Quantum Entanglement

We will see in this lesson that if the state vector of the combined two-qubit system can be written as a product of the quantum state vectors of the initial subsystems (mine and Bob’s) as illustrated in this equationEq_9_2, then the probabilities of my observing spin-up or spin-down for my qubit are independent of what Bob observes for his qubit. You might ask if things could be otherwise. Hold onto your state space vectors because we are now going to show that there are system states for which Bob’s and my observations are not independent. We will use the spin-up spin-down basis states to illustrate the issues, but you could use the computational basis states $|0⟩$ and $|1⟩$ as well.

To be quite general, we will write the system state vector for the two-qubit system as a superposition of the four basis states listed in this tableTable_9_2 as follows:

$$\ket S = c\ket{\uparrow_A\uparrow_B} + d\ket{\uparrow_A\downarrow_B} + e\ket{\downarrow_A\uparrow_B} + f\ket{\downarrow_A\downarrow_B}.$$

Here $c$, $d$, $e$, and $f$ are numbers that satisfy $c^2+d^2+e^2+f^2 = 1$; so, the probabilities add up to $1$. We will show (spoiler alert!) that if $cf = de$ (the product of the amplitudes of the two outer terms is equal to the product of the amplitudes of the two inner terms), then the state vector can be written, as shown in this equationEq_9_3 , as a product of the state vector for Bob’s qubit and the state vector for my qubit. In that case, Bob’s and my observations are independent. Remember that when we use “outer” and “inner” coefficients in this way, we need to use our standard ordering of the state labels with “outer” corresponding to $|00⟩$ and $|11⟩$ in the third column of this tableTable_9_2 and “inner” corresponding to $|01⟩$ and $|10⟩$.

The derivation of the criterion is not too difficult. Essentially, we want to write the state vector in the above equation as a product of two state vectors: one for me and one for Bob. In other words, we want to find the four amplitudes $g$, $h$, $m$, and $n$ that satisfy

$$\begin{align*} \ket S &= c\ket{\uparrow_A\uparrow_B} + d\ket{\uparrow_A\downarrow_B} + e\ket{\downarrow_A\uparrow_B} + f\ket{\downarrow_A\downarrow_B}\\ &= \underbrace{\{g\ket{\uparrow_A} + h\ket{\downarrow_A}\}}_{\text{Alice's qubit state}} \underbrace{\{m\ket{\uparrow_B} + n\ket{\downarrow_B}\}}_{\text{Bob's qubit state}}. \end{align*}$$

Notice that the second line came about merely by factoring out my states as required for independent observations. Multiplying out the second line of the above equation with the new variables and comparing that result to the first line, we find that we have four equations to satisfy:

$$gm = c \quad gn = d \quad hm = e \quad hn = f.$$

While this seems to get us nowhere, look carefully at the equations for $c$ and $d$. They have a factor of $g$ in common; so, dividing them will cancel out that variable. Similarly, $e$ and $f$ have a factor of $h$ in common. This calculation yields $m/n = c/d$ and $m/n = e/f$. Further, both of those pairs have $m/n$ in common so we have a criterion for independent states:

$$cf=de.$$

The mathematics can also be done using the other pairs of variables, $c~\&~e$ and $d~\&~f$, leading to an identical result. Thus, our criterion for the state in this equationEq_9_11 to be a product state is $cf = de$; the product of the two “outer” state amplitudes equals the product of the two “inner” state amplitudes.

Conversely—and this is the critical point—if $cf \neq de$, then the state vectors cannot be written as a simple product state and, as we shall see, my observations in general depend on what Bob observes and vice versa. The observations on the two qubits are not independent! In that case, the overall state vector of the system is called an entangled state vector. Or more succinctly, the system state is entangled.

This example shows that all entangled states are superposition states, but not all superposition states are entangled. The properties of entangled state vectors turn out to be rather weird (or at least very counter-intuitive) but they are exceedingly important for QIS and QC.