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.,

$$H \psi_i = E_i \psi_i.$$ (855)

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

$$\int_{-\infty}^\infty \psi_i^\ast \psi_j dx = \delta_{ij}.$$ (856)

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

$$\int_{-\infty}^\infty \int_{-\infty}^\infty \int_{-\infty}^\infty \psi_i^\ast \psi_j dx dy dz = \delta_{ij}.$$ (857)

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

$$\psi_i^\dag \psi_j = \delta_{ij}.$$ (858)

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

$$\langle \psi_i\vert\psi_j\rangle \equiv \langle i\vert j\rangle = \delta_{ij}.$$ (859)

Here, the term in angle brackets represents the integrals in Eqs. (856) and (857) in one- and three-dimensional regular space, respectively, and the spinor product (858) in spin-space. The advantage of our new notation is its great generality: i.e., it can deal with one-dimensional wavefunctions, three-dimensional wavefunctions, spinors, etc.

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

$$\psi_a = \sum_i c_i \psi_i.$$ (860)

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

$$c_i = \int_{-\infty}^\infty\psi_i^\ast \psi_a dx,$$ (861)

whereas in three dimensions we get

$$c_i = \int_{-\infty}^\infty\int_{-\infty}^\infty\int_{-\infty}^\infty\psi_i^\ast \psi_a dx dy dz.$$ (862)

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

$$c_i = \psi_i^\dag \psi_a.$$ (863)

We can represent all of the above possibilities by writing

$$c_i =\langle\psi_i\vert\psi_a\rangle\equiv \langle i\vert a\rangle.$$ (864)

The expansion (860) thus becomes

$$\psi_a = \sum_i\langle\psi_i\vert\psi_a\rangle \psi_i\equiv \sum_i \langle i\vert a\rangle \psi_i.$$ (865)

Incidentally, it follows that

$$\langle i\vert a\rangle^\ast=\langle a\vert i\rangle.$$ (866)

Finally, if $A$ is a general operator, and the wavefunction $\psi_a$ is expanded in the manner shown in Eq. (860), then the expectation value of $A$ is written (see Sect. 4.9)

$$\langle A\rangle = \sum_{i,j} c_i^\ast c_j A_{ij}.$$ (867)

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

$$A_{ij} = \int_{-\infty}^\infty\psi_i^\ast A \psi_j dx,$$ (868)

whereas in three dimensions we get

$$A_{ij} = \int_{-\infty}^\infty\int_{-\infty}^\infty\int_{-\infty}^\infty\psi_i^\ast A \psi_j dx dy dz.$$ (869)

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

$$A_{ij}=\psi_i^\dag A \psi_j.$$ (870)

We can represent all of the above possibilities by writing

$$A_{ij}=\langle \psi_i\vert A\vert\psi_j\rangle \equiv \langle i\vert A\vert j\rangle.$$ (871)

The expansion (867) thus becomes

$$\langle A\rangle \equiv\langle a\vert A\vert a\rangle= \sum_... ...rangle \langle i\vert A\vert j\rangle \langle j\vert a\rangle.$$ (872)

Incidentally, it follows that [see Eq. (194)]

$$\langle i\vert A\vert j\rangle^\ast=\langle j\vert A^\dag \vert i\rangle.$$ (873)

Finally, it is clear from Eq. (872) that

$$\sum_{i} \vert i\rangle \langle i\vert \equiv 1,$$ (874)

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