From ket vectors to wavefunction 2

In our previous post Position Operator: From Ket vectors to wavefunctions we have seen that wavefunctions are the scalar product of ket vectors with position eigenkets. More precisely we call these position space wavefunctions. We can also have momentum space wavefunctions which are the scalar product of the ket vectors with momentum operator eigenkets. We will look at these in some other post. Today we wish to see how do we write different operators and equations in position space.

Most of the introductory books on quantum mechanics introduce the theory in the position space making use of wavefunctions.

Consider the eigenvalue equation for the Hamiltonian operator

$latex \hat H \left|\psi\right\rangle=E\left|\psi\right\rangle &s=2$

Here for the time being we are not using the subscript on energy $latex E&s=2$ and the eigenket $latex \left|\psi\right\rangle&s=2$. Now we take the scalar product of the above equation with position eigenket $latex \left|x\right\rangle&s=2$ as

$latex \left\langle x |\hat H|\psi\right\rangle=E\left\langle x|\psi\right\rangle &s=2$

The right hand side we recognize as the wavefunction

$latex \left\langle x|\psi\right\rangle=\psi(x)&s=2$

The left side $latex \left\langle x |\hat H|\psi\right\rangle&s=2$ can be evaluated using the following rules

$latex \left\langle x|\hat x|\psi\right\rangle= x \left\langle x|\psi\right\rangle= x \psi(x)&s=2$

$latex \left\langle x|\hat p|\psi\right\rangle= -i\hbar \frac{\partial}{\partial x} \left\langle x|\psi\right\rangle= -i\hbar \frac{\partial}{\partial x}\psi(x)&s=2$

The first rule tells us that the position operator in position space is multiplicative and the momentum operator in position space is differential. Using these rules and remembering that the Hamiltonian is the sum of kinetic and potential energy operators

$latex \hat H=\frac {\hat p^2}{2m}+ \hat V(\hat x)&s=2$

we finally write the eigenvalue equation for the Hamiltonian operator in position space as

$latex \left(-\frac{\hbar^2}{2m}\frac{d^2}{dx^2}+V(x)\right)\psi(x)=E\psi(x)&s=2$.

This is Schrodinger time independent equation. See if you can do the mathematics to write this equation.


Leave a Reply

Your email address will not be published. Required fields are marked *