Question

(a) Suppose that $f\left(x\right)$and $g\left(x\right)$are two eigenfunctions of an operator$\hat{Q}$ , with the same eigenvalue q . Show that any linear combination of f andgis itself an eigenfunction of $\hat{Q}$, with eigenvalue q .

(b) Check that $f\left(x\right)=exp\left(x\right)$and$g\left(x\right)=exp(-x)$ are eigenfunctions of the operator$d^2/d{x}^2$ , with the same eigenvalue. Construct two linear combinations of and that are orthogonal eigenfunctions on the interval$(-1.1)$ .

Short answer

(a) The linear combination of f and g is an eigenfunction of the operator$\hat{Q}$ .

(b) Both of $f\left(x\right)$ and $g\left(x\right)$ are eigenfunctions of the operator $d^2/d{x}^2$, with the same eigenvalue of l . The two linear combinations of $f\left(x\right)$ and $g\left(x\right)$ are orthogonal.

Step-by-step solution

Definition of eigenfunction.

The eigenfunction is a function f of an operator A that provides the same function f when operator A is applied on f. Mathematically, it can be written as,

$\hat{A}f=kf$

Here, k is the constant known as eigen value.

Prove for part (a).

(a)

Assuming $f\left(x\right)$and $g\left(x\right)$are two eigenfunctions of an operator $\hat{Q}$, with the same eigenvalue q , we must demonstrate that any linear combination of f and g is an eigenfunction of the operator $\hat{Q}$$\hat{Q}f=qf$

And

$\hat{Q}g=qg$

And let us consider a function:

$\$h\left(x\right)=af\left(x\right)+bg\left(x\right)\$$

Where a and b are an arbitrary constant.

On solving the Therefore:

$\begin{array}{rcl}\hat{Q}h& =& \hat{Q}(af+bg)\\ & =& a\left(\hat{Q}f\right)+b\left(\hat{Q}g\right)\\ & =& a\left(qf\right)+b\left(qg\right)\\ & =& q(af+bg)\\ & =& qh\\ & & \end{array}$

So, the linear combination of and is an eigenfunction of the operator $\hat{Q}$.

Check for the given function and construction of linear combination

(b)

Check that $f\left(x\right)=e^x$ and $g\left(x\right)=e^{-x}$are eigenfunctions of the operator $d^2/d{x}^2$, as:

$\begin{array}{rcl}\frac{d^{2}f}{d{x}^2}& =& \frac{d^2}{d{x}^2}\left(e^x\right)\\ & =& \frac{d}{dx}\left(e^x\right)\\ & =& e^x\\ & =& f\\ & & \end{array}$

And for $g\left(x\right)$as:

$\begin{array}{rcl}\frac{d^{2}g}{d{x}^2}& =& \frac{d^2}{d{x}^2}\left(e^{-x}\right)\\ & =& \frac{d}{dx}\left(-e^{-x}\right)\\ & =& e^{-x}\\ & =& g\\ & & \end{array}$

Therefore, both of $f\left(x\right)$ and $g\left(x\right)$are eigenfunctions of the operator $d^2/d{x}^2$, with the same eigenvalue of 1, the two linear combinations of the functions are:

$\sinh \left(x\right)=\frac{1}{2}\left(e^x-e^{-x}\right)=\frac{1}{2}(f-g)$

And,

$\cosh \left(x\right)=\frac{1}{2}\left(e^x+e^{-x}\right)=\frac{1}{2}(f+g)$

Since is odd and is even, then the two linear combinations of $f\left(x\right)$and $g\left(x\right)$ are orthogonal in the interval (-1,1).