Section 6-9
Section 6-9 Eigenfunctions of a Hermitian Operator Form an Orthogonal Set
conjugate of Eq. (6-16) causes the imaginary parts to reverse sign, but they remain equal. Therefore, we may write
ˆ
A∗ψ∗ = a∗ψ∗
(6-17)
We multiply Eq. (6-16) from the left by ψ∗ and integrate over all spatial variables:
ψ∗ ˆ
Aψ dv = a ψ∗ψ dv
(6-18)
Similarly, we multiply Eq. (6-17) from the left by ψ and integrate:
ψ ˆ
A∗ψ∗dv = a∗ ψψ∗ dv
(6-19)
Since ˆ A is hermitian, the left-hand sides of Eqs. (6-18) and (6-19) are equal by definition (Eq. 6-10). Therefore, the right-hand sides are equal, and their difference is zero:
(a − a∗) ψ∗ψ dv = 0
(6-20)
Since ψ is square integrable the integral cannot be zero. Therefore, a − a∗ is zero, which requires that a be real.
6-9 Proof That Nondegenerate Eigenfunctions of aHermitian Operator Form an Orthogonal Set Let ψ and φ be two square-integrable eigenfunctions of the hermitian operator ˆ
A:
ˆ
Aψ = a1ψ
(6-21)
ˆ
A∗φ∗ = a2φ∗
(6-22)
Multiplying Eq. (6-21) from the left by φ∗ and Eq. (6-22) from the left by ψ, and integrating gives
φ∗ ˆ
Aψ dv = a1 φ∗ψ dv
(6-23)
ψ ˆ
A∗φ∗ dv = a2 ψφ∗ dv
(6-24)
The left sides of Eqs. (6-23) and (6-24) are equal by (6-10), and (a1 − a2) φ∗ψ dv = 0.
(6-25)
If a1 = a2, the integral vanishes. This proves that nondegenerate eigenfunctions are orthogonal.
Chapter 6 Postulates and Theorems of Quantum MechanicsEXAMPLE 6-2 It has been shown (Section 6-7) that i(d/dx) is a hermitian operator.
We know that it has eigenfunctions exp(±ikx) with eigenvalues ±k, which are real. So far, so good. However, this operator also has eigenfunctions exp(±kx), with eigenvalues ±ik, which are imaginary. This appears to violate the proof that eigenvalues of hermitian operators are real. Explain why neither of these eigenfunction sets is covered by the proof of section 6-8, and how one of them manages to obey the rule anyway.
SOLUTION The test for Hermiticity requires that iψ∗φ|+∞ −∞ = 0. If φ is ψ, and if ψ is square-integrable, this condition is satisfied, because ψ∗ψ vanishes at ±∞, giving 0 − 0 = 0. But neither of the exponential functions given above is square-integrable: They are both unequal to zero at ±∞, so they both fall outside of the proof as given. Despite this, exp(±ikx) does have real eigenvalues, leading us to look more closely. Is it the case that iψ∗ψ|+∞ −∞ = 0 for this set of functions, even though they do not vanish at infinity? It is indeed, since ψ∗ψ = 1, giving i − i = 0 for this term. Thus we see that our requirement that ψ be square integrable is more restrictive than what is necessary, namely that iψ∗ψ|+∞ −∞ = 0. Note that the other set of exponentials, exp(±kx), leads to iψ∗ψ = i exp(±2kx), which does not produce a value of zero when values at x = ∞ and x = −∞ are subtracted. Note also that the functions exp(±ikx) are orthogonal for different values of k, whereas the functions exp(±kx) are not.
The point of the above example is that all of our proofs about eigenvalues or eigenfunctions of hermitian operators refer to cases where the eigenfunctions satisfy the requirement that iψ∗ ψ|+∞ −∞ = 0. Square-integrability guarantees this, but some nonsquare-integrable sets of functions can satisfy it too. A hermitian operator can have eigenfunctions that are associated with complex or imaginary eigenvalues, but these must result from eigenfunctions that do not satisfy the requirement.
6-10 Demonstration That All Eigenfunctionsof a Hermitian Operator May Be Expressed
as an Orthonormal Set If a1 = a2, Eq. (6-25) is satisfied even when the integral is finite. Therefore, degenerate eigenfunctions need not be orthogonal. But they must be linearly independent or else they are the self-same function (to within a multiplicative constant), and if they are linearly independent, they can be converted to an orthogonal pair. Hence, it is always possible to express the degenerate eigenfunctions of a hermitian operator as an orthogonal set (and, as we have just proved, it is necessary that nondegenerate
eigenfunctions be orthogonal). Furthermore, the functions must be square integrable, hence normalizable. In general, then, we are able to assume that all of the eigenfunctions of a hermitian operator can be expressed as an orthonormal set.
