Or, given more room, quantum particles enjoy more company.
(Warning: This post is incomplete)
Assume reader background knowledge: quantum mechanics, 2-state system
In condensed matter or quantum information textbooks, you may have encountered the concept entanglement monogamy, which states that there can be at most two qubits that are maximally entangled; i.e. if two qubits are maximally entangled, you cannot entangle them with additional qubits.
Now, it is easy to think that quantum particles in general exhibit entanglement monogamy; this turns out to be wrong. I made this mistake, that’s why I am writing this post to repend for my sins. In short, entanglement monogamy only applies to qubits, which are particles with two possible states. To circumvent entanglement monogamy, we will instead use qudits, which are particles with \(d\) possible states. Then, it is in fact possible to maximally entangled arbitrary number of particles. I call this entanglement polygamy. This is the result we will arrive at by the end of this blogpost. For the experts, feel free to directly skip to the Maximally entangle 3 quantum particles section.
Here are the convention I will use (feel free to skip this part now and come back to this when you see it used in the article.
What does it mean for 2 qubits to be maximally entangled? Let \(\ket{\psi}\) be the state of representing the 2 qubits. We can measure the spin-spin correlation function \(\bra{\psi} S_i\otimes S_i \ket{\psi}\) in all 3 spatial directions (\(i=1,2,3\)) and see if the correlation function is maximized in all 3 spatial directions. If all 3 are maximized, the 2 qubits are maximally entangled.
I will now show that one of the Bell states (sometimes also called the EPR pairs)
\begin{equation} \ket{EPR}= \frac{1}{\sqrt{2}}\left(\ket{00}+\ket{11}\right) \end{equation}
is a maximally entangled state (in fact, all 4 Bell states are maximally entangled).
Working out the correlation function \(\langle S_x^{(1)}\otimes S_x^{(2)} \rangle\) explicitly:
\[\begin{align*} \langle S_x^{(1)}\otimes S_x^{(2)} \rangle = \bra{EPR} S_x^{(1)}\otimes S_x^{(2)} \ket{EPR} &= \text{Tr} (S_x^{(1)}\otimes S_x^{(2)} \rho_{EPR})\\ &= \text{Tr} \left[\frac{1}{\sqrt{2}}\left(S_x^{(1)}\ket{0}\otimes S_x^{(2)}\ket{0} + S_x^{(1)}\ket{1}\otimes S_x^{(2)}\ket{1}\right)\bra{EPR}\right]\\ &= \frac{\hbar^2}{4} \text{Tr} \left[\frac{1}{\sqrt{2}}\left(\ket{1}\otimes \ket{1} + \ket{0}\otimes \ket{0}\right)\bra{EPR}\right]\\ &= \frac{\hbar^2}{4} \text{Tr} \left[\ket{EPR}\bra{EPR}\right]\\ &= \frac{\hbar^2}{4} \text{Tr} [\rho_{EPR}]\\ &= \frac{\hbar^2}{4} \end{align*}\]The spin-spin correlation function is maximized, since it reports the value \(\frac{\hbar^2}{4}\), which means whenever one measure the 1st spin to have \(S_x\) expectation value \(\pm \frac{\hbar}{2}\), the 2nd spin would also have expectation value \(\pm \frac{\hbar}{2}\).
The correlation function in the other 2 directions can be obtained in the same way, so I shall leave it as an exercise for the reader ;) What’s important is the result: in all 3 directions, the spin-spin correlation function is maximized. This is the indicator for a maximcally entangled spin state.
As a side remark, having maximized spin-spin correlation functions in all spatial directions is not possible for any classical state. The best a classical state can do is a mixed state, e.g. \(\rho = \frac{1}{2} \ket{00}\bra{00} +\frac{1}{2} \ket{00}\bra{00}\), which maximized the spin-spin correlation function in 1 direction: \(\langle S_z^{(1)}\otimes S_z^{(2)} \rangle\); however, the x- and y- component of spin will be completely uncorrelated, i.e. , \(\langle S_x^{(1)}\otimes S_x^{(2)} \rangle = \langle S_y^{(1)}\otimes S_y^{(2)} \rangle = 0\).
Another way to see that $\ket{EPR}$ is maximally entangled is that even though it is a pure state, the reduced density matrix is \(\rho = \frac{1}{2}\ket{0}\bra{0} + \frac{1}{2}\ket{1}\bra{1}\), which is a classical mixed state with maximal Von Neumann entropy.
In order to maximally entangle 3 particles, my idea is to not restrict ourselves to just qubits (spin-1/2 particles), but use spin-1 particle. My claim is that the following state is maximally entangled
\[\ket{\psi} = \frac{1}{\sqrt{3}} (\ket{abc} + \ket{bca} + \ket{cab})\]where \(a=-1, b=0, c=1\).
Now, as a first indicator that it is maximally entangled, we note that the reduced density matrix for the 1-particle subsystem is \(\rho_1 = \frac{1}{3}\ket{a}\bra{a} + \frac{1}{3}\ket{b}\bra{b} + \frac{1}{3}\ket{c}\bra{c}\), which is a classically maximally mixed state.
We then calculate the spin-spin correlation function.
To be continued…
Addendum: For anyone who is interested in learning more about these materials, I highly recommend checking out Quantum Computation and Quantum Information by Michael Nielson and Isaac Chuang (or sometimes affectionately refered to as Mike & Ike); it is very pedagogical on a lot of fascinating topics.