Lecture 6- TISE

Review of S.E

$$p=\frac{h}{\lambda };E=\hslash \omega$$ $$\Psi =\text{complex}$$ $$e^{i\omega t}$$

Has to be continuous and normalize-able

$$j\equiv \frac{\hslash }{2mi}\left({\Psi }^{\ast }\frac{d\Psi }{dx}-\frac{\Psi d{\Psi }^{\ast }}{dx}\right)$$

S.E

$$-\frac{{\hslash }^2}{2m}\frac{d^2\Psi }{d{x}^2}+V\Psi =\frac{\hslash }{i}\frac{d\Psi }{pt}$$

We can express any wavefunction as a sum of pure momentum states

2 ways to calculate momentum

1. Hard way. math major way. Use the wave function to compute ${\Phi }_p(p)$.
2. Easy way. (better way) Express p as a function of x Using momentum operator

Correspondence principle

• Anything calculated from quantum physics in the right limits should correspond to classical physics in the right limits
• We should find a momentum operator that corresponds

Pure momentum state

$$\psi (x,t)=A{e}^{i(px-Et)/\hslash }=A{e}^{i(kx-\omega t)}$$

And using the momentum operator:

$$p_{\text{op}}\Psi =\frac{\hslash }{i}\frac{\partial \Psi }{\partial x}=\frac{\hslash }{i}\frac{Aip}{\hslash }{e}^{i(px-E)/\hslash }$$

General operator construction

Solving S.E (time independent)

$$-\frac{{\hslash }^2}{2m}f(t)\frac{d^2\psi (x)}{d{x}^2}+V\psi (x)f(t)=i\hslash \psi (x)\frac{df(t)}{dt}$$

divide everything by

$$\psi (x)f(t)$$

You get

$$-\frac{i{\hslash }^2}{f}\frac{d^2\psi (x)}{d{x}^2}+V(x)=\frac{ib\hslash }{f}\frac{df(t)}{dt}=E$$ $$\frac{i\hslash }{f}\frac{df}{dt}=E$$ $$\frac{df}{f}=\frac{1}{i\hslash }Edt$$ $$\frac{{\hslash }^2}{2m}\frac{1}{\psi }\frac{d^2\psi }{d{x}^2+v}-V\psi =-E\psi$$ $$\frac{{\hslash }^2}{2m}\frac{d^2\psi }{d{x}^2}+(E-V)\psi =0$$

Time independent S.E Time independent Schrodinger equation

TISE

$$\frac{d^2\psi }{d{x}^2}+\frac{2m}{\hslash }(E-V)\psi =0$$