Confusion over basis sets in quantum mechanics

Question

In a sense, I feel like I understand every step in the following (fairly typical) calculation, but I don't understand the conclusion.

I consider a vector space spanned by two basis vectors $|\phi_A\rangle$ and $|\phi_B\rangle$. I can use this basis to decompose an arbitrary state:

$$ |\psi\rangle=\sum_{i=A,B}\langle\phi_i|\psi\rangle|\phi_i\rangle=\sum_{i=A,B}\phi_i|\phi_i\rangle, $$

where I have defined $\phi_i=\langle\phi_i|\psi\rangle$.

Now I assume that $|\psi\rangle$ is an eigenvector of the Hamiltonian, so that (invoking the resolution of the identity in the first step):

$$ \langle\phi_m|H|\psi\rangle=\sum_{i=A,B}\langle\phi_m|H|\phi_i\rangle\langle\phi_i|\psi\rangle=\sum_{i=A,B}H_{m,i}\phi_i=E\langle\phi_m|\psi\rangle=E\phi_m $$

where I have defined $H_{m,i}=\langle\phi_m|H|\phi_i\rangle$.

Equating terms we see that $\sum_{i=A,B}H_{m,i}\phi_i=E\phi_m$, which reminds us of the rule for matrix multiplication. Therefore, we represent the operator $H$ by a matrix acting on a vector which represents the state $|\psi\rangle$:

$$ \left[\begin{matrix} H_{A,A} & H_{A,B}\\ H_{B,A} & H_{B,B} \end{matrix} \right] \left[ \begin{matrix} \phi_{A}\\ \phi_{B} \end{matrix} \right]=E\left[ \begin{matrix} \phi_{A}\\ \phi_{B} \end{matrix} \right]. $$

The column vector could instead be written:

$$ |\psi\rangle=\left[\begin{matrix} \phi_{A}\\ \phi_{B} \end{matrix} \right]=\phi_A\left[\begin{matrix} 1\\ 0 \end{matrix} \right]+\phi_B\left[\begin{matrix} 0\\ 1 \end{matrix} \right] $$.

But hang on, that expansion is exactly the same as the expansion I made in the very first equation, if we make the identification:

$$ |\phi_A\rangle=\left[\begin{matrix} 1\\ 0 \end{matrix} \right];|\phi_B\rangle=\left[\begin{matrix} 0\\ 1 \end{matrix} \right]. $$

I don't understand this: I thought the basis I chose in the beginning was arbitrary, but now I seem to know what my basis vectors are. That is, they are not arbitrary at all, and I at least think I know what they are specifically.

My feeling is that I have at some point (without realizing) assumed that this is my basis. I think this was likely the step in which I assumed that $|\psi\rangle$ is an eigenvector of $H$. Is that correct?

Answer 1

My feeling is that I have at some point (without realizing) assumed that this is my basis. I think this was likely the step in which I assumed that $|\psi\rangle$ is an eigenvector of $H$. Is that correct?

That was not the step where this happened. There were two crucial steps: in the first you invoked "resolution of the identity" which implicitly assumes $\langle\phi_a|\phi_b\rangle = \delta_{ab},$ so the vectors became orthogonal and of unit size, precisely the sort of vectors which could be a basis for a 2D vector space.

The step where this then more directly happened was when you decided that you would define:$$H_{ab} = \langle\phi_a|\hat H|\phi_b\rangle, ~~\psi_a = \langle \phi_a|\psi\rangle.$$

At this very point you decided that if $|\psi\rangle$ were $|\phi_0\rangle$ then you were going to assign it coordinates $\psi_0 = 1, \psi_1 = 0,$ thus being the $\begin{bmatrix}1\\0\end{bmatrix}$ element of the vector space in that basis.

Answer 2

In order to understand the manipulations you made, you should first of all understand that the column vector notation is a relative one, in the sense that it is defined with respect to some (arbitrary) fixed basis (at least in the context of formal vector spaces). Every finite-dimensional vector space admits a basis; so you can always decompose one of its elements in terms of that basis:

$$ x=\sum_{i}\ c_{i}\ e_{i}\qquad \qquad y=\sum_{i}\ d_{i}\ e_{i} $$ and so on, where the $e_{i}$'s are the elements that make up the basis. As for the sum of $x$ and $y$, you can set

$$ x+y=\sum_{i}(c_{i}+d_{i})\ e_{i} $$ This sum has precisely the same algebraic properties as the sum defined between two column vectors: if you define $x$ and $y$ to be $$ x=\begin{pmatrix}c_{1}\\\vdots\\c_{n}\end{pmatrix}\qquad\qquad y=\begin{pmatrix}d_{1}\\\vdots\\d_{n}\end{pmatrix} $$ then $$ x+y=\begin{pmatrix}c_{1}+d_{1}\\\vdots\\c_{n}+d_{n}\end{pmatrix} $$ This algebraic property does not depend upon the basis you chose. On the other hand, the coefficients $c_{i}$ and $d_{i}$ do. Let's say, however, that you have fixed a basis. Then the column vector representation for the elements of that basis is $$ e_{1}=\begin{pmatrix}1\\0\\\vdots\\0\end{pmatrix}\qquad\qquad e_{2}=\begin{pmatrix}0\\1\\\vdots\\0\end{pmatrix}\qquad \qquad\dots\qquad\qquad e_{n}=\begin{pmatrix}0\\0\\\vdots\\1\end{pmatrix} $$ The elements of other bases can be expressed in this representation too. As we have $$ e_{j}'=\sum_{i}\ b_{ji}\ e_{i} $$ for some set of coefficients $b_{ji}$ (here $\{e_{j}'\}$ is the second basis), then in the representation you chose $$ e'_{1}=\begin{pmatrix}b_{11}\\b_{12}\\\vdots\\b_{1n}\end{pmatrix}\qquad\qquad e'_{2}=\begin{pmatrix}b_{21}\\b_{22}\\\vdots\\b_{2n}\end{pmatrix}\qquad \qquad\dots\qquad\qquad e'_{n}=\begin{pmatrix}b_{n1}\\b_{n2}\\\vdots\\b_{nn}\end{pmatrix} $$ In conclusion, the column vector representation is always given in relation to some specific basis; the elements of the basis through which you decided to represent the elements of the vector space (and only them) then must be represented by column vectors with one $1$ and $n-1$ $0$'s in their rows. What you ask in your question does not depend on QM or on the Hamiltonian; its answer is a general fact of (elementary) representation theory in linear algebra.