Context. I was recently reading a passage from a book on Quantum Mechanics where the author discusses a technique to determine the components of a vector in terms of an alternate basis. The technique seems to be, for vector $|f\rangle$ and basis elements $|k\rangle$ and $|x\rangle$ (from two distinct basis sets), that the way to find the $|k\rangle$ component of $|f\rangle$ in terms of the $|k\rangle$ basis is to compute
$$
\langle k|f \rangle
$$
and likewise for
$$
\langle x|f \rangle
$$
Here is the fuller context from page 68 of Principles of Quantum Mechanics:
Problem: When I try to apply this general pattern against a simpler vector space example, I don't get the answer I expect.
Example setup.
- Suppose we're operating in the $\mathbb{R}^2(\mathbb{R})$ vector space.
- Let us denote by $e_i$ the standard basis. That is, $e_1 = (1, 0)$ and $e_2 = (0, 1)$.
- Let us denote by $d_i$ the "diagonal" basis, with $d_1 = (1/\sqrt{2}, 1/\sqrt{2})$ and $d_2 = (-1/\sqrt{2}, 1/\sqrt{2})$ (but note here I'm denoting these alternative basis elements in the original $e_i$ basis!).
- Let $\vec{v} = (-1, -1)$ (again in terms of the standard $e_i$ basis).
Setting up some inner product calculations. Let our inner product be the standard dot product. Since both $e_i$ and $d_i$ are orthogonal bases, the formula
$$
\langle v | w \rangle = \sum_i v_i w_i
$$
holds under both bases.
Inner products in terms of the $e_i$ basis.
- $\langle d_1 | v \rangle = (1/\sqrt{2}*-1) + (1/\sqrt{2}*-1) = -\sqrt{2}$
- $\langle e_1 | v \rangle = (1*-1) + (0*-1) = -1$
- $\langle d_2 | v \rangle = (-1/\sqrt{2}*-1) + (1/\sqrt{2}*-1) = 0$
- $\langle e_2 | v \rangle = (0*-1) + (1*-1) = -1$
Inner products in terms of the $d_i$ basis. Under this new basis I'm assuming that $|d_1\rangle = (1, 0)$, $|d_2\rangle = (0, 1)$, $|e_1\rangle = (1, -1)$, $|e_2\rangle = (1, 1)$, and $|v\rangle = (-\sqrt{2}, 0)$:
- $\langle d_1 | v \rangle = (1*-\sqrt{2}) + (0*0) = -\sqrt{2}$
- $\langle e_1 | v \rangle = (1*-\sqrt{2}) + (-1*0) = -\sqrt{2}$
- $\langle d_2 | v \rangle = (0*-\sqrt{2}) + (1*0) = 0$
- $\langle e_2 | v \rangle = (1*-\sqrt{2}) + (1*0) = -\sqrt{2}$
Result. When operating under the standard basis,
$$
(\langle d_1 | v \rangle, \langle d_2 | v \rangle) = (-\sqrt{2}, 0)
$$
as desired. But when operating under the diagonal basis
$$
(\langle e_1 | v \rangle, \langle e_2 | v \rangle) = (-\sqrt{2}, -\sqrt{2}) \ne (-1, -1)
$$
I am able to go from $e_i$ to $d_i$ but not from $d_i$ to $e_i$.
Best Answer
I'm not sure what you are trying to do, and I fear you are misreading the text you are quoting. Changing basis amounts to going from the x-basis to the k-basis, which you wisely chose to replicate/parallel on your $\mathbb{R}^2(\mathbb{R})$ vector space.
The normalized x-basis is then simulated by $|e_1\rangle = (1, 0)^T$ and $|e_2\rangle = (0, 1)^T$. The k-basis is simulated by $|d_1\rangle = (1, 1 )^T/\sqrt{2}$ and $|d_2\rangle = (-1, 1)/\sqrt{2}$. The quantum state $|f\rangle$ is simulated by $|\vec{v}\rangle = -(1, 1)^T=-\sqrt{2} |d_1\rangle $.
To change basis from x to k, as in your (1.10.34), you utilize the (orthogonal; in QM it is unitary) transformation matrix T, $$ |d_i\rangle = |e_1\rangle \langle e_1 |d_i\rangle + |e_2\rangle \langle e_2 |d_i\rangle = \sum_j T_{ij} |e_j\rangle, \\ T_{ij}=\langle e_j|d_i\rangle, \\ |\vec d\rangle \langle \vec e|= {\mathbb T}= \begin{pmatrix} 1&1 \\ -1 & 1 \end{pmatrix}/\sqrt{2}, \tag{1} $$ where ${\mathbb T}^{-1} = {\mathbb T}^{T}$ taking you back from the d to the e basis.
Importantly, the two bases are complete, so, as above, (and in your text, where their continuous analogs are inserted), $$ {\mathbb I}= |e_1\rangle \langle e_1|+ |e_2\rangle \langle e_2| = |d_1\rangle \langle d_1| + |d_2\rangle \langle d_2|. \tag{2} $$
You may now compute the "wavefunctions", $$ \langle e_i|\vec v\rangle = -(1,1)^T \\ \langle d_i|\vec v\rangle = \sum_j \langle d_i|e_j\rangle \langle e_j|\vec v\rangle = - \sqrt{2} \delta_{i1}. \tag{3} $$
I've no idea what the latter part of your questions purports to be doing. The bases' relation is valid for all representations. You already changed bases. This generalizes naturally to QM.