Question

The eigenvalues and eigenfunctions of a quantum mechanical operator A are denoted
by an and ψn,respectively. If f(x) denotes a function that can be expanded in the
powers of x, show that:
f (A)ψ n = f (an)ψ n

Answer 1

Answer on Question #62449 - Physics - Quantum Mechanics

The eigenvalues and eigenfunctions of a quantum mechanical operator $A$ are denoted by $a_{n}$ and $\psi_{n}$ , respectively. If $f(x)$ denotes a function that can be expanded in the powers of $x$ , show that:

$$f (A) \psi_ {n} = f (a _ {n}) \psi_ {n}.$$

Solution:

$\psi_{n}$ are eigenfunctions and $a_{n}$ are eigenvalues of $A$

$$A \psi_ {n} = a _ {n} \psi_ {n}.$$

For some power of the operator $A$ :

$$A ^ {m} \psi_ {n} = A ^ {m - 1} A \psi_ {n} = A ^ {m - 1} a _ {n} \psi_ {n} = a _ {n} A ^ {m - 2} A \psi_ {n} = a _ {n} ^ {2} A ^ {m - 2} \psi_ {n} = \dots = a _ {n} ^ {m - 1} A \psi_ {n} = a _ {n} ^ {m} \psi_ {n}.$$

Function $f(x)$ can be expanded in the powers of $x$ , that is,

$$f (x) = \sum_ {k = 0} ^ {\infty} \frac {1}{k !} f ^ {(k)} (0) x ^ {k}.$$

We can define function of an operator as

$$f (A) = \sum_ {k = 0} ^ {\infty} \frac {1}{k !} f ^ {(k)} (0) A ^ {k}.$$

Then

$$\begin{array}{l} f (A) \psi_ {n} = \sum_ {k = 0} ^ {\infty} \frac {1}{k !} f ^ {(k)} (0) A ^ {k} \psi_ {n} = \sum_ {k = 0} ^ {\infty} \frac {1}{k !} f ^ {(k)} (0) (A ^ {k} \psi_ {n}) = \sum_ {k = 0} ^ {\infty} \frac {1}{k !} f ^ {(k)} (0) a _ {n} ^ {k} \psi_ {n} \\ = \left(\sum_ {k = 0} ^ {\infty} \frac {1}{k !} f ^ {(k)} (0) a _ {n} ^ {k}\right) \psi_ {n} = f (a _ {n}) \psi_ {n}. \\ \end{array}$$

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