Properties of the Eigenvalues and Eigenvectors of a Hermitian Operator | Quantum Science Philippines
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Properties of the Eigenvalues and Eigenvectors of a Hermitian Operator


 

Properties of the Eigenvalues and Eigenvectors of a Hermitian Operator

 



 

We can say that an operator A is a Hermitian Operator if it satisfies that: A^\dagger = A.

Given the eigenvalues and eigenvectors of the Hermitian Operators, the following properties apply:

  1. The eigenvalues of a Hermitian operator are real.
  2. Two eigenvectors of a Hermitian operator corresponding to two different eigenvalues are orthogonal. 

In this section,we are to prove these statements.

 

Proof:

(1.)  The eigenvalues of a Hermitian operator are real.

Suppose that we have a Hermitian operator H, then we have

H|u\rangle=\lambda |u\rangle                            [1]

where |u\rangle is the eigen vector and \lambda is the eigenvalue of the operator H.

Taking the scalar product of [1] with \langle u| yields,

\langle u| H |u\rangle = \langle u|\lambda|u\rangle                 [2]

Here, we have \langle u|H|u\rangle to be real because we assumed our operator H to be a Hermitian operator already. And hence this implies that \lambda is also real.

Now, we note that H is a Hermitian operator and by definition, H^\dagger=H.

Then for the conjugate we can have,

\langle u| H^\dagger |u\rangle = \langle u|\lambda^\star|u\rangle              [3]

Since the definition establishes that H^\dagger=H, then we equate the scalar products of these 2 operators.

  \langle u|H|u\rangle = \langle u| H^\dagger |u\rangle               [4]

Using equations [2] and [3], we have

\langle u|\lambda|u\rangle = \langle u|\lambda^\star|u\rangle                  [5]

\lambda \langle u|u\rangle = \lambda^\dagger \langle u|u \rangle                    [6]

We know already that \langle u|u \rangle =1. So we are left with,

\mathbf{\lambda} = \mathbf{\lambda^\star}

Since we have seen that the eigenvalues of the conjugate of the Hermitian operator to be  real.

Therefore we can say that the eigenvalues of a Hermitian operator are real. \Box

 

 

(2.) Two eigenvectors of a Hermitian operator corresponding to two different eigenvalues are orthogonal. 

Let |u \rangle and |v \rangle be eigenvectors of a Hermitian operator H, corresponding to the eigenvalues \lambda and \mu respectively.

For both eigenvectors to be orthogonal, it must satisfy that: \langle u|v \rangle = \langle v|u \rangle =0.

Now, we have

H |u \rangle = \lambda |u \rangle                  [7]

H |v \rangle = \mu |v \rangle                  [8]

Since H is Hermitian, we can write [8] in the form

\langle v| H = \langle v| \mu                  [9]

If we take the scalar product of [7] with \langle v| and [9] with |u \rangle , then

\langle v| H |u \rangle= \langle v | \lambda |u \rangle          [10]

\langle v| H |u \rangle= \langle v | \mu |u \rangle          [11]

Equating equations [10] and [11] yields.

\langle v | \lambda |u \rangle = \langle v | \lambda |u \rangle           [12]

\lambda \langle v | u \rangle = \mu \langle v | u \rangle             [13]

(\lambda - \mu) \langle v | u \rangle = 0           [14]

Since (\lambda - \mu) \neq 0 , then the only way for equation [14] to be true is to set \mathbf{\langle v | u \rangle} = \mathbf{0} .

Thus, two eigenvectors of a Hermitian operator corresponding to two different eigenvalues are orthogonal\Box

 

By: Joshua J. Ordeniza, MS Physics Student, MSU-IIT




	

				
	

				

	

			

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