Two qubits and quantum entanglement

A two-qubit state can be parametrized as

(12)\[\left|\psi\right> = \alpha_0\left|00\right> + \alpha_1\left|01\right> +\alpha_2\left|10\right> +\alpha_3\left|11\right>\]

where \(\alpha_i\) are complex numbers. The probabilistic interpretation requires \(\sum_i \|\alpha_i\|^2=1\). And again we can get rid of the overall phase factor by requiring \(\alpha_0\) to be real. Thus there are six independent real parameters for two-qubit pure states.

Often times, the binary state labels are converted to decimal labels to simplify the notation

\[\left|\psi\right> = \alpha_0\left|0\right> + \alpha_1\left|1\right> +\alpha_2\left|2\right> +\alpha_3\left|3\right>\]

This is helpful when we deal with many qubits.

two-qubit gates

It’s possible to make two-qubit gates out of single-qubit gates. For example,

\[\begin{split}\begin{align}X\otimes X =& \begin{bmatrix} 0 & 1 \\ 1 & 0\end{bmatrix}\otimes \begin{bmatrix} 0 & 1 \\ 1 & 0\end{bmatrix} \\ =& \begin{bmatrix} 0 & 0 & 0 & 1\\ 0 & 0 & 1 & 0\\ 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0\end{bmatrix}\end{align}\end{split}\]

where \(X\) is the Pauli X gate. However, such gates are less interesting than the ones that cannot be expressed as tensor products, because they don’t bring the two qubits into interaction.

As an example of truly two-qubit gates, let’s generalize the XOR gate in classical computing to the quantum case. The truth table of XOR gate is

Inputs Output
A B A XOR B
0 0 0
0 1 1
1 0 1
1 1 0

In classical computing, three bits are involved in XOR gate: two input bits and one output bit. In quantum computing, all (single-qubit and multiple-qubit) gates are unitary matrices that map the input states to the output states. In other words, the input and output qubits are the same qubits.

Thus to extend XOR gate to the quantum case, we need to expand the output a two-bit string. The easiest way is to overwrite the output to qubit B and let qubit A remain its input state. Writing out explicitly, we have

\[\begin{split}\left|00\right>\rightarrow \left|00\right> \\ \left|01\right>\rightarrow \left|01\right> \\ \left|10\right>\rightarrow \left|11\right> \\ \left|11\right>\rightarrow \left|10\right>\end{split}\]

If we choose the representation

\[\begin{split}\left|00\right> = \begin{bmatrix}1\\0\\0\\0\end{bmatrix}, \left|01\right> = \begin{bmatrix}0\\1\\0\\0\end{bmatrix}, \left|10\right> = \begin{bmatrix}0\\0\\1\\0\end{bmatrix}, \left|11\right> = \begin{bmatrix}0\\0\\0\\1\end{bmatrix},\end{split}\]

then the quantum equivalent of the classical XOR gate takes the form

(13)\[\begin{split}\begin{align} CNOT =& \begin{bmatrix} 1& 0 & 0 & 0\\ 0& 1 & 0 & 0\\ 0& 0 & 0 & 1\\ 0& 0 & 1 & 0 \end{bmatrix}\\ =& \left|0\right>\left<0\right|\otimes I + \left|1\right>\left<1\right|\otimes X \end{align}\end{split}\]

In fact, it is an important 2-qubit quantum gate called Controlled-NOT or CNOT gate, and it will be used many times later in this book. Its name comes from the following interpretation: if the first qubit is in state \(\left|0\right>\), don’t do anything to the second qubit; if the first qubit is in state \(\left|1\right>\), flip the second qubit state.

Here we pick XOR gate as an example because it is a reversible gate. Other classical two-bit gates such as AND gate and OR gate are irreversible, and their extension to the quantum region will involve three qubits. We will see that in the next chapter.

CNOT gate belongs to an important class of quantum gates: the controlled-unitary gates. In the two-qubit case, controlled-unitary gates have an explicit matrix form

\[\begin{split}C(U) = \begin{bmatrix} 1& 0 & 0 & 0\\ 0& 1 & 0 & 0\\ 0& 0 & u_{00} & u_{01}\\ 0& 0 & u_{10} & u_{11} \end{bmatrix}\end{split}\]

where \(u_{ij}\) are the matrix components of the single-qubit unitary gate.

separable states and entangled states

Some of the two-qubit states can be constructed from two single-qubit states, i.e.,

(14)\[\left|\psi\right> = \left|\psi_1\right> \otimes \left|\psi_2\right>\]

These states are called separable states. The states that cannot be decomposed to tensor product of single-qubit states are called entangled states. Recall from the previous chapter that each single-qubit pure state has only two real parameters. Thus two-qubit separable states live in a four dimensional subspace of the six dimensional two-qubit state space. In other words, there are a lot more entangled states than separable states.

See also

Here I only talk about pure state. The identification and quantification of entanglement in mixed states is more complicated. In short, a separable two-qubit state can be put in the form of

\[\rho = \sum_{i=1}^K p_i \rho_A \otimes \rho_B\]

for some \(K\) and \(\{p_i\}\). You can see that it’s a much harder problem than the pure state separability, because of the extra parameters \(\{p_i\}\).

