Pauli matrices for a single qubit system. Oct 28, 2023 at 23:53
• You can get all $X(x)$ by just doing cyclic rotations of coloums of the $4 \times 4$ identity matrix. The operator $Z(z)$ works as follows on our basis states- $$Z(z)|j\rangle = e^{\frac{i 2 \pi z j}{4}}|j\rangle\,,$$ and also $$X(x)|j\rangle = |j \oplus x \rangle\,.$$ Oct 29, 2023 at 0:26
• Clear image, in case it is not visible clearly on your screen. Oct 29, 2023 at 0:27