Exercise 2.3 Discussion / Discussion

This is the discussion related to the wiki page Exercise 2.3 Discussion.

Solution to Exercise 2.3

FlippingBits 30 May 2014 23:59

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.

Last edited on 31 May 2014 01:01 by FlippingBits Show more

bernardobl 19 Jul 2017 14:11

I 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 trivial linear 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 FlippingBit

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

"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

Last edited on 19 Jul 2017 15:41 by bernardobl Show more

rshaffer 19 Feb 2019 15:51

We use the standard convention:

(1)

\begin{align} \ket{+} &= \frac{1}{\sqrt2}(\ket{0} + \ket{1}) \qquad \ket{\mathbf{i}} = \frac{1}{\sqrt2}(\ket{0} + i \ket{1}) \\ \ket{-} &= \frac{1}{\sqrt2}(\ket{0} - \ket{1}) \qquad \ket{\mathbf{-i}} = \frac{1}{\sqrt2}(\ket{0} - i \ket{1}) \end{align}

a.
$\ket{+} = \frac{1}{\sqrt2}(\ket{0} + \ket{1})$ is a superposition with respect to the standard basis.
It is not a superposition with respect to the Hadamard $\{\ket{+}, \ket{-}\}$ basis.

b.
$\frac{1}{\sqrt2}(\ket{+} + \ket{-}) = \ket{0}$ is not a superposition with respect to the standard basis.

c.
$\frac{1}{\sqrt2}(\ket{+} - \ket{-}) = \ket{1}$ is not a superposition with respect to the standard basis.

d.
$\frac{\sqrt3}{2} \ket{+} - \frac{1}{2} \ket{-}$ is a superposition with respect to the standard basis.
To find a basis in which this is not a superposition, we need to find a vector which is orthogonal. Let $\ket{v} = \frac{\sqrt3}{2} \ket{+} - \frac{1}{2} \ket{-}$. We are looking for $\ket{v^\perp} = \alpha \ket{+} + \beta \ket{-}$ such that $\braket{v}{v^\perp} = 0$ and $|\alpha|^2 + |\beta|^2 = 1$:

(2)

\begin{align} \braket{v}{v^\perp} &= \Bigg[ \frac{\sqrt3}{2} \bra{+} - \frac{1}{2} \bra{-} \Bigg]\Bigg[ \alpha \ket{+} + \beta \ket{-} \Bigg] \\ &= \frac{\sqrt3}{2} \alpha - \frac{1}{2} \beta = 0 \implies \alpha = \frac{1}{2},\ \beta = \frac{\sqrt3}{2} \end{align}

Thus, a basis in which $\ket{v}$ is not a superposition is $\{ \ket{v}, \ket{v^\perp} \}$, where these are defined as above.

e.
$\frac{1}{\sqrt2}(\ket{\mathbf{i}} - \ket{\mathbf{-i}}) = i \ket{1}$ is not a superposition with respect to the standard basis.

f.
$\frac{1}{\sqrt2}(\ket{0} - \ket{1}) = \ket{-}$ is a superposition with respect to the standard basis.
It is not a superposition with respect to the Hadamard $\{\ket{+}, \ket{-}\}$ basis.