# Matrix representation of a unitary operator, change of basis

• A
• Kashmir
In summary, the expansion coefficients in the new basis can be obtained by multiplying (1.5.10) by the column matrix representing ##U^{\dagger}## .f

#### Kashmir

If ##U## is an unitary operator written as the bra ket of two complete basis vectors :

##U=\sum_{k}\left|b^{(k)}\right\rangle\left\langle a^{(k)}\right|##

##U^\dagger=\sum_{k}\left|a^{(k)}\right\rangle\left\langle b^{(k)}\right|##

And we've a general vector ##|\alpha\rangle## such that ##|\alpha\rangle=\sum_{a^{\prime}}\left|a^{\prime}\right\rangle\left\langle a^{\prime} \mid \alpha\right\rangle##

Sakurai writes at pg 50 :
"how can we obtain ##\left\langle b^{\prime} \mid \alpha\right\rangle##, the expansion coefficients in the new basis? answer is very simple: Just multiply (1.5.9) by ##\left\langle b^{(k)}\right|##
##
\left\langle b^{(k)} \mid \alpha\right\rangle=\sum_{l}\left\langle b^{(k)} \mid a^{(l)}\right\rangle\left\langle a^{(l)} \mid \alpha\right\rangle=\sum_{l}\left\langle a^{(k)}\left|U^{\dagger}\right| a^{(l)}\right\rangle\left\langle a^{(l)} \mid \alpha\right\rangle .
##
##(1.5 .1##
In matrix notation, (1.5.10) states that the column matrix for ##|\alpha\rangle## in the new basis can be obtained just by applying the square matrix ##U^{\dagger}## to the colum matrix in the old basis:

So if the matrix representing ##U^\dagger## is applied on to the matrix representing ##|\alpha\rangle## ,it gives the vectors representation in the new basis.

But when I apply ##U^\dagger## onto say an basis vector ##\left|a_{1}\right\rangle## ,it doesn't give me the vectors representation in new basis as shown below :

##\begin{aligned} U^{\dagger}\left|a_{1}\right\rangle &=\sum_{k}\left|a^{k}\right\rangle\left\langle b^{k} \mid a_{1}\right\rangle \\ &=\sum_{k}\left(\left\langle b^{k} \mid a_{1}\right\rangle\right) \cdot\left|a^{k}\right\rangle \end{aligned}##

Why should it? You rather have
$$\ket{b^k}=\hat{U} \ket{a^k}.$$
Sakurai in the quoted text uses the adjoint of this
$$\bra{b^k}=\bra{a^k} \hat{U}^{\dagger}.$$