week 5

states and operators vs. column vectors and matrices (with applications to degenerate perturbation theory)

One of the basic tasks in quantum mechanics is the calculation of eigenstates $\ket{a_1},\ket{a_2},\ldots$ and eigenvalues $a_1, a_2, \ldots$ associated with some operator $\hat{A}$, i.e. $\hat{A}\ket{a_n} = a_n \ket{a_n}$. To accomplish this task it is often convenient to first "convert" the operator into a matrix, i.e. $$ \hat{A} \to \begin{bmatrix} A_{11} & A_{12} & \cdots \\ A_{21} & A_{22} & \cdots \\ \vdots & \vdots & \ddots \end{bmatrix} $$ since we know the rules / techniques for finding the eigenvectors and eigenvalues of matrices.

It can however be unclear how in practice we convert operators into matrices and, even more so, how to convert the eigenvectors of the matrix $A_{mn}$ into the desired eigenstates of the original operator $\hat{A}$. In these notes I will do a slow walkthrough of the correspondence between states and column vectors and operators and matrices.

states and column vectors

Quantum states are vectors in some hilbert space $V$. Since hilbert spaces are vector spaces, there will be certain collections $\{ \ket{\phi_1}, \ket{\phi_2},\ldots, \ket{\phi_N}\}$ of vectors in $V$ that together constitute something called a basis. A basis has the property that any vector $\ket{ \psi }$ in $V$ can be written as a unique linear combination of vectors in the basis, i.e.
We call a collection of vectors $\{ \ket{\phi_1}, \ket{\phi_2},\ldots, \ket{\phi_N}\}$ in $V$ a basis for $V$ if:
For any vector $\ket{\psi}$ in $V$, there exists a unique $N$-tuple of numbers $(c_1,c_2,\ldots,c_N)$ such that $$ \ket{\psi} = \sum_{n=1}^N c_n \ket{\phi_n} $$
A couple notes about bases (plural of basis): Q1: Show that if a basis $\{ \ket{\phi_1}, \ket{\phi_2},\ldots, \ket{\phi_N}\}$ is ortho-normal, then the "expansion coefficients" $c_1, c_2, \ldots, c_N$ of some state $\ket{\psi}$ with respect to this basis are given by: $$ c_n = \ip{ \phi_n }{ \psi } $$ A1: $$ c_n = \sum_{m=1}^N c_m \delta_{nm} = \sum_{m=1}^N c_m \ip{\phi_n}{\phi_m} = \ip{\phi_n}{ \sum_{m=1}^N c_m \phi_m} = \ip{\phi_n}{ \psi } $$ If we make a choice of basis, it then follows from the above discussion that there is a correspondence between states (e.g. $\ket{\psi}$) in an $N$-dimensional vector space $V$ and $N$-tuples of numbers (e.g. $\left( c_1, c_2, \ldots, c_N \right)$). Moreover, we can arrange these $N$-tuples into a suggestive column-form $\begin{bmatrix} c_1 \\ c_2 \\ \ldots \\ c_N \end{bmatrix} \equiv \bm{c}$ (bold face).

Denote the correspondence just described by the symbol $\EQ$, so that $ \ket{\psi} \EQ \bm{c} $ if $\ket{\psi} = \sum_{n=1}^N c_n \ket{\phi_n}$. By arranging the expansion coefficients in the same way as column vectors, we are suggesting that
the expansion coefficients $\left( c_1, c_2, \ldots, c_N \right)$ can themselves be thought of as vectors in the vector space $\C^N$
Q2: Let's elaborate on the statement we just made. Let $\ket{\psi}$ and $\ket{\psi'}$ be two vectors in $V$, and let $\bm{c}$ and $\bm{d}$ be their respective expansion coefficients with respect to a basis $\{ \ket{\phi_1}, \ket{\phi_2},\ldots, \ket{\phi_N}\}$, so that $\ket{\psi} \EQ \bm{c}$ and $\ket{\psi'} \EQ \bm{d}$. Further let $\lambda$ be some complex number. Show that the correspondence $\EQ$ is linear, i.e. it commutes with addition and scalar multiplication in the sense that:

Q3: If the $\{\ket{\phi_N}\}$ are an orthonormal basis then show that the expansion coefficients also commute with the inner product, in the sense that:

operators and matrices

It is not usually pointed out explicitly, but the set of all operators on a vector space $V$ are themselves a vector space, because we can add operators as well as scale them by constant multiples. (Note that although operators are vectors they are not elements of $V$, i.e. the space of vectors they operate on1.) It follows then that there is some collection of operators which form a basis. Given a basis $\{\ket{\phi_N}\}$ on the original space $V$, it turns out we can construct such a basis of operators from all possible outer products $\ket{\phi_m} \bra{\phi_n}$, so that, given any operator $\hat{A}$, there exists a set of coefficients $A_{mn}$, with $n,m=1,2,\ldots,N$ such that $$ \hat{A} = \sum_{m,n=1}^N A_{mn} \ket{\phi_m}\bra{\phi_n} $$