  • Schmacher

Bell states

The nominal examples of two-qubit entangled states are the so-called Bell states:

(15)\[\begin{split}\left|\Phi^+\right> & = \left|00\right> + \left|11\right> \\ \left|\Phi^-\right> & = \left|00\right> - \left|11\right> \\ \left|\Psi^+\right> & = \left|01\right> + \left|10\right> \\ \left|\Psi^-\right> & = \left|01\right> - \left|10\right>\end{split}\]

Here I omit the normalization factor \(1/\sqrt{2}\) to save typing. Note that the four Bell states are orthogonal to each other thus form a basis set. We will see later that it is a very useful basis set to use.

Bell states can be transformed to the computational basis states \(\left|00\right>\), \(\left|01\right>\), \(\left|10\right>\), \(\left|11\right>\) by two gate operations

(16)\[\begin{split}\left|00\right> =H\otimes I \cdot CNOT \left|\Phi^+\right> \\ \left|01\right> =H\otimes I \cdot CNOT \left|\Psi^+\right> \\ \left|10\right> =H\otimes I \cdot CNOT \left|\Phi^-\right> \\ \left|11\right> =H\otimes I \cdot CNOT \left|\Psi^-\right>\end{split}\]

and equivalently,

\[\begin{split}CNOT \cdot H\otimes I \left|00\right> = \left|\Phi^+\right> \\ CNOT \cdot H\otimes I \left|01\right> = \left|\Psi^+\right> \\ CNOT \cdot H\otimes I \left|10\right> = \left|\Phi^-\right> \\ CNOT \cdot H\otimes I \left|11\right> = \left|\Psi^-\right>\end{split}\]

The significance of the computational bases is that they represent states on hardware and in reality one can only measure these states. Thus if we know some computation results are in one of the four Bell states, we can apply CNOT and Hadamart gate to convert the results to computational basis states first, and then make the measurement.

Separable state is the quantum analogy of statistical independent random variables. Entangled states have the peculiar feature that the total system is less complex than the subsystems. To fully understand this statement, one needs to know the concept of entropy in classical information theory. For simplicity, I will use \(\left|\Phi^+\right>\) as example.

If we measure \(\left|\Phi^+\right>\) in the two-qubit computational bases, the measurement outcome follows Bernoulli distribution with \(p=0.5\), just like tossing a fair coin.

See also

Here is a short introduction to entropy. The essense of entropy is state counting. The more states the bigger the entropy. And entropy quantifies the amount of information, or uncertainty, or possibilities. Suppose the system could be in \(N\) states with equal probability, then its entropy is

\[S = \log N.\]

Here the logrithm function is a historical convention, any monotonic function could be used. Its base is also arbitrary as long as used consistently. In classical information theory, \(N=2\) is of special importance because that is the number of states for a bit. In that case, base 2 is used and we have \(S=\log_2 2= 1\), i.e., one bit of information. This is also the entropy of a fair coin.

What if the coin is not fair? Here we cannot count the states directly since they are not of equal probability. Instead, we can count something else that

  • both characterizes the coin bias and
  • is made of individual events of equal probabilities.

This quantity is the number of configurations with certain heads given a fixed number of coin tosses. For example, suppose we make four tosses and focus on configurations with two heads. There are six such configurations: HHTT, HTHT, HTTH, THHT, THTH, TTHH. Here HTHT means the first and third tosses end up with heads. Note that these configurations are of the same probability, no matter what the coin bias is.

There are still the questions of how many tosses to do and how many heads to focus on. Obviously any finite number of tosses doesn’t make sense. And the law of large numbers automatically fixes the head counts for us, i.e., with \(M\) tosses, the number of heads is

\[M_H = Mp_H\]

in the limit of large \(M\), and \(p_H\) is the head probability. This is where the coin bias enters the picture.

This idea can be expressed as

\[S = \lim_{M\rightarrow \infty} \frac{\log_2 C(M, M_H)}{M}\]

where \(C(n, k)\) is the n-choose-k function, and the denominator is there to prevent the numerator to blow up.

Using Stirling’s approximation,

\[\log M! = M\log M -M + O(\log M)\]

we get the binary entropy function

\[S = -p_H\log p_H - p_T\log p_T\]

It is consistent with the fair coin case. Also the fair coin case has the largest entrpoy, which is \(1\).

It turns out that entanglement is a useful resource for quantum communication. I will give two examples here - quantum superdense coding and quantum teleportation. Usually people think entanglement is also important to get speedup in quantum algorithms than their classical counterparts. But this topic is a bit controversial. It turns out that entanglement is not the only non-classical correlation possessed by qubits (see quantum discord for example), and there are quantum algorithms providing exponential speedup over best-known classical algorithms and the computation processes do not contain entanglement.

quantum superdense coding

The purpose of quantum superdense coding is to send one qubit which carries four messages, i.e., two bits of classical information. This work is published in

      1. Bennett and Stephen J. Wiesner, Phys. Rev. Lett. 69, 2881 (1992)

It may appear that one qubit could encode infinite messages since its state space is continuous (recall the qubit state is a 2D complex vector in terms of wave function, or a 3D real vector on unit sphere in terms of Bloch vector). However, encoding messages in non-orthogonal quantum states messes up the decoding, due to the wave function collapse of measurement.

