3.2.5 Entangled states

In Eq. 3.50 we used the tensor product to build a two-qubit state from two single-qubit states. In Section 3.2.3, we saw that these product states are exactly those states that can be constructed by applying local quantum operations to $\left|00\right\rangle=\left|0\right\rangle\otimes\left|0\right\rangle$ (a product state itself). Now, there also exist two-qubit states that are not product states. We call these states entangled, and as we will see they are very important for quantum computing.

How can we determine if a state is a product state or not? Even though quantum states are specified by amplitudes and not by probabilities, we can use the same method that we used in Section 3.1.8 for detecting whether a probability distribution is correlated. Given a two-qubit state in the form (3.31), we first use Eq. 3.29 to compute

\displaystyle\Delta(\left|\psi\right\rangle)=\psi_{00}\psi_{11}-\psi_{01}\psi_% {10}.

(3.67)

Then, $\left|\psi\right\rangle$ is a product state if and only if $\Delta(\left|\psi\right\rangle)=0$ . In this way, entangled states are analogous to correlated probability distributions.

A simple but important example of an entangled two-qubit state is

\displaystyle\left|\Phi^{+}\right\rangle=\frac{1}{\sqrt{2}}\left|00\right% \rangle+\frac{1}{\sqrt{2}}\left|11\right\rangle.

(3.68)

This state is analogous to a pair of perfectly correlated random bits from Eq. 3.28. It is entangled since

\displaystyle\Delta(\left|\Phi^{+}\right\rangle)=\frac{1}{\sqrt{2}}\cdot\frac{% 1}{\sqrt{2}}-0\cdot 0=\frac{1}{2}\neq 0.

We call $\left|\Phi^{+}\right\rangle$ the maximally entangled state of two qubits (although neither the reason for this name nor the peculiar notation are very clear at this point).

How can we build entangled states? Just like when we wanted to build correlated states of two bits, we can use the controlled-NOT operation to make two quantum bits interact. For example, the following sequence of operations prepares the maximally entangled state $\left|\Phi^{+}\right\rangle$ :

Let us quickly verify this:

\displaystyle\mathrm{CNOT}_{1\to 2}\left(H\otimes I\right)\left|00\right% \rangle=\mathrm{CNOT}_{1\to 2}\left(\frac{1}{\sqrt{2}}\left|00\right\rangle+% \frac{1}{\sqrt{2}}\left|10\right\rangle\right)=\frac{1}{\sqrt{2}}\left|00% \right\rangle+\frac{1}{\sqrt{2}}\left|11\right\rangle.

Note that applying the CNOT directly to $\left|00\right\rangle$ , or for that matter to any other basis state, would have not worked (see Eq. 3.63).

Homework 3.6 (Another entangled state).

1.

Verify that the state $\left|\psi\right\rangle=\frac{1}{2}\left|01\right\rangle+\frac{\sqrt{3}}{2}% \left|10\right\rangle$ is entangled.

Hint: Compute Eq. 3.67.
2.

Find a sequence of operations in Quirky that prepares the state $\left|\psi\right\rangle$ .

Hint: You may need to create a suitable rotation.
3.

What are the probabilities of measurement outcomes when measuring both qubits of $\left|\psi\right\rangle$ ? Use Quirky to confirm your result.

Hack.

1.

We calculate $\Delta(\left|\psi\right\rangle)=-\frac{\sqrt{3}}{4}\neq 0$ . The fact that this is non-zero implies that $\left|\psi\right\rangle$ is entangled.

We can get the two desired amplitudes from $\cos(\pi/3)=\frac{1}{2}$ and $\sin(\pi/3)=\frac{\sqrt{3}}{2}$ . This suggests that we use a rotation $U(\theta)$ with angle $\theta=\pi/3$ in the following way:

This does indeed the job:

	$\displaystyle\mathrm{CNOT}_{1\to 2}\,\big{\lparen}U(\pi/3)\otimes\mathrm{NOT}% \big{\rparen}\left\|00\right\rangle$	$\displaystyle=\mathrm{CNOT}_{1\to 2}\left(\frac{1}{2}\left\|01\right\rangle+% \frac{\sqrt{3}}{2}\left\|11\right\rangle\right)$
		$\displaystyle=\frac{1}{2}\left\|01\right\rangle+\frac{\sqrt{3}}{2}\left\|10% \right\rangle.$

3.

