Time-Independent Schrodinger Equation and Hermitian Operators

_{Disclaimer: I lay no claim to the originality of the basic ideas expounded here. Its distinctness lies more in its scope and in the detail of its exposition.}

Okay, the previous section's was littered with a little math. Now comes more math.

Stationary states and the time-independent Schrodinger equation

Previously, we have derived the Schrodinger equation by transforming the classical energy into operators. Today, we will specialize to the case when the particle is in a state with definite energy E. From the first section, we've derived that following form,

where specifies the space-dependent part of the function. Let's take the absolute modulus of this wave function to obtain the probability amplitude,

It is obvious that our time amplitude, is time independent, such a state is called stationary state.

Now, lets substitute the time-dependent solution we have to the Schrodinger equation,

after some cancellation of the factor , we obtain,

But note that our potential is still a time dependent. It turns out that for a definite energy to be attained, our potential-energy function must be time independent. We can replace . And we have,

This equation is the time-independent Schrodinger equation, and is an eigenvalue equation of the form

with

This operator is known as the Hamiltonian operator.

Eigenvalue spectra and the results of measurements

Recall the following equation of the previous sections,

The set of all possible eigenvalues a of an operator is called the spectrum of the operator . And for any physically realizable measurement of the observable A, it will always yield a value belonging to this spectrum.

For this eigenvalue problem there are two cases that we should consider.

Case 1

The wave function describing the state of the particle is an eigenfunction of , that is,

Then the result of measuring A will certainly be

Case 2

The wavefunction describing the state of the particle is not eigenfunction of that is, the action of on gives a function that is not simply a scaled version of .

But since the eigenfunctions of form a complete set, in the sense that any normalized function can be expanded in terms of them, we may write as such an expansion:

Then we have,

For this case, the result of measuring A yields the result with probability

Hermitian Operators

One might wonder if all observable have an equivalent operators. It turns out there is a condition for an observable to have an equivalent operator, that is, the physical observable operator must be Hermitian.

Consider the textbook definition of a Hermitian operator.

Definition An operator is said to be hermitian if for any pair of normalizable wave function and , the relation

always hold.

Two important properties of Hermitian operators:

1. The eigenvalues of Hermitian operators are always real.

Proof:

Choose a that are of the same with the wave function of the operator, with corresponding eigenvalue that is, we let

​ Using our definition of hermitian, we have

that is, , or, in other words, the eigenvalue is real.

2. Hermitian operators corresponding to different eigenvalues has eigenfunctions that are orthogonal to each other

Proof:

Now, choose to be different eigenfunctions of with the corresponding eigenvalues,

Again, using the definition of hermitian,

Utilizing the first fact that , we have

since we must have:

We say that is orthogonal with the wave function .

Remember this two facts about Hermitian operators they will come handy in almost all problems in quantum mechanics. For the moment, we will use the idea of orthogonality to prove an inequality:

(we will use this to prove the general uncertainty relation in future sections). To prove this inequality, let's divide it into two case.

Case 1

The case where the function corresponds to the equality.

Case 2

For the case where we will express it as linear combination in terms of each other.

1. 

   If the function is not only proportional to but also a part proportional to a normalized function orthogonal to , with (where ), we have:
   

   
   but then , so we have,
   

2. 

   Similarly, expressing in terms of with the function orthogonal to and with both function normalized to unity so that , we have
   

   

   but then so we have,
   

   

Multiplying the results of 1 and 2, we have,

Therefore, we obtain the inequality. Q.E.D.