2.4.3 Reflections

Any qubit operation is either a rotation or a reflection. We are already familiar with the most general rotation, U⁢(θ)U(\theta), defined in Eq. 2.27. However, in terms of reflections we have encountered only two of them so far: ZZ and NOT\mathrm{NOT}, see Eqs. 2.26 and 2.25. But what does the most general reflection look like?

One way of obtaining any reflection is by taking some fixed reflection (say, the NOT\mathrm{NOT} reflection) and composing it with suitable rotations so that the axis of the reflection is adjusted by the right amount. In the following exercise, you will show how to obtain the ZZ reflection from the NOT\mathrm{NOT} reflection in two different ways.

Homework 2.4 (Z from NOT\mathrm{NOT}).

Let ZZ, NOT\mathrm{NOT}, and U⁢(θ)U(\theta) be the qubit operations defined in Eqs. 2.26, 2.25 and 2.27.

  1. 1.
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    Find an angle θ\theta such that Z=U⁢(θ)⁢NOT⁢U⁢(−θ)Z=U(\theta)\,\mathrm{NOT}\,U(-\theta).

  2. 2.
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    Find an angle θ\theta such that Z=NOT⁢U⁢(θ)Z=\mathrm{NOT}\,U(\theta).

Can you visualize these two sequences of transformations on the circle?

Hint: Take a look at Fig. 2.4 and the figure you drew for Homework 2.2.

Hack.
  1. 1.
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    Choose θ=−π/4\theta=-\pi/4. We check that the resulting operation U⁢(−π/4)⁢NOT⁢U⁢(π/4)U(-\pi/4)\,\mathrm{NOT}\,U(\pi/4) is the same as ZZ by checking that it acts the same on the states |0⟩\left|0\right\rangle and |1⟩\left|1\right\rangle:

    U⁢(−π4)⁢NOT⁢U⁢(π4)⁢|0⟩\displaystyle U\left\lparen-\frac{\pi}{4}\right\rparen\,\mathrm{NOT}\,U\left% \lparen\frac{\pi}{4}\right\rparen\left|0\right\rangle =U⁢(−π4)⁢NOT⁢|+⟩=U⁢(−π4)⁢|+⟩=|0⟩,\displaystyle=U\left\lparen-\frac{\pi}{4}\right\rparen\,\mathrm{NOT}\,\left|+% \right\rangle=U\left\lparen-\frac{\pi}{4}\right\rparen\left|+\right\rangle=% \left|0\right\rangle,
    U⁢(−π4)⁢NOT⁢U⁢(π4)⁢|1⟩\displaystyle U\left\lparen-\frac{\pi}{4}\right\rparen\,\mathrm{NOT}\,U\left% \lparen\frac{\pi}{4}\right\rparen\left|1\right\rangle =−U⁢(−π4)⁢NOT⁢|−⟩=U⁢(−π4)⁢|−⟩=−|1⟩.\displaystyle=-U\left\lparen-\frac{\pi}{4}\right\rparen\,\mathrm{NOT}\,\left|-% \right\rangle=U\left\lparen-\frac{\pi}{4}\right\rparen\left|-\right\rangle=-% \left|1\right\rangle.

    By linearity, this implies that the operation acts exactly the same on all states.

  2. 2.
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    Choose θ=π/2\theta=\pi/2. Then

    NOT⁢U⁢(π2)⁢|0⟩\displaystyle\mathrm{NOT}\,U\left\lparen\frac{\pi}{2}\right\rparen\left|0\right\rangle =NOT⁢|1⟩=|0⟩,\displaystyle=\mathrm{NOT}\,\left|1\right\rangle=\left|0\right\rangle,
    NOT⁢U⁢(π2)⁢|1⟩\displaystyle\mathrm{NOT}\,U\left\lparen\frac{\pi}{2}\right\rparen\left|1\right\rangle =−NOT⁢|0⟩=−|1⟩.\displaystyle=-\mathrm{NOT}\,\left|0\right\rangle=-\left|1\right\rangle.

It turns out that you can in fact obtain any reflection by using a similar trick. The most general reflection is of the form

V⁢(θ)=NOT⁢U⁢(θ)=U⁢(−θ)⁢NOT.V(\theta)=\mathrm{NOT}\,U(\theta)=U(-\theta)\,\mathrm{NOT}. (2.33)

For example, a very useful operation is the Hadamard transformation that acts on the basis states as follows (see Fig. 2.6):

H⁢|0⟩=12⁢(|0⟩+|1⟩)=|+⟩,H⁢|1⟩=12⁢(|0⟩−|1⟩)=|−⟩.H\left|0\right\rangle=\frac{1}{\sqrt{2}}\left\lparen\left|0\right\rangle+\left% |1\right\rangle\right\rparen=\left|+\right\rangle,\qquad H\left|1\right\rangle% =\frac{1}{\sqrt{2}}\left\lparen\left|0\right\rangle-\left|1\right\rangle\right% \rparen=\left|-\right\rangle. (2.34)

It is obtained as the following special case of the general reflection:

H=V⁢(π/4).H=V(\pi/4). (2.35)

In summary, any qubit operation is either a rotation U⁢(θ)U(\theta) or a reflection V⁢(θ)V(\theta), for some angle θ\theta.

Figure 2.6: The Hadamard operation HH on a qubit amounts to a reflection about the 45/245/2 degree (or π/8\pi/8) axis (dotted). Depicted are also the states |0⟩\left|0\right\rangle, |1⟩\left|1\right\rangle, |+⟩\left|+\right\rangle, and |−⟩\left|-\right\rangle from Eq. 2.34.