We get $[01]$ with probability $\frac{1}{4}$ and $[10]$ with probability $\frac{3}{4}$ . Indeed:

The maximally entangled state in Eq. 3.68 is one of a family of four states, called the Bell states. The Bell states are defined as follows:

$\displaystyle\left\|\Phi^{+}\right\rangle$	$\displaystyle=\frac{1}{\sqrt{2}}\left\|00\right\rangle+\frac{1}{\sqrt{2}}\left\|% 11\right\rangle,$	(3.69)
$\displaystyle\left\|\Phi^{-}\right\rangle$	$\displaystyle=\frac{1}{\sqrt{2}}\left\|00\right\rangle-\frac{1}{\sqrt{2}}\left\|% 11\right\rangle,$	(3.70)
$\displaystyle\left\|\Psi^{+}\right\rangle$	$\displaystyle=\frac{1}{\sqrt{2}}\left\|01\right\rangle+\frac{1}{\sqrt{2}}\left\|% 10\right\rangle,$	(3.71)
$\displaystyle\left\|\Psi^{-}\right\rangle$	$\displaystyle=\frac{1}{\sqrt{2}}\left\|01\right\rangle-\frac{1}{\sqrt{2}}\left\|% 10\right\rangle.$	(3.72)

They are named after John Steward Bell, who was one of the first to recognize the remarkable features of quantum entanglement. How can we create these four Bell states? We saw above that we can obtain $\left|\Phi^{+}\right\rangle$ by applying a Hadamard and a CNOT operation to the basis state $\left|00\right\rangle$ . It is an easy exercise to verify that the other three Bell states can be constructed similarly, that is, by applying the same sequence of operations to the other three basis states. In other words, if we define

U_{\text{Bell}}=\mathrm{CNOT}_{1\to 2}\left(H\otimes I\right)

(3.73)

then

	$\displaystyle\left\|\Phi^{+}\right\rangle=U_{\text{Bell}}\left\|00\right\rangle,% \qquad\left\|\Phi^{-}\right\rangle=U_{\text{Bell}}\left\|10\right\rangle,$
	$\displaystyle\left\|\Psi^{+}\right\rangle=U_{\text{Bell}}\left\|01\right\rangle,% \qquad\left\|\Psi^{-}\right\rangle=U_{\text{Bell}}\left\|11\right\rangle.$

Exercise 3.12 (Preparing Bell states).

Draw how you would create the other three Bell states in Quirky: $\left|\Phi^{-}\right\rangle$ , $\left|\Psi^{+}\right\rangle$ , and $\left|\Psi^{-}\right\rangle$ .

Solution.

•

$\left|\Phi^{-}\right\rangle$ :
•

$\left|\Psi^{+}\right\rangle$ :
•

$\left|\Psi^{-}\right\rangle$ :

Exercise 3.13 (Discriminating Bell states).

Alice’s donkey robot got lost while on an exploration mission! It wants to urgently let Alice know of its position so that she can come to its rescue. The donkey is located in one of the four neighborhoods surrounding the school. To communicate which one, the donkey sends a two-qubit quantum message $\left|x,y\right\rangle$ , where $x\in\{0,1\}$ tells the $x$ coordinate and $y\in\{0,1\}$ tells the $y$ coordinate of its location:

Unfortunately, Alice’s evil classmate Eve is jamming the signal and what Alice receives instead is one of the four Bell states as shown above. Help Alice to correctly decode the signal and locate her donkey robot! That is, find a sequence of operations that maps each of the four Bell states back onto the corresponding basis state $\left|x,y\right\rangle$ .

Solution.

Note that this is the same as inverting the operation

U_{\text{Bell}}

in Eq. 3.73. Recall that

U_{\text{Bell}}=\mathrm{CNOT}_{1\to 2}\left(H\otimes I\right)

. This is a composite operation, so recall from 2.5 that

U_{\text{Bell}}^{{-1}}=(H\otimes I)^{{-1}}\,\mathrm{CNOT}_{1\to 2}^{{-1}}=(H^{% {-1}}\otimes I)\,\mathrm{CNOT}_{1\to 2}^{{-1}}.

Recall from Eq. 3.66 that

\mathrm{CNOT}_{1\to 2}

is the inverse of itself, i.e.,

\mathrm{CNOT}_{1\to 2}^{{-1}}=\mathrm{CNOT}_{1\to 2}

. It is also true that

H^{{-1}}=H

(this holds for any reflection, in fact). This implies that we can undo

U_{\text{Bell}}

by applying the same two operations but in reverse order:

U_{\text{Bell}}^{{-1}}=(H\otimes I)\,\mathrm{CNOT}_{1\to 2}.

That is, we first apply

\mathrm{CNOT}_{1\to 2}

and then

H

on the first qubit, as in:

$\displaystyle\left\|\Phi^{+}\right\rangle$	$\displaystyle=\frac{1}{\sqrt{2}}\left\|00\right\rangle+\frac{1}{\sqrt{2}}\left\|% 11\right\rangle,$	(3.69)
$\displaystyle\left\|\Phi^{-}\right\rangle$	$\displaystyle=\frac{1}{\sqrt{2}}\left\|00\right\rangle-\frac{1}{\sqrt{2}}\left\|% 11\right\rangle,$	(3.70)
$\displaystyle\left\|\Psi^{+}\right\rangle$	$\displaystyle=\frac{1}{\sqrt{2}}\left\|01\right\rangle+\frac{1}{\sqrt{2}}\left\|% 10\right\rangle,$	(3.71)
$\displaystyle\left\|\Psi^{-}\right\rangle$	$\displaystyle=\frac{1}{\sqrt{2}}\left\|01\right\rangle-\frac{1}{\sqrt{2}}\left\|% 10\right\rangle.$	(3.72)