Schrodinger Time Dependent Equation PDF

# Schrodinger Time Dependent Equation The equation is given by Austrian physicist Erwin Schrodinger in 1926. It follows the de Broglie's hypothesis. The wave is associated with subatomic particles called matter waves. - It is a second order differential equation. - It tells us about the energy and...

# Schrodinger Time Dependent Equation The equation is given by Austrian physicist Erwin Schrodinger in 1926. It follows the de Broglie's hypothesis. The wave is associated with subatomic particles called matter waves. - It is a second order differential equation. - It tells us about the energy and other properties of a given system. - It gives information about the behavior of an electron bound to a nucleus. We know that, $H = T + V$ $E = \frac{p^2}{2m} + V(x)$ ### **Hamiltonian = K.E. + P.E.** (Take any axis (x,y, or z)) According to wave mechanics, $\psi = Ae^{i(wt-kx)}$ Here - w = angular frequency - k = angular wave no. k = It is no. of waves or cycles per unit distance. Now, $ w = 2\pi \nu$ We also know that, $ E = h\nu$ $E = \frac{h\nu}{h}$ ## **Derivation** So, w = $2\pi\nu= \frac{E}{h}$ Now, K = $2\pi \nu = \frac{h}{λ}$ We know that, $\lambda=\frac{h}{p}$ Therefore, $\psi = Ae^{i(wt-px)}$ ## **Differentiating the equation w.r.t t** $\frac{d\psi}{dt} = \frac{d}{dt}[Ae^{i(wt-px)}]$ $\frac{d\psi}{dt} = Ae^{i(wt-px)} \frac{d}{dt}[i(wt-px)]$ $d\psi = Ae^{i(wt-px)} \cdot i\omega \cdot e^{i(wt-px)}$ $\frac{d\psi}{dt} = i\omega Ae^{i(wt-px)} \cdot e^{i(wt-px)}$ $\frac{d\psi}{dt} = i\omega \psi$ We see above term is equivalent to, $\frac{d\psi}{dt}=i\omega \psi = i\frac{E}{h}\psi = E\psi$ ## **Differentiating the equation w.r.t x** $\frac{d\psi}{dx} = \frac{d}{dx}[Ae^{i(wt-px)}]$ $\frac{d\psi}{dx} = Ae^{i(wt-px)} \frac{d}{dx}[i(wt-px)]$ $\frac{d\psi}{dx} = Ae^{i(wt-px)} \cdot (-ip) \cdot e^{i(wt-px)}$ $\frac{d\psi}{dx}= -ip Ae^{i(wt-px)} \cdot e^{i(wt-px)}$ $\frac{d\psi}{dx} = -ip \psi$ $\frac{d^2 \psi}{dx^2}= \frac{d}{dx} [ -ip \psi ]= -ip \frac{d}{dx}\psi$ $\frac{d^2 \psi}{dx^2}= -ip^2 \psi$ Now, going to equation no. 1 $E= \frac{p^2}{2m} + V(x)$ Multiplying & on both sides $ E\psi=\frac{p^2}{2m}\psi+V(x)\psi$ $i\hbar \frac{d\psi}{dt} = -\frac{\hbar^2}{2m} \frac{d^2 \psi}{dx^2} + V(x) \psi$ This is the Schrodinger Time Dependent Equation. ## **Schrodinger Time Independent Equation** We know that, $H = T + V$ $E = K.E + P.E$ $E = \frac{p^2}{2m} + V(x)$ Multiply with $\psi$ on both side: $E\psi(x) = \frac{p^2}{2m}\psi(x) + V(x)\psi(x)$ The general form of $\psi$ is $e^{i(kx-wt)}$. ## **Differentiating the equation twice w.r.t x** $\frac{d\psi(x)}{dx} = \frac{d}{dx}[e^{i(kx-wt)}]$ $\frac{d\psi(x)}{dx} = ik e^{i(kx-wt)}$ ## **Differentiating again w.r.t x** $\frac{d^2\psi(x)}{dx^2} = ik(ik)e^{i(kx-wt)}$ $\frac{d²\psi(x)}{dx²} = ik²e^{i(kx-wt)}$ We know that, $K = \frac{2\pi}{λ}$ $K = \frac{2\pi}{h} p$ $K = \frac{2\pi \hbar}{h}$ $K = \frac{2\pi \ ħ }{h}$ $K = \frac{2\pi h}{2\pi \hbar}$ $K = \frac{h^2}{2\pi \hbar}$ $K^2 = (\frac{h^2}{2\pi h})^2$ $K^2 = \frac{h^4}{4\pi^2 h}$ $K^2 = \frac{h^2}{4\pi^2}$ $K^2 = \frac{p^2}{h^2}$ Therefore, $\frac{d²\psi(x)}{dx²} = \frac{p²}{h²}e^{i(kx-wt)}$ $\frac{d²\psi(x)}{dx²} = -\frac{p²}{h^2}\psi(x)$ ## **Rearranging the equation** $E Ψ(x) = -\frac{h^2}{2m}\frac{d^2 \psi(x)}{dx^2} + V(x)\psi(x)$ $E\psi(x) - V(x)\psi(x) = -\frac{h^2}{2m} \frac{d^2 \psi(x)}{dx^2}$ ## **Multiply the equation by 2m** $2m E\psi(x) - 2m V(x) \psi(x) = -h^2 \frac{d² \psi(x)}{dx²}$ $2m [E\psi(x) - V(x) \psi(x)] + h² \frac{d² \psi(x)}{dx²} = 0$ $h² \frac{d² \psi(x)}{dx²} + 2m [E\psi(x) - V(x) \psi(x)] = 0$ ## **Zeroth Law of Thermodynamics** If two systems are in thermal equilibrium with each other, they have the same temperature. ## **First Law of Thermodynamics** Energy cannot be created or destroyed, only transformed from one form to another. - The change in internal energy (ΔE) is equal to the heat (q) added to the system plus the work (w) done on the system. - ΔE = q + w ## **Second Law of Thermodynamics** The total entropy of an isolated system can never decrease over time. It can only remain constant or increase. ## **Third Law of Thermodynamics** As the temperature of a system approaches absolute zero (0 Kelvin), its entropy approaches a constant value. ## **Gibbs Free Energy** Gibbs free energy (G) is a thermodynamic potential that can be used to calculate the maximum amount of reversible work that may be performed by a thermodynamic system at constant temperature and pressure. It is defined as: $G = H - TS$ where: - G is the Gibbs free energy - H is the enthalpy - T is the temperature - S is the entropy **The Change in Gibbs Free Energy** $ΔG = ΔH - TΔS$ $ΔG = ΔU + Δ(PV) - TΔS$ $ΔG = ΔU + PΔV - TΔS$ ## **Helmholtz Free Energy** Helmholtz free energy (A) is a thermodynamic potential that can be used to calculate the maximum amount of reversible work that may be performed by a thermodynamic system at constant temperature and volume. It is defined as: $A = U - TS$ where: - A is the Helmholtz free energy - U is the internal energy - T is the temperature - S is the entropy **The Change in Helmholtz Free Energy** $ΔA = ΔU - TΔS$ $ΔA = -W_{rev}$ **Significance of Helmholtz free energy** - It is a measure of the maximum amount of useful work that can be obtained from a closed system at constant temperature and volume. - It is inversely related to the spontaneity of a process. If ΔA is negative, then the process is spontaneous. If ΔA is positive, then the process is non-spontaneous. ## **Work function** The work function (A) is a thermodynamic function that is defined as the difference between the internal energy (U) and the product of the temperature (T) and the entropy (S) of a system. $A = U - TS$ **Importance of work function** - It is a measure of the maximum amount of work that can be obtained from a system at constant temperature and volume. - It is often used in conjunction with the Gibbs free energy to determine the spontaneity of a process. ## **Gibbs-Helmholtz Equation** The Gibbs-Helmholtz equation relates the Gibbs free energy (G) to the enthalpy (H), the temperature (T), and the entropy (S). The Gibbs-Helmholtz equation is given by: $d(ΔG) / dT = (ΔH - ΔG) / T$ **Derivation of Gibbs-Helmholtz Equation** $dG = -SdT + VdP$ $d(ΔG)/dT = -S$ $d(ΔG)/dT = ΔS$ $d(ΔG)/dT = (ΔH - ΔG)/T.$