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Eigenstates and Eigenvalues
Consider a general realspace operator . When this operator
acts on a general wavefunction the result is usually a wavefunction
with a completely different shape. However, there are certain special
wavefunctions which are such that when 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

(245) 
where is a complex number, then is called an eigenstate of
corresponding to the eigenvalue .
Suppose that is an Hermitian operator corresponding to some physical dynamical
variable.
Consider a particle whose wavefunction is . The expectation of
value in this state is simply [see Eq. (192)]

(246) 
where use has been made of Eq. (245) and the normalization
condition (140). Moreover,

(247) 
so
the variance of is [cf., Eq. (160)]

(248) 
The fact that the variance is zero implies that every measurement of is bound to
yield the same result: namely, . Thus, the eigenstate is a state which is
associated with a unique value of the dynamical variable corresponding to . 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

(249) 
Hence, if
then

(250) 
which reduces to [see Eq. (245)]

(251) 
assuming that is properly normalized.
Two wavefunctions, and , are said to be orthogonal
if

(252) 
Consider two eigenstates of , and , which
correspond to the two different eigenvalues and , respectively. Thus,
Multiplying the complex conjugate of the first equation by ,
and the second equation by , and then integrating over all
, we obtain
However, from Eq. (249), the lefthand sides of the above two
equations are equal. Hence, we can
write

(257) 
By assumption, , yielding

(258) 
In other words, eigenstates of an Hermitian operator corresponding to
different eigenvalues are automatically orthogonal.
Consider two eigenstates of , and , which
correspond to the same eigenvalue, . 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
and is also an eigenstate of corresponding
to the eigenvalue . Thus, even if and are not
orthogonal, we can always choose two linear combinations
of these eigenstates which are orthogonal. For instance,
if and are properly normalized, and

(259) 
then it is easily demonstrated that

(260) 
is a properly normalized eigenstate of , corresponding to the
eigenvalue , which is orthogonal to . 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 , any general wavefunction, , can be written

(261) 
where the are complex weights, and the are the properly
normalized (and mutually orthogonal) eigenstates of : i.e.,

(262) 
where is the eigenvalue corresponding to the eigenstate ,
and

(263) 
Here, is called the Kronecker deltafunction, and
takes the value unity when its two indices are equal, and zero otherwise.
It follows from Eqs. (261) and (263)
that

(264) 
Thus, the expansion coefficients in Eq. (261) are easily determined,
given the wavefunction and the eigenstates .
Moreover, if is a properly normalized wavefunction then Eqs. (261) and (263)
yield

(265) 
Next: Measurement
Up: Fundamentals of Quantum Mechanics
Previous: Heisenberg's Uncertainty Principle
Richard Fitzpatrick
20100720