Eigenstates and Eigenvalues

Consider a general real-space operator $A(x)$. When this operator acts on a general wavefunction $\psi(x)$ the result is usually a wavefunction with a completely different shape. However, there are certain special wavefunctions which are such that when $A$ acts on them the result is just a multiple of the original wavefunction. These special wavefunctions are called eigenstates, and the multiples are called eigenvalues. Thus, if

$$A \psi_a(x) = a \psi_a(x),$$ (245)

where $a$ is a complex number, then $\psi_a$ is called an eigenstate of $A$ corresponding to the eigenvalue $a$.

Suppose that $A$ is an Hermitian operator corresponding to some physical dynamical variable. Consider a particle whose wavefunction is $\psi_a$. The expectation of value $A$ in this state is simply [see Eq. (192)]

$$\langle A\rangle = \int_{-\infty}^\infty \psi_a^{\ast} A \... ...a dx = a \int_{-\infty}^\infty \psi_a^{\ast} \psi_a dx =a,$$ (246)

where use has been made of Eq. (245) and the normalization condition (140). Moreover,

$$\langle A^2\rangle = \int_{-\infty}^\infty \psi_a^{\ast} A^... ...dx =a^2 \int_{-\infty}^\infty \psi_a^{\ast} \psi_a dx =a^2,$$ (247)

so the variance of $A$ is [cf., Eq. (160)]

$$\sigma_A^{ 2} = \langle A^2\rangle - \langle A\rangle^2 = a^2-a^2 = 0.$$ (248)

The fact that the variance is zero implies that every measurement of $A$ is bound to yield the same result: namely, $a$. Thus, the eigenstate $\psi_a$ is a state which is associated with a unique value of the dynamical variable corresponding to $A$. This unique value is simply the associated eigenvalue.

It is easily demonstrated that the eigenvalues of an Hermitian operator are all real. Recall [from Eq. (222)] that an Hermitian operator satisfies

$$\int_{-\infty}^\infty \psi_1^\ast (A \psi_2) dx = \int_{-\infty}^\infty (A \psi_1)^\ast \psi_2 dx.$$ (249)

Hence, if $\psi_1=\psi_2=\psi_a$ then

$$\int_{-\infty}^\infty \psi_a^\ast (A \psi_a) dx = \int_{-\infty}^\infty (A \psi_a)^\ast \psi_a dx,$$ (250)

which reduces to [see Eq. (245)]

$$a=a^\ast,$$ (251)

assuming that $\psi_a$ is properly normalized.

Two wavefunctions, $\psi_1(x)$ and $\psi_2(x)$, are said to be orthogonal if

$$\int_{-\infty}^{\infty}\psi_1^\ast \psi_2 dx = 0.$$ (252)

Consider two eigenstates of $A$, $\psi_a$ and $\psi_{a'}$, which correspond to the two different eigenvalues $a$ and $a'$, respectively. Thus,

$$A \psi_a = a \psi_a,$$ (253)

$$A \psi_{a'} = a' \psi_{a'}.$$ (254)

Multiplying the complex conjugate of the first equation by $\psi_{a'}$, and the second equation by $\psi_a^\ast$, and then integrating over all $x$, we obtain

$$\int_{-\infty}^\infty (A \psi_a)^\ast \psi_{a'} dx = a \int_{-\infty}^\infty\psi_a^\ast \psi_{a'} dx,$$ (255)

$$\int_{-\infty}^\infty \psi_a^\ast (A \psi_{a'}) dx = a' \int_{-\infty}^{\infty}\psi_a^\ast \psi_{a'} dx.$$ (256)

However, from Eq. (249), the left-hand sides of the above two equations are equal. Hence, we can write

$$(a-a') \int_{-\infty}^\infty\psi_a^\ast \psi_{a'} dx = 0.$$ (257)

By assumption, $a\neq a'$, yielding

$$\int_{-\infty}^\infty\psi_a^\ast \psi_{a'} dx = 0.$$ (258)

In other words, eigenstates of an Hermitian operator corresponding to different eigenvalues are automatically orthogonal.

Consider two eigenstates of $A$, $\psi_a$ and $\psi_a'$, which correspond to the same eigenvalue, $a$. Such eigenstates are termed degenerate. The above proof of the orthogonality of different eigenstates fails for degenerate eigenstates. Note, however, that any linear combination of $\psi_a$ and $\psi_a'$ is also an eigenstate of $A$ corresponding to the eigenvalue $a$. Thus, even if $\psi_a$ and $\psi_a'$ are not orthogonal, we can always choose two linear combinations of these eigenstates which are orthogonal. For instance, if $\psi_a$ and $\psi_a'$ are properly normalized, and

$$\int_{-\infty}^\infty \psi_a^\ast \psi_a' dx = c,$$ (259)

then it is easily demonstrated that

$$\psi_a'' = \frac{\vert c\vert}{\sqrt{1-\vert c\vert^2}}\left(\psi_a - c^{-1} \psi_a'\right)$$ (260)

is a properly normalized eigenstate of $A$, corresponding to the eigenvalue $a$, which is orthogonal to $\psi_a$. It is straightforward to generalize the above argument to three or more degenerate eigenstates. Hence, we conclude that the eigenstates of an Hermitian operator are, or can be chosen to be, mutually orthogonal.

It is also possible to demonstrate that the eigenstates of an Hermitian operator form a complete set: i.e., that any general wavefunction can be written as a linear combination of these eigenstates. However, the proof is quite difficult, and we shall not attempt it here.

In summary, given an Hermitian operator $A$, any general wavefunction, $\psi(x)$, can be written

$$\psi = \sum_{i}c_i \psi_i,$$ (261)

where the $c_i$ are complex weights, and the $\psi_i$ are the properly normalized (and mutually orthogonal) eigenstates of $A$: i.e.,

$$A \psi_i = a_i \psi_i,$$ (262)

where $a_i$ is the eigenvalue corresponding to the eigenstate $\psi_i$, and

$$\int_{-\infty}^\infty \psi_i^\ast \psi_j dx = \delta_{ij}.$$ (263)

Here, $\delta_{ij}$ is called the Kronecker delta-function, and takes the value unity when its two indices are equal, and zero otherwise.

It follows from Eqs. (261) and (263) that

$$c_i = \int_{-\infty}^\infty \psi_i^\ast \psi dx.$$ (264)

Thus, the expansion coefficients in Eq. (261) are easily determined, given the wavefunction $\psi$ and the eigenstates $\psi_i$. Moreover, if $\psi$ is a properly normalized wavefunction then Eqs. (261) and (263) yield

$$\sum_i \vert c_i\vert^2 =1.$$ (265)