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.

ReplyOptionsI agree with FlippingBits. Just complementing the anwser

A single qubit state $\ket \psi$ is a superposition in respect to a given basis $\{\ket K , \ket{K_\perp}\}$ if it can be written as a

non triviallinear combination of the basis (as stated in the second paragraph of FlippingBit)To find a basis for which $\ket \psi$ is not a superposition, all one needs to do is find a orthogonal state $\ket {\psi_\perp}$

$\braket \psi {\psi_\perp} = 0$, and the new basis is of course $\{\ket \psi , \ket{\psi_\perp}\}$

a)As explained by FlippingBitb)$\frac{1}{\sqrt{2}}(\ket + + \ket -) \Rightarrow \frac{1}{\sqrt{2}}( \frac{1}{\sqrt{2}}(\ket 0 + \ket 1) + \frac{1}{\sqrt{2}}(\ket 0 - \ket 1)) \Rightarrow \frac{1}{2}(\ket 0 + \ket 1) + \frac{1}{2}(\ket 0 - \ket 1)$

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

"b)" is also not a superposition (non trivial linear combination)

c)$\frac{1}{\sqrt{2}}(\ket + - \ket -) \Rightarrow \frac{1}{\sqrt{2}}( \frac{1}{\sqrt{2}}(\ket 0 + \ket 1) - \frac{1}{\sqrt{2}}(\ket 0 - \ket 1)) \Rightarrow \frac{1}{2}(\ket 0 + \ket 1) - \frac{1}{2}(\ket 0 - \ket 1)$

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

"c)" is not a superposition (non trivial linear combination)

d)$\frac{\sqrt3}{2}\ket + - \frac{1}{2}\ket - \Rightarrow \frac{\sqrt3}{2}( \frac{1}{\sqrt{2}}(\ket 0 + \ket 1) ) - \frac{1}{2} (\frac{1}{\sqrt{2}}(\ket 0 - \ket 1))$

$\frac{1}{\sqrt{2}}(\ket + - \ket -) = \frac{\sqrt3 - 1}{2\sqrt{2}}\ket 0 + \frac{\sqrt{3} + 1}{2\sqrt{2}}\ket 1$ which is a trivial case,

therefore "d) is a superposition"

In the basis $\{\frac{\sqrt3 - 1}{2\sqrt{2}}\ket 0 + \frac{\sqrt{3} + 1}{2\sqrt{2}}\ket 1 ,\frac{\sqrt3 - 1}{2\sqrt{2}}\ket 0 - \frac{\sqrt{3} + 1}{2\sqrt{2}}\ket 1\}$ this state is not in superposition. as explained by FlippingBit

One may check by doing $\braket {(\frac{\sqrt3 - 1}{2\sqrt{2}}\ket 0 + \frac{\sqrt{3} + 1}{2\sqrt{2}}\ket 1)} {(\frac{\sqrt3 - 1}{2\sqrt{2}}\ket 0 - \frac{\sqrt{3} + 1}{2\sqrt{2}}\ket 1\})} = 0$

e)$\frac{1}{\sqrt{2}}(\ket i - \ket{-i}) \Rightarrow \frac{1}{\sqrt{2}}( \frac{1}{\sqrt{2}}(\ket 0 + i\ket 1) - \frac{1}{\sqrt{2}}(\ket 0 - i\ket 1)) \Rightarrow \frac{1}{2}(\ket 0 + i\ket 1) - \frac{1}{2}(\ket 0 - i\ket 1)$

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

"e)" is not a superposition (non trivial linear combination)

f)as explained by FlippingBit

ReplyOptions