Real Hilbert Space 1 | M.E. Irizarry-Gelpí

Fri 09 February 2018
Physics
#quantum

The qubit is the workhorse in quantum computation. Since it is based on an actual physical system, it uses the full technology of quantum mechanics: complex numbers and a 2-dimensional complex Hilbert space. A 1-qubit quantum state is given by

$| \psi \rangle = a | 0 \rangle + b | 1 \rangle.$

Here $a$ and $b$ are complex numbers that satisfy

$|a|^{2} + |b|^{2} = 1.$

So instead of four real degrees of freedom, you have three. Actually, if you also allow quantum states to be equivalent up to a constant phase factor, then you only have two real degrees of freedom.

Consider restricting the coefficients to be real instead of complex. Then, instead of unitary matrices (members of the group $U(2)$ ), then you have orthogonal matrices (members of the group $O(2)$ ). These are matrices $M$ that satisfy the identity

$M M^{T} = M^{T} M = I.$

The 2-by-2 orthogonal matrices come in two kinds:

$O(2) = O_{+}(2) \oplus O_{-}(2).$

Members of $O_{+}(2)$ have determinant $+1$ , and have the form:

$R_{+}(\alpha) = \begin{pmatrix} \cos{(\alpha)} & -\sin{(\alpha)} \\ \sin{(\alpha)} & \cos{(\alpha)} \end{pmatrix} = \cos{(\alpha)} I + \sin{(\alpha)} Q, \qquad I \equiv \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}, \qquad Q \equiv \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}.$

Similarly, members of $O_{-}(2)$ have determinant $-1$ , and have the form:

$R_{-}(\beta) = \begin{pmatrix} \sin{(\beta)} & \cos{(\beta)} \\ \cos{(\beta)} & -\sin{(\beta)} \end{pmatrix} = \cos{(\beta)} J + \sin{(\beta)} P, \qquad J \equiv \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, \qquad P \equiv \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}.$

Note that

$J^2 = P^2 = I, \qquad Q^2 = -I.$

Also

$JP + PJ = PQ + QP = QJ + JQ = 0.$

However,

$JP = Q, \qquad JQ = P, \qquad QP = J.$

You can write a member of $O_{+}(2)$ in exponential form

$R_{+}(\alpha) = \exp{\left( \alpha Q \right)}.$

In this form, it is clear that

$R_{+}(\alpha) R_{+}(\beta) = R_{+}(\alpha + \beta).$

This also means that

$R_{+}(\alpha) R_{+}(\beta) - R_{+}(\beta) R_{+}(\alpha) = 0.$

That is, the elements of $O_{+}(2)$ commute with each other.

You can also check that

$R_{+}(\alpha) R_{-}(\beta) = R_{-}(\beta - \alpha), \qquad R_{-}(\alpha) R_{+}(\beta) = R_{-}(\alpha + \beta), \qquad R_{-}(\alpha)R_{-}(\beta) = R_{+}(\beta - \alpha).$

The last identity implies that the elements of $O_{-}(2)$ do not commute with each other:

$R_{-}(\alpha)R_{-}(\beta) - R_{-}(\beta)R_{-}(\alpha) = 2 \sin{(\beta - \alpha)} Q.$

Indeed, you can use $J$ to write elements of one kind in terms of elements of the other kind:

$R_{-}(\alpha) = J R_{+}(\alpha), \qquad R_{+}(\alpha) = J R_{-}(\alpha).$

Instead of two real degrees of freedom, a (normalized) state vector can only have one real degree of freedom:

$| \theta \rangle = \cos{(\theta)} |0^{\circ} \rangle + \sin{(\theta)} | 90^{\circ} \rangle.$

Instead of being equivalent up to a complex phase factor, now states are equivalent up to a sign. Thus, the angle $\theta$ above is restricted to

$0 \leq \theta < \pi.$

Note that

$R_{+}(\alpha) | \theta \rangle = | \theta + \alpha \rangle, \qquad R_{-}(\alpha) J | \theta \rangle = | \theta - \alpha \rangle.$

The inner product between two state vectors is

$\langle \alpha | \beta \rangle = \cos{(\alpha - \beta)}.$

Other interesting invariants are

$\langle \alpha | J | \beta \rangle = \sin{(\alpha + \beta)}, \qquad \langle \alpha | P | \beta \rangle = \cos{(\alpha + \beta)}, \qquad \langle \alpha | Q | \beta \rangle = \sin{(\alpha - \beta)}.$

In particular, the last one means that $Q | \theta \rangle$ is orthogonal to $| \theta \rangle$ .