Physics Vidyapith ✍️

A non-relativistic free particle of mass $m$ moving in the positive $x$-direction with speed $v_{x}$ has kinetic energy $E=\frac{1}{2} m v^{2}_{x}$ and momentum $p_{x}=mv_{x}$ The energy and momentum are associated with a wave of wavelength $\lambda$ and frequency $\nu$ given by $\lambda = \frac{h}{p_{x}}$ and $\nu=\frac{E}{h}$ The propagation constant $k_{x}$ of the wave is $k_{x}= \frac{2\pi}{\lambda}=\frac{2\pi}{\left(\frac{h}{p_{x}} \right)}=\frac{p_{x}}{\left(\frac{h}{2\pi} \right)}=\frac{p_{x}}{\hbar}$ and the angular frequency $\omega$ is $\omega = 2\pi \nu = \frac{2\pi E}{\hbar}=\frac{E}{\hbar}$ A plane wave traveling along the $x$ axis in the positive direction may be represented by $\psi(x,t)=A e^{-i\left(k_{x} \: x - \omega t\right)}$ Now subtitute the value of $\omega$ and $k_{x}$ in above equation then we get $\psi(x,t)=A e^{i\left( \frac{p_{x}}{\hbar} \: x - \frac{E}{\hbar} \: t\right)}$ $\psi(x,t)=A e^{\frac{i}{\hbar}\l

Derivation: The time dependent Schrodinger quation for the wave function function $psi(x,y,z,t)$ is $-\frac{\hbar^{2}}{2m}\nabla^{2} \psi + V\psi=i\hbar\frac{\partial \psi}{\partial t} \qquad(1)$ Since wave function, $\psi$ is complex quantity i.e. $\psi=\psi_{1}+i \: \psi_{2} \qquad(2)$ Where $\psi_{1}$ and $\psi_{2}$ are real functions of $x,y,z,t$. Substituting this form for $\psi$ in equation $(1)$, we get $-\frac{\hbar^{2}}{2m}\nabla^{2} \left( \psi_{1}+i \: \psi_{2} \right) + V\left( \psi_{1}+i \: \psi_{2} \right) \\ \qquad = i\hbar\frac{\partial }{\partial t} \left( \psi_{1}+i \: \psi_{2} \right)$ Equation real and imaginary parts on either side of this equation, we obtain the following two equations: $-\frac{\hbar^{2}}{2m}\nabla^{2} \psi_{1} + V\psi_{1}=-\hbar\frac{\partial \psi_{2}}{\partial t} \qquad(3)$ $-\frac{\hbar^{2}}{2m}\nabla^{2} \psi_{2} + V\psi_{2}=\hbar\frac{\partial \psi_{1}}{\partial t} \qquad(4)$ Mutiplying equation $(4)$ by $-

1.) Derivation of Probability Current Density for a free particle: Let a particle of mass $m$ is moving in the positive $x$- direction in the region from $x_{1}$ to $x_{2}$. For the one-dimensional motion of the particle, the wave function is $psi(x,t)$ Let $dA$ be the area of the cross-section of the region. The probability of finding a particle in the region is $\int_{x_{1}}^{x_{2}} \psi(x,t) \: \psi^{*}(x,t) \: dx \: dA \qquad(1)$ and the probability density of finding the particle in the region is $P=\psi(x,t) \: \psi^{*}(x,t) \qquad(2)$ If the probability of finding the particle in the region decreases with time, the rate of decrease of the probability that the particle is in the region from $x_{1}$ to $x_{2}$ per unit area is called the probability current density out of the region. Therefore, the probability current density $S_{2} - S_{1}$ out of the region in the positive $x$-direction is given by $S_{2} - S_{1} = - \frac{1}{dA} \left[- \frac{d}{dt

Ehrenfest's Theorem Statement: The theorem states that Quantum mechanics gives the same results as classical mechanics for a particle for which average or expectation values of dynamical quantities are involved. Proof of theorem: The proof of the theorem for one-dimensional motion of a particle by showing that 1) $\frac{d \left < x \right >}{dt} = \frac{\left < p_{x} \right > }{m}$ 2) $\frac{d \left < p_{x} \right >}{dt} = \left < F_{x} \right >$ 1.) To Show that: $\frac{d \left < x \right > }{dt} = \frac{\left < p_{x} \right > }{m}$ Let $x$ is the position coordinate of a particle of mass $m$, at time $t$ The expectation value of position $x$ of a particle is given by $\left < x \right > = \int_{- \infty}^{+ \infty} \psi^{*} (x,t) . x \: \psi (x,t) dx \qquad (1)$ Differentiating the above equation $(1)$ with respect to $t$ $\frac{d \left < x \right > }{dt} = \int_{- \infty}^{+ \infty} x \frac{