Improved notation

Before commencing our investigation, it is helpful to introduce some improved notation. Let the $\psi_i$ be a complete set of eigenstates of the Hamiltonian, $H$, corresponding to the eigenvalues $E_i$: i.e.,

$$H\,\psi_i = E_i\,\psi_i.$$ (834)

Now, we expect the $\psi_i$ to be orthonormal (see Sect. 4.9). In one dimension, this implies that

$$\int_{-\infty}^\infty \psi_i^\ast\,\psi_j \,dx = \delta_{ij}.$$ (835)

In three dimensions (see Sect. 7), the above expression generalizes to

$$\int_{-\infty}^\infty \int_{-\infty}^\infty \int_{-\infty}^\infty \psi_i^\ast\,\psi_j\,dx\,dy\,dz = \delta_{ij}.$$ (836)

Finally, if the $\psi_i$ are spinors (see Sect. 10) then we have

$$\psi_i^\dag\,\psi_j = \delta_{ij}.$$ (837)

The generalization to the case where $\psi$ is a product of a regular wave-function and a spinor is fairly obvious. We can represent all of the above possibilities by writing

$$\langle \psi_i\vert\psi_j\rangle \equiv \langle i\vert j\rangle = \delta_{ij}.$$ (838)

Here, the term in angle brackets represents the integrals in Eqs. (835) and (836) in one- and three-dimensional regular space, respectively, and the spinor product (837) in spin-space. The advantage of our new notation is its great generality: i.e., it can deal with one-dimensional wave-functions, three-dimensional wave-functions, spinors, etc.

Expanding a general wave-function, $\psi_a$, in terms of the energy eigenstates, $\psi_i$, we obtain

$$\psi_a = \sum_i c_i\,\psi_i.$$ (839)

In one-dimension, the expansion coefficients take the form (see Sect. 4.9)

$$c_i = \int_{-\infty}^\infty\psi_i^\ast\,\psi_a\,dx,$$ (840)

whereas in three-dimensions we get

$$c_i = \int_{-\infty}^\infty\int_{-\infty}^\infty\int_{-\infty}^\infty\psi_i^\ast\,\psi_a\,dx\,dy\,dz.$$ (841)

Finally, if $\psi$ is a spinor then we have

$$c_i = \psi_i^\dag\,\psi_a.$$ (842)

We can represent all of the above possibilities by writing

$$c_i =\langle\psi_i\vert\psi_a\rangle\equiv \langle i\vert a\rangle.$$ (843)

The expansion (839) thus becomes

$$\psi_a = \sum_i\langle\psi_i\vert\psi_a\rangle\,\psi_i\equiv \sum_i \langle i\vert a\rangle\,\psi_i.$$ (844)

Incidentally, it follows that

$$\langle i\vert a\rangle^\ast=\langle a\vert i\rangle.$$ (845)

Finally, if $A$ is a general operator, and the wave-function $\psi_a$ is expanded in the manner shown in Eq. (839), then the expectation value of $A$ is written (see Sect. 4.9)

$$\langle A\rangle = \sum_{i,j} c_i^\ast\,c_j\,A_{ij}.$$ (846)

Here, the $A_{ij}$ are unsurprisingly known as the matrix elements of $A$. In one-dimension, the matrix elements take the form

$$A_{ij} = \int_{-\infty}^\infty\psi_i^\ast\,A\,\psi_j\,dx,$$ (847)

whereas in three-dimensions we get

$$A_{ij} = \int_{-\infty}^\infty\int_{-\infty}^\infty\int_{-\infty}^\infty\psi_i^\ast\,A\,\psi_j\,dx\,dy\,dz.$$ (848)

Finally, if $\psi$ is a spinor then we have

$$A_{ij}=\psi_i^\dag\,A\,\psi_j.$$ (849)

We can represent all of the above possibilities by writing

$$A_{ij}=\langle \psi_i\vert A\vert\psi_j\rangle \equiv \langle i\vert A\vert j\rangle.$$ (850)

The expansion (846) thus becomes

$$\langle A\rangle \equiv\langle a\vert A\vert a\rangle= \sum_... ...rangle \langle i\vert A\vert j\rangle \langle j\vert a\rangle.$$ (851)

Incidentally, it follows that [see Eq. (176)]

$$\langle i\vert A\vert j\rangle^\ast=\langle j\vert A^\dag \vert i\rangle.$$ (852)

Finally, it is clear from Eq. (851) that

$$\sum_{i} \vert i\rangle \langle i\vert \equiv 1,$$ (853)

where the $\psi_i$ are a complete set of eigenstates, and 1 is the identity operator.