In matrix notation, say I have the vector $\begin{bmatrix} 1 \\ 0 \end{bmatrix}$. It is currently represented in the computational basis $\{\begin{bmatrix} 1 \\ 0\end{bmatrix}, \begin{bmatrix} 0 \\ 1\end{bmatrix}\}$. I want to now represent it in the basis $\{\begin{bmatrix} \frac{1}{\sqrt 2} \\ \frac{1}{\sqrt 2}\end{bmatrix}, \begin{bmatrix} \frac{1}{\sqrt 2} \\ \frac{-1}{\sqrt 2}\end{bmatrix}\}$. To accomplish this, I use the correct change of basis matrix:

$$ \begin{bmatrix} \frac{1}{\sqrt 2} & \frac{1}{\sqrt 2} \\ \frac{1}{\sqrt 2} &\frac{-1}{\sqrt 2}\end{bmatrix} \begin{bmatrix} 1 \\ 0\end{bmatrix} = \begin{bmatrix} \frac{1}{\sqrt 2} \\ \frac{1}{\sqrt 2}\end{bmatrix} $$

When I see that final vector, I know to read it in the latter basis. And I can compute that the RHS in the second basis is in fact the LHS in the first basis.

Now, when I do the same thing with bra-ket notation, I have:

$$ \left(|0\rangle \langle + | + |1 \rangle \langle -|\right)|0\rangle = \frac{|0\rangle + |1\rangle}{\sqrt 2} $$

When I see the final result here, do I internally read $|0\rangle$ as $|+\rangle$ and $|1\rangle$ as $|-\rangle$?

The explicit writing of bases in the bra-ket notation I find slightly confusing.

  • $\begingroup$ I wonder why matrices are even used in quantum theory, knowing that they are a headache in classical high-dimensional statistics due to curse of dimensionality, ill-conditioning and computational complexity $\endgroup$
    – develarist
    Nov 4, 2020 at 13:55

1 Answer 1


Expressing your first equality in bra-ket notation is simply $$H \vert 0 \rangle = \vert + \rangle.$$ In the spirit of your second equality, $H$ can be expressed as $$H \equiv \vert + \rangle \langle 0 \vert + \vert - \rangle \langle 1 \vert.$$ The advantage of this more verbose expression of $H$ is that it makes it very clear how $H$ transforms the computational basis states: $$H \vert 0 \rangle = \vert + \rangle \langle 0 \vert 0 \rangle + \vert - \rangle \langle 1 \vert 0 \rangle = \vert + \rangle,$$ since the value of the inner products, $\langle 0 \vert 0 \rangle = 1$ and $\langle 1 \vert 0 \rangle = 0$, should be immediately recognized. Note that the RHS of this equation gives the state in terms of the $\lbrace \vert + \rangle, \, \vert - \rangle \rbrace$ basis. In the RHS of your second equality, the state is still represented over the computational basis.


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