# Change of basis, matrix and operators

If $$U$$ is an unitary operator written as the bra ket of two complete basis vectors i.e

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

Then

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

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: $$\quad(\mathrm{New})=\left(U^{\dagger}\right)($$ old $$)$$

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}

• Can downvoters please comment what should be improved about the question? Aug 17, 2022 at 10:36
• I didn't downvote but $\left < b^k | a_1 \right > \neq b^k \left | a_1 \right >$. Aug 17, 2022 at 10:44
• I also didn't downvote, but you are using a multitude of notations which makes the question very difficult to read. For example, Is $\left|a_1\right>=\left|a^{(1)}\right>$? $\left|b^k\right>=\left|b^{(k)}\right>$? Why do you change mid-question? What is $\left|a'\right>$? $\left|b'\right>$? Aug 17, 2022 at 11:14
• @ConnorBehan, where did i write that? Aug 17, 2022 at 12:22
• @ɪdɪətstrəʊlə, edited it. Aug 17, 2022 at 12:26

$$|a^1\rangle =\sum_{j}(\langle b^{j}|a^1\rangle) |b^{j}\rangle=\sum_{j}(\langle a^j|U^{\dagger}|a^{1}\rangle) |b^j\rangle$$
I think that at the end you are looking for something different. Following the sakurai page, it states that $$U\vec{a} =\vec{b}$$ without the transpose symbol. In fact if you apply only U you obtain the correct matrix. the U trasposed is useful to obtain the "old" vectors $$\vec{a_{k}}$$ from the new basis formed by $$\vec{b_{k}}$$ vector