Summing Qubit States

Computer Science Level 1

Write the qubit state

∣ Ψ ⟩ = i ∣ 0 ⟩ − ( 1 + i ) ( ∣ 0 ⟩ + ∣ 1 ⟩ ) |\Psi\rangle = i|0\rangle - (1+i) \big(|0\rangle + |1\rangle\big)

as a two-dimensional vector.

( i − 1 − 1 − i ) \begin{pmatrix} i-1 \\ -1-i \end{pmatrix} ( i − 1 − i ) \begin{pmatrix} i \\ -1-i \end{pmatrix} ( − 1 − 1 − i ) \begin{pmatrix} -1 \\ -1-i \end{pmatrix} ( 1 − 1 − i ) \begin{pmatrix} 1 \\ -1-i \end{pmatrix}

Matt DeCross by Matt DeCross , 27, USA

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Matt DeCross
Jan 19, 2016

The conventional basis for representing qubit states is ∣ 0 ⟩ = ( 1 0 ) |0\rangle = \begin{pmatrix} 1 \\ 0 \end{pmatrix} , ∣ 1 ⟩ = ( 0 1 ) |1\rangle = \begin{pmatrix} 0 \\ 1\end{pmatrix} . We can thus write the state ∣ Ψ ⟩ |\Psi\rangle as:

∣ Ψ ⟩ = ( i − ( 1 + i ) ) ∣ 0 ⟩ − ( 1 + i ) ∣ 1 ⟩ = − 1 ∣ 0 ⟩ − ( 1 + i ) ∣ 1 ⟩ = − 1 ∗ ( 1 0 ) − ( 1 + i ) ∗ ( 0 1 ) = ( − 1 − 1 − i ) . \begin{aligned} |\Psi\rangle &= (i-(1+i))|0\rangle -(1+i)|1\rangle = -1|0\rangle - (1+i)|1\rangle \\ &= -1*\begin{pmatrix} 1 \\ 0 \end{pmatrix} - (1+i)*\begin{pmatrix} 0 \\ 1 \end{pmatrix} = \begin{pmatrix} -1 \\ -1-i\end{pmatrix}.\end{aligned}

13 Helpful 3 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...