Derivation of time dependent Schrodinger's wave equation

مرسلة بواسطة Physics Vidyapith

Time-dependent Schrodinger wave equation:

Let a particle of mass $m$ is moving along the positive $x$-direction. So the total energy $E$ of the particle is:

$E=\frac{1}{2}mv^{2}+V(x)$

$E=\frac{(mv)^{2}}{2m}+V(x)$

$E=\frac{(P_{x}^{2})^{2}}{2m}+V(x) \qquad(1)$

Since moving particles are associated with the wave function $\psi(x,t)$. So multiply $\psi(x,t)$ on both sides of equation$(1)$

$E\psi(x,t) =\frac{(P_{x}^{2})^{2}}{2m} \psi(x,t)+V(x) \psi(x,t) \qquad(2)$

The wave function $\psi(x,t)$ representing the plane wave associated with the particle is given by:

$\psi(x,t)=A e^{\frac{i}{\hbar}(P_{x}.x-Et)} \qquad(3)$

Differentiate with respect to $x$ the above equation $(3)$

$\frac{\partial \psi(x,t)}{\partial x}= A e^{\frac{i}{\hbar}(P_{x}.x-Et)} (\frac{i}{\hbar})P_{x} \qquad(4)$

Again differentiate the above equation$(4)$

$\frac{\partial^{2} \psi(x,t)}{\partial x^{2}}= A e^{\frac{i}{\hbar}(P_{x}.x-Et)} (\frac{i}{\hbar})^{2}P_{x}^{2}$

$\frac{\partial^{2} \psi(x,t)}{\partial x^{2}}= - \frac{P_{x}^{2}}{\hbar ^{2}} \psi(x,t)$

$P_{x}^{2}\psi(x,t) =- \hbar^{2} \frac{\partial^{2} \psi(x,t)}{\partial x^{2}} \qquad(5)$

Now differentiate the equation $(3)$ with respect to $t$

$\frac{\partial \psi(x,t)}{\partial t}= A e^{\frac{i}{\hbar}(P_{x}.x-Et)} (\frac{i}{\hbar})(-E)$

$\frac{\partial \psi(x,t)}{\partial t}= - (\frac{i}{\hbar})E\psi(x,t)$

$ E\psi(x,t)= -(\frac{\hbar}{i}) \frac{\partial \psi(x,t)}{\partial t}$

$ E\psi(x,t)= i^{2}(\frac{\hbar}{i}) \frac{\partial \psi(x,t)}{\partial t} \qquad (\because i^{2}=-1)$

$ E\psi(x,t)= i \hbar \frac{\partial \psi(x,t)}{\partial t} \qquad(6)$

Now substitute the value $E\psi(x,t)$ and $P_{x}^{2} \psi(x,t)$ in equation $(2)$. Then

$i \hbar \frac{\partial \psi(x,t)}{\partial t}= -\frac{\hbar^{2}}{2m} \frac{\partial^{2} \psi(x,t)}{\partial x^{2}}+ V(x) \psi(x,t)$

Thus above equation is time-dependent Schrodinger equation. Another form of the above equation can be written as:

$(i \hbar \frac{\partial}{\partial t}) \psi(x,t)= [-\frac{\hbar^{2}}{2m} \frac{\partial^{2}}{\partial x^{2}}+ V(x) ] \psi(x,t)$

Where
$i \hbar \frac{\partial}{\partial t}$→ Energy operator (E)

$-\frac{\hbar^{2}}{2m} \frac{\partial^{2}}{\partial x^{2}}+ V(x)$→ Hamiltonian operator(H)

Now above equation:

$E \psi(x,t)=H \psi(x,t)$

Physics Vidyapith ✍️