Q4: Show that if $\{\ket{\phi_N}\}$ is orthonormal then the expansion coefficents $A_{mn}$ of an operator $\hat{A}$ are given by $$ A_{mn} = \ip{\phi_m}{\hat{A} \phi_n} $$ A4: \begin{align} & A_{mn} \\ = & \sum_{i,j=1}^N A_{ij} \delta_{im} \delta_{jn} \\ = & \sum_{i,j=1}^N A_{ij} \ip{ \phi_m }{\phi_i} \ip{ \phi_j }{\phi_n} \\ = & \bra{\phi_m} \left( \sum_{i,j=1}^N A_{ij} \ket{\phi_i}\bra{\phi_j} \right) \ket{\phi_n} \\ = & \ip{\phi_m}{\hat{A}\phi_n} \end{align} Let's organize our full list of "basis operators" in the following form: $$ \left\{ \begin{array}{cccc} \ket{\phi_1} \bra{\phi_1},\, & \ket{\phi_1} \bra{\phi_2},\, & \ldots,\, & \ket{\phi_1} \bra{\phi_N},\, \\ \ket{\phi_2} \bra{\phi_1},\, & \ket{\phi_2} \bra{\phi_2},\, & \ldots,\, & \ket{\phi_2} \bra{\phi_N},\, \\ \vdots & & & \\ \ket{\phi_N} \bra{\phi_1},\, & \ket{\phi_N} \bra{\phi_2},\, & \ldots,\, & \ket{\phi_N} \bra{\phi_N} \end{array} \right\} $$ This suggests a possible connection between operators and matrices, namely
the expansion coefficients $$ \left( \begin{array}{cccc} A_{11},\, & A_{12},\, & \ldots,\, & A_{1N},\, \\ A_{21},\, & A_{22},\, & \ldots,\, & A_{2N},\, \\ \vdots & & & \\ A_{N1},\, & A_{N2},\, & \ldots,\, & A_{NN} \end{array} \right) $$ can be thought of as $N \times N$ matrices
By complete analogy with the state$\leftrightarrow$column vector correspondence, we define a correspondence $\EQ$ between an operator $\hat{A}$ and a matrix $A_{mn}$ so that $\hat{A} \EQ A_{mn}$ if $\hat{A} = \sum^N_{m,n=1} A_{mn} \ket{\phi_m} \bra{\phi_n}$.

Q4: Let: Check that the correspondence $\EQ$ is linear, i.e. it commutes with addition and scalar multiplication in the sense that: Q4: Retain the same assumptions we made in Q4 and further assume that the basis $\{ \ket{\phi_n} \}$ is orthonormal. Show then that the correspondence $\EQ$ commutes with (in addition to scalar multiplication and addition) operator/matrix multiplication in the sense that $$ \hat{A} \hat{B} \EQ \left[ A B \right]_{mn} $$ where $\left[ AB \right]_{mn}$ is the usual rule for matrix multiplication, i.e. $$ \left[ AB \right]_{mn} = \sum_{l=1}^N A_{ml}B_{ln} $$ This is in fact where the rule for matrix multiplication comes from! The elements from matrices $A_{mn}$ and $B_{mn}$ are combined in special way to form a product matrix $\left[AB\right]_{mn}$ that corresponds to the "product" operator $\hat{A}\hat{B}$ that operates on vectors by applying the operators $\hat{B}$ and $\hat{A}$ in succession, i.e. "composing" them.
A4: \begin{align} & \hat{A}\hat{B} \\ = & \left( \sum^N_{m,k=1} A_{mk} \ket{\phi_m}\bra{\phi_k} \right) \left( \sum_{l,n=1}^N B_{ln} \ket{\phi_l} \bra{\phi_n} \right) \\ = & \sum^N_{m,n,k,l=1} A_{mk} B_{ln} \ket{\phi_m} \bra{\phi_n} \ip{ \phi_k }{ \phi_l } \\ = & \sum^N_{m,n,k,l=1} A_{mk} B_{ln} \ket{\phi_m} \bra{\phi_n} \delta_{kl} \\ = & \sum^N_{m,n} \left( \sum_{l=1}^N A_{ml} B_{ln} \right) \ket{\phi_m} \bra{\phi_n} \end{align}