One way to orthogonalize two nonorthogonal, linearly independent functions (which may or may not be eigenfunctions) will now be demonstrated. Let the functions be ψ and φ (assumed normalized) and the integral of their product have the value S: ψ∗φ dv = S
(6-26)
conjugate of Eq. (6-16) causes the imaginary parts to reverse sign, but they remain equal. Therefore, we may write
ˆ
A∗ψ∗ = a∗ψ∗
(6-17)
We multiply Eq. (6-16) from the left by ψ∗ and integrate over all spatial variables:
ψ∗ ˆ
Aψ dv = a ψ∗ψ dv
(6-18)
Similarly, we multiply Eq. (6-17) from the left by ψ and integrate:
ψ ˆ
A∗ψ∗dv = a∗ ψψ∗ dv
(6-19)
Since ˆ A is hermitian, the left-hand sides of Eqs. (6-18) and (6-19) are equal by definition (Eq. 6-10). Therefore, the right-hand sides are equal, and their difference is zero:
(a − a∗) ψ∗ψ dv = 0
(6-20)
Since ψ is square integrable the integral cannot be zero. Therefore, a − a∗ is zero, which requires that a be real.
6-9 Proof That Nondegenerate Eigenfunctions of aHermitian Operator Form an Orthogonal Set Let ψ and φ be two square-integrable eigenfunctions of the hermitian operator ˆ
A:
ˆ
Aψ = a1ψ
(6-21)
ˆ
A∗φ∗ = a2φ∗
(6-22)
Multiplying Eq. (6-21) from the left by φ∗ and Eq. (6-22) from the left by ψ, and integrating gives
φ∗ ˆ
Aψ dv = a1 φ∗ψ dv
(6-23)
ψ ˆ
A∗φ∗ dv = a2 ψφ∗ dv
(6-24)
The left sides of Eqs. (6-23) and (6-24) are equal by (6-10), and (a1 − a2) φ∗ψ dv = 0.
(6-25)
If a1 = a2, the integral vanishes. This proves that nondegenerate eigenfunctions are orthogonal.
Chapter 6 Postulates and Theorems of Quantum MechanicsEXAMPLE 6-2 It has been shown (Section 6-7) that i(d/dx) is a hermitian operator.
We know that it has eigenfunctions exp(±ikx) with eigenvalues ±k, which are real. So far, so good. However, this operator also has eigenfunctions exp(±kx), with eigenvalues ±ik, which are imaginary. This appears to violate the proof that eigenvalues of hermitian operators are real. Explain why neither of these eigenfunction sets is covered by the proof of section 6-8, and how one of them manages to obey the rule anyway.
SOLUTION The test for Hermiticity requires that iψ∗φ|+∞ −∞ = 0. If φ is ψ, and if ψ is square-integrable, this condition is satisfied, because ψ∗ψ vanishes at ±∞, giving 0 − 0 = 0. But neither of the exponential functions given above is square-integrable: They are both unequal to zero at ±∞, so they both fall outside of the proof as given. Despite this, exp(±ikx) does have real eigenvalues, leading us to look more closely. Is it the case that iψ∗ψ|+∞ −∞ = 0 for this set of functions, even though they do not vanish at infinity? It is indeed, since ψ∗ψ = 1, giving i − i = 0 for this term. Thus we see that our requirement that ψ be square integrable is more restrictive than what is necessary, namely that iψ∗ψ|+∞ −∞ = 0. Note that the other set of exponentials, exp(±kx), leads to iψ∗ψ = i exp(±2kx), which does not produce a value of zero when values at x = ∞ and x = −∞ are subtracted. Note also that the functions exp(±ikx) are orthogonal for different values of k, whereas the functions exp(±kx) are not.
The point of the above example is that all of our proofs about eigenvalues or eigenfunctions of hermitian operators refer to cases where the eigenfunctions satisfy the requirement that iψ∗ ψ|+∞ −∞ = 0. Square-integrability guarantees this, but some nonsquare-integrable sets of functions can satisfy it too. A hermitian operator can have eigenfunctions that are associated with complex or imaginary eigenvalues, but these must result from eigenfunctions that do not satisfy the requirement.
6-10 Demonstration That All Eigenfunctionsof a Hermitian Operator May Be Expressed
as an Orthonormal Set If a1 = a2, Eq. (6-25) is satisfied even when the integral is finite. Therefore, degenerate eigenfunctions need not be orthogonal. But they must be linearly independent or else they are the self-same function (to within a multiplicative constant), and if they are linearly independent, they can be converted to an orthogonal pair. Hence, it is always possible to express the degenerate eigenfunctions of a hermitian operator as an orthogonal set (and, as we have just proved, it is necessary that nondegenerate
eigenfunctions be orthogonal). Furthermore, the functions must be square integrable, hence normalizable. In general, then, we are able to assume that all of the eigenfunctions of a hermitian operator can be expressed as an orthonormal set.
One way to orthogonalize two nonorthogonal, linearly independent functions (which may or may not be eigenfunctions) will now be demonstrated. Let the functions be ψ and φ (assumed normalized) and the integral of their product have the value S: ψ∗φ dv = S
(6-26)