Quantum Information HT24, Bra-ket notation

Created: January 15, 2024 | Read markdown | About these notes | View in context | Study these flashcards |

Flashcards

How are $ \vert 0\rangle$ and $ \vert 1\rangle \in \mathbb C^2$ defined in bra-ket notation, and more generally, how is $ \vert k\rangle \in \mathbb C^d$ defined?

bra-ket-computational-basis-vectors
\[\begin{aligned} \vert 0\rangle &:= {1 \choose 0} \in \mathbb C^2 \\ \vert 1\rangle &:= {0 \choose 1} \in \mathbb C^2 \\ \vert k\rangle &:= e _ k \in \mathbb C^d \\ \end{aligned}\]

where $e _ k$ is the $k$-th unit vector.

Suppose $\psi = {\alpha \choose \beta} \in \mathbb C^2$. What would this equal in bra-ket notation?

column-vector-to-ket
\[\psi = \alpha \vert 0\rangle + \beta \vert 1\rangle\]

Suppose $\psi = {\alpha \choose \beta} \in \mathbb C^2$. What would $\langle \psi \vert $ equal in ordinary vector notation?

column-vector-to-bra
\[\langle\psi \vert = \overline \alpha \langle 0 \vert + \overline \beta\langle 1 \vert = \overline \alpha \begin{pmatrix} 1 & 0 \end{pmatrix} + \overline \beta \begin{pmatrix} 0 & 1 \end{pmatrix}\]

Suppose $\psi = {\alpha \choose \beta}, \psi' = {\alpha' \choose \beta'} \in \mathbb C^2$. What is $\langle\psi \vert \psi'\rangle$

bra-ket-inner-product
\[\begin{pmatrix}\overline \alpha & \overline \beta\end{pmatrix} \begin{pmatrix}\alpha' \\ \beta'\end{pmatrix} = \overline \alpha \alpha' + \overline \beta \beta'\]

Suppose $\psi = {\alpha \choose \beta}$. How can you write $\langle \psi \vert $ more familiarly in terms of $\psi$?

bra-is-conjugate-transpose
\[\psi^\dagger\]

where $\dagger$ represents the conjugate transpose.

What are column vectors called in bra-ket notation?

ket-definition

Kets.

What are row vectors called in bra-ket notation?

bra-definition

Bras.

Suppose you have

\[ \vert \alpha \rangle = \begin{pmatrix}\alpha _ 0 \\ \alpha _ 1 \\\vdots \\ \alpha _ {d-1}\end{pmatrix}\]

and

\[ \vert \beta \rangle = \begin{pmatrix}\beta _ 0 \\ \beta _ 1 \\\vdots \\ \beta _ {d-1}\end{pmatrix}\]

What does $ \vert \alpha \rangle \langle \beta \vert $ equal?

outer-product-matrix
\[\begin{pmatrix} \alpha _ 0 \overline \beta _ 0 & \alpha _ 0\overline{\beta _ 1} & \cdots \\ \alpha _ 1 \overline \beta _ 0& \alpha _ 1 \overline \beta _ 1 & \cdots \\ \vdots & \vdots & \ddots \end{pmatrix}\]

Suppose $A = \vert \alpha \rangle \langle \beta \vert $. What is $A^\dagger$?

outer-product-conjugate-transpose
\[ \vert \beta\rangle\langle \alpha \vert \]

Suppose you have an orthonormal basis $\{ \vert \psi _ k\rangle : k = 0, \cdots, d-1 \}$. What is

\[\sum^{d-1} _ {m = 0} \vert \psi _ m\rangle\langle\psi _ m \vert \]
completeness-relation
\[I\]

@Define $ \vert 0, \theta \rangle$ and $ \vert 1, \theta\rangle$.

theta-rotated-basis-definition
\[\begin{aligned} \vert 0, \theta\rangle &:= \cos \frac \theta 2 \vert 0\rangle + \sin \frac \theta 2 \vert 1\rangle \\ \vert 1, \theta\rangle &:= \sin \frac \theta 2 \vert 0\rangle - \cos \frac \theta 2 \vert 1\rangle \end{aligned}\]

Proofs



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