← Exercise library ← All exercises for this theme

Theme 2 — Hilbert Spaces and Dirac Notation

Change of orthonormal basis

Keywords: change of basis · unitary matrix · change-of-basis matrix · representation of an operator · Hadamard · invariants

Exercise 1 : Change of orthonormal basis

In $\C^2$ , consider the canonical orthonormal basis $(\ket0,\ket1)$ and the basis

\ket+=\frac{\ket0+\ket1}{\sqrt2},\qquad\ket-=\frac{\ket0-\ket1}{\sqrt2}.

The change-of-basis matrix is defined as the matrix $S$ whose columns are the components of the new basis vectors in the old basis, that is, $S_{ij}=\braket{e_i}{f_j}$ if $(\ket{e_i})$ is the old basis and $(\ket{f_j})$ the new one.

Question 1

Verify that

(\ket+,\ket-)

is an orthonormal basis. Write down the change-of-basis matrix

S

and show that it is unitary.

Solution

We have

\braket++=\frac12(1+1)=1

\braket—=1

and

\braket+-=\frac12(1-1)=0

: the basis is orthonormal. The columns of

S

are the components of

\ket+

and

\ket-

\begin{aligned} S=\frac{1}{\sqrt2}\begin{pmatrix}1&1\\1&-1\end{pmatrix}. \end{aligned}

This matrix, called the Hadamard matrix, is real and symmetric, so $S^\dagger=S$ , and direct calculation gives $S^\dagger S=S^2=\frac12\begin{pmatrix}2&0\\0&2\end{pmatrix}=\mathbf{1}$ . It is unitary.

Question 2

Show that, in general, the change-of-basis matrix between two orthonormal bases is unitary.

Solution

Let us calculate the elements of

S^\dagger S

using the closure relation in the old basis:

(S^\dagger S)_{jk}=\sum_i(S^\dagger)_{ji}S_{ik}=\sum_i\braket{e_i}{f_j}^*\braket{e_i}{f_k}=\sum_i\braket{f_j}{e_i}\braket{e_i}{f_k}=\braket{f_j}{f_k}=\delta_{jk}.

Thus $S^\dagger S=\mathbf{1}$ . In other words, the columns of $S$ are orthonormal, which simply expresses the fact that the new basis is orthonormal. In finite dimension, this also implies $SS^\dagger=\mathbf{1}$ .

Question 3

A vector

\ket\psi

has components

(\alpha,\beta)

in the basis

(\ket0,\ket1)

. What are its components in the basis

(\ket+,\ket-)

? Express the result using

S

Solution

The new components are the projections

\braket+\psi

and

\braket-\psi

\braket+\psi=\frac{\alpha+\beta}{\sqrt2},\qquad\braket-\psi=\frac{\alpha-\beta}{\sqrt2}.

In general, $\braket{f_j}{\psi}=\sum_i\braket{f_j}{e_i}\braket{e_i}{\psi}=\sum_i(S^\dagger)_{ji}\psi_i$ : the column vector of the new components is obtained by applying $S^\dagger$ to the column vector of the old components. Take care with the direction: it is $S^\dagger=S^{-1}$ , not $S$ , that transforms the components.

Question 4

Show that the matrix of an operator

\hat A

in the new basis is

A'=S^\dagger AS

. Calculate the matrices of

\sigma_z=\mathrm{diag}(1,-1)

and

B=\begin{pmatrix}2&1\\1&2\end{pmatrix}

in the basis

(\ket+,\ket-)

. Comment on the results.

Solution

The matrix elements in the new basis are

A'_{jk}=\bra{f_j}\hat A\ket{f_k}

. Inserting two closure relations in the old basis,

A'_{jk}=\sum_{i,l}\braket{f_j}{e_i}\bra{e_i}\hat A\ket{e_l}\braket{e_l}{f_k}=\sum_{i,l}(S^\dagger)_{ji}A_{il}S_{lk}=(S^\dagger AS)_{jk}.

For $\sigma_z$ , we find

\begin{aligned} S^\dagger\sigma_zS=\frac12\begin{pmatrix}1&1\\1&-1\end{pmatrix}\begin{pmatrix}1&0\\0&-1\end{pmatrix}\begin{pmatrix}1&1\\1&-1\end{pmatrix}=\frac12\begin{pmatrix}1&-1\\1&1\end{pmatrix}\begin{pmatrix}1&1\\1&-1\end{pmatrix}=\begin{pmatrix}0&1\\1&0\end{pmatrix}=\sigma_x . \end{aligned}

In the new basis, $\sigma_z$ has the matrix of $\sigma_x$ : its diagonal elements are zero, since $\ket\pm$ are not eigenstates of $\sigma_z$ . For $B$ , note that $B\ket+=3\ket+$ and $B\ket-=\ket-$ , so that

\begin{aligned} S^\dagger BS=\begin{pmatrix}3&0\\0&1\end{pmatrix}. \end{aligned}

The basis $(\ket+,\ket-)$ is an eigenbasis of $B$ : diagonalising a matrix consists precisely in finding a basis in which its matrix is diagonal. We verify that the trace ( $4$ ) and determinant ( $3$ ) are the same in both bases: they are invariants, which depend only on the operator and not on the basis.