For example, suppose we encode four messages in

\[\begin{split}\begin{align}\left|0\right> =& \begin{bmatrix}1 \\ 0 \end{bmatrix} \\ \left|1\right> =& \begin{bmatrix}0 \\ 1 \end{bmatrix} \\ \left|+\right> =& \left|0\right> + \left|1\right> \\ \left|-\right> =& \left|0\right> - \left|1\right> \end{align}\end{split}\]

and we receive an encoded qubit from someone. No matter how we measure it, we won’t be sure about which state it is in. In fact, there is a so-called Holevo’s bound which says one qubit can at most carry the information of one classical bit.

The trick of quantum superdense coding is to first share a Bell state before transmitting the message-carrying qubit. For example, share \(\left|\Psi^-\right>\) between Alice and Bob.

The encoding can be done by applying one of the four gates \(\{I, X, Y, Z\}\) on Alice’s qubit,

\[\begin{split}\begin{align} I \left|\Psi^-\right> =& \left|\Psi^-\right> \\ X \left|\Psi^-\right> =& \left|\Phi^-\right> \\ Y \left|\Psi^-\right> =& i\left|\Phi^+\right> \\ Z \left|\Psi^-\right> =& \left|\Psi^+\right> \\ \end{align}\end{split}\]

Then Alice can send her qubit to Bob, and Bob can decode the message by converting the Bell state to computational basis state as in Eq. (16) and making a measurement.

Note that quantum superdense coding does not break Holevo’s bound because two qubits are sent. It’s just that one qubit can be sent at an earlier time and it alone does not carry information.

quantum teleportation

The purpose of quantum teleportation is to instantaneously destroy a quantum state on Alice’s side and create it on Bob’s side without sending any qubits between Alice and Bob. This work is published in

  • C. H. Bennett, G. Brassard, C. Crépeau, R. Jozsa, A. Peres, W. K. Wootters Teleporting an Unknown Quantum State via Dual Classical and Einstein-Podolsky-Rosen Channels, Phys. Rev. Lett. 70, 1895 (1993).

Again, the prerequisite is to have Alice and Bob share an entangled state, say \(\left|\Phi+\right>\). The initial state can be written as

\[\left|\xi\right>_A\otimes\left(\left|0\right>_A\left|0\right>_B + \left|1\right>_A\left|1\right>_B \right) = (a\left|0\right> + b\left|1\right>)_A \otimes \left(\left|0\right>_A\left|0\right>_B + \left|1\right>_A\left|1\right>_B \right)\]

where the subscripts denote the qubits ownership.

Expressing Alice’s two qubits in Bell state basis, we get

\[\begin{split}\begin{align}&\left|\Psi^+\right>_A \otimes (a\left|0\right> + b \left|1\right>)_B\\ +& \left|\Psi^-\right>_A \otimes (a\left|0\right> - b \left|1\right>)_B \\ +& \left|\Phi^+\right>_A \otimes (b\left|0\right> + a \left|1\right>)_B\\ + &\left|\Phi^-\right>_A \otimes (-b\left|0\right> + a \left|1\right>)_B\end{align}\end{split}\]

Thus by measuring her qubits in the Bell state basis, Alice can project Bob’s qubit into one of the four states. So if Alice further informs Bob about her measurement result, say from classical communication such as phone call, Bob can recover the original state \(\left|\xi\right>_A\). For example, if Alice’s measurement result is \(\left|\Psi-\right>\), then Bob can apply \(Z\) gate on his qubit.

One may wonder why not sending \(\left|\xi\right>\) directly. One potential usage of teleportation is to send half of the Bell state when quantum communication cost is low and when the cost is high, one can use cheap classical communication to effectively transfer quantum state.

entanglement measure

Since quantum entanglement can be used as communication resource, it is of practical interest to quantify entanglement.

For two-qubit pure states, verifying whether a state is separable is simple. Given state in Eq. (12), we can check the validity of the following equalities

\[\begin{split}\left\{\begin{array}{ll} \frac{\alpha_0}{\alpha_2} = \frac{\alpha_1}{\alpha_3}\\ \frac{\alpha_0}{\alpha_1} = \frac{\alpha_2}{\alpha_3} \end{array} \right.\end{split}\]

A natural question to ask is: are some entangled states more entangled than others? For example, it seems plausible that the state

\[\left|\psi_\epsilon\right> = \epsilon\left|00\right> + (1-\epsilon)\left|\Psi^- \right>\]

is less entangled than the Bell state and more entangled than \(\left|00\right>\).

There are a lot of open questions in entanglement measure. And many interesting results exist for small systems: two-qubits (dimension 4), qubit-qutrit (dimension 6), three-qubits (dimension 8). Since this is an introductory book, maybe I stop here.