Phy5645/HydrogenAtomProblem2

(Problem written by team 5. Based on problem 8.6 in Schaum's QM)

Consider a particle in a central field and assume that the system has a discrete spectrum. Each orbital quantum number Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle l } has a minimum energy value. Show that this minimum value increases as Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle l } increases.

We begin by writing the Hamiltonian of the system.

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle H=\frac{-\hbar^2}{2mr^2}\frac{\partial}{\partial r}(r^2 \frac{\partial}{\partial r}) + \frac{\hbar^2}{2m}\frac{l(l+1)}{r^2} + V(r)}

Using Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle H_1 =\frac{-\hbar^2}{2mr^2}\frac{\partial}{\partial r}(r^2 \frac{\partial}{\partial r}) + V(r)} we have that

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle H = H_1 + \frac{\hbar^2}{2m}\frac{l(l+1)}{r^2}}

The minimum value of the energy in the state Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle l } is

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E_{min}^{l} = \int \psi_{l}^{*} [H_1 + \frac{\hbar^2}{2m}\frac{l(l+1)}{r^2}] \psi_{l} \, \mathrm{d}^3r}

The minimum value of the energy in the state Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle l+1 } is given by

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E_{min}^{l+1} = \int \psi_{l+1}^{*} [H_1 + \frac{\hbar^2}{2m}\frac{(l+1)(l+2)}{r^2}] \psi_{l+1} \, \mathrm{d}^3r}

This equation for the Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle l+1 } state can then be written in the form

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E_{min}^{l+1} = \int \psi_{l+1}^{*} \frac{\hbar^2}{m}\frac{l+1}{r^2} \psi_{l+1} \, \mathrm{d}^3r + \int \psi_{l+1}^{*} [H_1 + \frac{\hbar^2}{2m}\frac{l(l+1)}{r^2}] \psi_{l+1} \, \mathrm{d}^3r}

Since Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mid\psi_{l+1}\mid^2 } and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{\hbar^2}{m}\frac{l+1}{r^2} } are positive, the first term in the above equation is always positive. Consider now the second term:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int \psi_{l+1}^{*} [H_1 + \frac{\hbar^2}{2m}\frac{l(l+1)}{r^2}] \psi_{l+1} \, \mathrm{d}^3r}

Note that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi_{l} } is an eigenfunction of the Hamiltonian Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle H = H_1 + \frac{\hbar^2}{2m}\frac{l(l+1)}{r^2}} and corresponds to the minimum eignevalue of this hamiltonian, therefore, by variational theorem

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int \psi_{l+1}^{*} [H_1 + \frac{\hbar^2}{2m}\frac{l(l+1)}{r^2}] \psi_{l+1} \, \mathrm{d}^3r > \int \psi_{l}^{*} [H_1 + \frac{\hbar^2}{2m}\frac{l(l+1)}{r^2}] \psi_{l} \, \mathrm{d}^3r }

Thus,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int \psi_{l}^{*} [H_0 + \frac{\hbar^2}{2m}\frac{l(l+1)}{r^2}] \psi_{l} \, \mathrm{d}^3r < \int \psi_{l+1}^{*} [H_0 + \frac{\hbar^2}{2m}\frac{l(l+1)}{r^2}] \psi_{l+1} \, \mathrm{d}^3r}

This proves that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E_{min}^{l} < E_{min}^{l+1} }