5.2 From one qubit to two

In classical physics, the transition from a single object to a composite system of many objects is trivial: in order to describe the state of, say, 42 objects at any given moment of time, it is sufficient to describe the state of each of the objects separately. Indeed, the classical state of 42 point-like particles is described by specifying the position and the momentum of each particle.

In the classical world, “the whole is exactly the sum of its parts”; in the quantum world, Aristotle had it right when he said “the whole is greater than the sum of its parts”.

Consider, for example, a pair of qubits. Suppose that each one is described by a state vector: the first one by |a\rangle, and the second one by |b\rangle. One might therefore think that the most general state of the two qubits should be represented by a pair of state vectors, |a\rangle|b\rangle, with one for each qubit. Indeed, such a state is certainly possible, but there are other states that cannot be expressed in this form. In order to write down the most general state of two qubits we first focus on the basis states.

For a single qubit we have been using the standard basis \{|0\rangle,|1\rangle\}. For two qubits we may choose the following as our standard basis states:104 \begin{array}{cc} |00\rangle \equiv |0\rangle|0\rangle & |01\rangle \equiv |0\rangle|1\rangle \\|10\rangle \equiv |1\rangle|0\rangle &|11\rangle \equiv |1\rangle|1\rangle. \end{array} Within each ket, the first symbol refers to the first qubit, and the second to the second, and we have tacitly assumed that we can distinguish the two qubits by their location, or some other means.

Now, the most general state of the two qubits (a bipartite state) is a normalised linear combination of these four basis states, i.e. a vector of the form |\psi\rangle = c_{00}|00\rangle + c_{01}|01\rangle + c_{10}|10\rangle + c_{11}|11\rangle. Physical interpretation aside, let us count how many real parameters are needed to specify this state. Six, right? We have four complex numbers (the c_{ij}), which gives eight real parameters; we then restrict by the normalisation condition, along with the fact that states differing only by a global phase factor are equivalent, which leaves us with six real parameters. Now, by the same line of argument, we need only two real parameters to specify the state of a single qubit, and hence need four real parameters to specify any state of two qubits of the form |a\rangle|b\rangle.

But four is less than six! So it cannot be the case that every state of two qubits can be expressed as a pair of states |a\rangle|b\rangle, simply for “dimension reasons”.

For example, compare the two states of two qubits, \frac{1}{\sqrt{2}}|00\rangle + \frac{1}{\sqrt{2}}|01\rangle \quad\text{and}\quad \frac{1}{\sqrt{2}}|00\rangle + \frac{1}{\sqrt{2}}|11\rangle. The first one is separable, i.e. we can view it as a pair of state vectors where each one pertains to one of the two qubits: \frac{1}{\sqrt{2}}|00\rangle + \frac{1}{\sqrt{2}}|01\rangle = \frac{1}{\sqrt{2}} \;\; \underbrace{\vphantom{\frac{1}{\sqrt{2}}}|0\rangle}_{\text{qubit 1}} \;\; \underbrace{(|0\rangle+|1\rangle)}_{\text{qubit 2}}, The second state, however, does not admit such a decomposition: there do not exist any \psi_1,\psi_2 such that \frac{1}{\sqrt{2}}|00\rangle + \frac{1}{\sqrt{2}}|11\rangle = |\psi_1\rangle|\psi_2\rangle and so we say that it is an entangled state.

Any bipartite state that cannot be viewed as a pair of two states pertaining to the constituent subsystems is said to be entangled.

We’ll give another, equivalent but more mathematical (and notational), definition of entanglement once we understand how tensor products work.

  1. It looks like we are defining some sort of “multiplication rule” for kets here, saying that |a\rangle|b\rangle:=|ab\rangle. This is indeed the case, but to talk about this properly we need to introduce the idea of tensor products (which we do very soon, in Section 5.3).↩︎