putting it together

Q5: Let $\{ \ket{\phi_n} \}$ be orthonormal, and suppose $\hat{A}$ is an operator with a corresponding matrix $A_{mn} \equiv \bm{A}\,$2 and $\ket{\psi}$ is a state with a corresponding column vector $\bm{c}$. Check that the correspondence $\EQ$ is compatible in the sense that: $$ A \ket{\psi} \EQ \bm{A} \bm{c} $$ A5: \begin{align} & A \ket{\psi} \\ = & \left( \sum_{j,k=1}^N A_{jk} \ket{\phi_j} \bra{\phi_k} \right) \sum_{l=1}^N c_l \ket{\phi_l} \\ = & \sum_{j,k,l=1}^N A_{jk} c_l \ip{\phi_k}{\phi_l} \ket{\phi_j} \\ = & \sum_{j,k,l=1}^N A_{jk} c_l \delta_{kl} \ket{\phi_j} \\ = & \sum_{j=1}^N \left( \sum_{k=1}^N A_{jk} c_k \right) \ket{\phi_j} \\ = & \sum_{j=1}^N \left( \bm{A} \bm{c} \right)_j \ket{\phi_j} \end{align} Q6: Show that if a column vector $\bm{c}$ is an eigenvector of the matrix $\bm{A}$ with eigenvalue $\lambda$, i.e. $$ \bm{A}\bm{c} = \lambda \bm{c} $$ then, given any orthonormal basis $\{ \ket{\phi_n} \}$, the corresponding state $\ket{\psi} \EQ \bm{c}$ is an eigenstate of the corresponding operator $\hat{A} \EQ A_{mn}$ with eigenvalue $\lambda$, i.e. $$ \hat{A} \ket{\psi} = \lambda \ket{\psi} $$ A6: $$ \hat{A} \ket{\psi} \EQ \bm{A} \bm{c} = \lambda \bm{c} \EQ \lambda \ket{\psi} $$ Since the correspondence $\EQ$ is unique, this implies $$ \hat{A} \ket{\psi} = \lambda \ket{\psi} $$ Notice how streamlined and compact the reasoning is when we exploit the linear and compatibility properties of the correspondence $\EQ$.

An equivalent way to phrase the previous result is
If we construct a matrix $\bm{A}$ from an operator $\hat{A}$ using an orthonormal basis $\{\ket{\phi_n}\}$ and the correspondence $\EQ$, i.e. $\hat{A} \EQ \bm{A}$, then we can construct an eigenstate $\ket{\psi}$ of $\hat{A}$ from an eigenvector $\bm{c}$ of $\bm{A}$ using the same correspondence $\EQ$, i.e. $\ket{\psi} \EQ \bm{c}$ or $\ket{\psi} = \sum_{n=1}^N c_n \ket{\phi_n}$.

degenerate perturbation theory

Suppose we have an unperturbed hamiltonian $\hat{H}_o$ operating on a hilbert space $V$ and an orthonormal energy eigenbasis $\{\ket{\phi_{nm}}\}$, where $\ket{\phi_{nm}}$ is the $m$th eigenstate of energy $E_n$.

When we apply a perturbation $\hat{H}'$, our general task is to find the "good" eigenstates of each energy level. Once we have the good eigenstates it is straightforward to calculate the corrections to the energy levels due to the perturbation (see Griffiths). To define the good eigenstates, we first point out that
the states $\{ \ket{\phi_{n1}},\ldots,\ket{\phi_{n2}},\ket{\phi_{nM_n}} \} \equiv \{ \ket{\phi_{n(\cdot)}} \}$ (where $M_n$ is the degeneracy of the $n$th energy level) make up an orthonormal basis of the linear subspace $V^{(n)}$ of eigenstates with energy eigenvalue $E_n$.
Then we define an operator $\hat{H}'_{(n)}$ that acts on vectors in $V^{(n)}$ by the formula $$ \hat{H}'_{(n)} := \sum_{m,m'=1}^{M_n} \ip{ \phi_{nm} }{ H' \phi_{nm'}} \ket{\phi_{nm}} \bra{\phi_{nm'}} $$ Note here that crucially the $n$ index is fixed (not summed over). We could equivalently define $\hat{H}'_{(n)}$ as the projection of $\hat{H}'$ onto $V^{(n)}$.

We now present the definition of the good eigenstates:
The "good" eigenstates of the $n$th energy level are defined to be the eigenstates of the operator $\hat{H}'_{(n)}$.
In practice we find these eigenstates using the correspondence principles we developed in the previous three sections. The basis $\{ \ket{\phi_{n(\cdot)}} \}$ (i.e. with $n$ fixed) defined earlier comprises an orthonormal basis of $V^{(n)}$. We can convert $\hat{H}'_{(n)}$ to an $M_n \times M_n$ matrix $\bm{H}'_{(n)}$ via $\EQP$, i.e. $$ \left[\bm{H}'_{(n)}\right]_{ij} = \ip{ \phi_{ni} }{ H'_{(n)} \phi_{nj} } = \ip{ \phi_{ni} }{ H' \phi_{nj} } $$ We then find the $M_n$ eigenvectors $\bm{c}_{n1},\bm{c}_{n2},\ldots,\bm{c}_{nM_n}$ of the matrix $\bm{H}'_{(n)}$, from which we can construct the $M_n$ eigenstates $\ket{\phi'_{n1}},\ket{\phi'_{n2}},\ldots,\ket{\phi'_{nM_n}}$ of $\hat{H}'_{(n)}$ (which are by definition the good eigenstates of the $n$th energy level) by the same correspondence $\EQP$, i.e. $$ \ket{\phi'_{nm}} = \sum_{l}^{M_n} \left[c_{nm}\right]_l \ket{\phi_{nl}} $$ Note that the $\ket{\phi_{nl}}$ are the original (unprimed) basis vectors.

footnotes

1 Kind of like how fantasy football is a sport but not the same sport as football.

2 It will be more convenient from now on to refer to a matrix using bold face $\bm{A}$ instead of general element $A_{mn}$