Exercise 2.3 Discussion

$\def\i{\mathbf {i}}\def\ket#1{|{#1}\rangle}\def\bra#1{\langle{#1}|}\def\braket#1#2{\langle{#1}|{#2}\rangle}\mathbf{Exercise\ 2.3}$

Which states are superpositions with respect to the standard basis, and which are not? For each state that is a superposition, give a basis with respect to which it is not a superposition.

a) $\ket +$

b) $\frac{1}{\sqrt 2}(\ket + + \ket -)$

c) $\frac{1}{\sqrt 2}(\ket + - \ket -)$

d) $\frac{\sqrt 3}{2}\ket + - \frac{1}{2} \ket -)$

e) $\frac{1}{\sqrt 2}(\ket\i - \ket{-\i} )$

f) $\frac{1}{\sqrt 2}(\ket 0 - \ket 1)$

page revision: 1, last edited: 12 Nov 2011 05:41

The states in (a), (d), and (f) are superpositions with respect to the standard basis.

A single-qubit state $\ket{\psi}$ is not a superposition with respect to a basis $\{\ket{\psi}, \ket{\psi_{\perp}} \}$ that contains the state $\ket{\psi}$ as well as a unit vector $\ket{\psi_{\perp}}$ that is orthogonal to $\ket{\psi}$.

Thus, the state in (a) is not a superposition with respect to the Hadamard basis $\{ \ket{+}, \ket{-} \}$.

The state in (d) is not a superposition with respect to the basis $\{\tfrac{\sqrt{3}}{2}\ket{+} - \tfrac{1}{2}\ket{-} \,,\, \tfrac{1}{2}\ket{+} + \tfrac{\sqrt{3}}{2}\ket{-} \}$.

The state in (f) is equal to $\ket{-}$, so it is not a superposition with respect to the Hadamard basis.

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