Electron on Helium Surface: Difference between revisions

From PhyWiki
Jump to navigation Jump to search
Line 120: Line 120:
\end{align}</math>
\end{align}</math>


<math> \begin{align}


   \langle \psi_{n=2}(z) | z | \psi_{n=1}(z) \rangle \right
<math>\begin{align}
 
   \langle \psi_{n=2}(z) | z | \psi_{n=1}(z) \rangle
    
    
   &=  
   &= \langle zR{20} | z | zR_{10} \rangle
    
   \\
  &= \int_{0}^{\infty} dz z^3 e^{ frac{3}{2} frac{z}{a_0} } \left( 2 - \frac{z}{a_0} \right)
    \left[ \frac{1}{ \left( \int_{0}^{\infty} dz z^2 e^{\frac{-2z}{a_0}} \right)^{1/2}
                      \left( \int_{0}^{\infty} dz z^2 \left(2 - \frac{z}{a_0} \right)^2 e^{\frac{-z}{a_0}} \right)^{1/2}
    } \right]
\end{align}</math>
\end{align}</math>
where the quantity inside the brackets is the normalization for <math> \psi_1 </math> and <math> \psi_2 </math>, respectively.
let <math> x = \frac{z}{a_0} </math>
<math>\begin{align}
\langle \psi_{n=2}(z) | z | \psi_{n=1}(z) \rangle
  &\rightarrow
  a_0 \int_{0}^{\infty} dx x^3 e^{ /frac{3}{2} x } \left( 2 - x \right)
    \left[ \frac{1}{ \left( \int_{0}^{\infty} dx x^2e^{-2x} \right)^{1/2}
                      \left( \int_{0}^{\infty} dx x^2 \left( 2 - x \right)^2 e^{ -x } \right)^{1/2}
    } \right]
  \\
  &= a_0 \frac{32 \sqrt{2}}{81}
\end{align}</math>
So the probablity of a transition from ground state to the first excited state is:
<math>
P_{1 \rightarrow 2} = \frac{e^2E_0^2}{\hbar^2} \frac{1}{ \omega_{21}^2 + \frac{1}{\tau^2} } \frac{2^{11}}{3^{8}}
</math>

Revision as of 15:42, 20 April 2010

An electron close to the surface of liquid helium experiences an attractive force due to the electrostatic polarization of the helium and a repulsive force due to the exclusion principle(hard core). To a reasonable approximation for the potential when helium fills the space where 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 z<0 } :

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 V(z) = \begin{cases} -\frac{Q^2e^2}{z} &\mbox{if} \qquad z>0\\ \infty &\mbox{if} \qquad else \end{cases} }

Note: the potential is infinite when 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 z<=0} because the cannot penetrate the helium surface.


(a) Solve the Schrödinger equation. Find the Eingenenergies and Eigenvalues.

(b) An electric field is turned on at t=0 which produces the perturbation:

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 V(z) = \begin{cases} eE_0ze^{t/\tau} &\mbox{if} \qquad t>=0\\ 0 &\mbox{if} \qquad else \end{cases} }

If the electron is initially in its ground state, find the probability makes a transition to its first excited state for times 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 t >> \tau } .

Solution...

(a) Solve the Schrödinger equation.

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}{2m}\nabla^2 + V(z) }

The Schrödinger equation for when 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 z>0} 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 \left[ \frac{\hbar^2}{2m} \left( \frac{\partial ^2}{\partial x^2} + \frac{\partial ^2}{\partial y^2} + \frac{\partial ^2}{\partial z^2} \right) -\frac{Q^2e^2}{z} \right] \psi (\mathbf{r}) = E\psi (\mathbf{r}) }

Using separation of variables:

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 (x,y,z) = X(x)Y(y)Z(z) }

For X and Y we get place waves.

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 X(x)Y(y) = Ae^{i(k_x x + k_y y)} }

This corresponds to motion parallel to the helium surface.

For z-component the Schroedinger equation becomes:

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 \left[ \frac{\hbar^2}{2m}\frac{\partial ^2}{\partial z^2} - \frac{Q^2e^2}{z} \right] Z(z) = }

This has the same form as the hydrogin atom with l=0 (s-wave). Since similar equations have similar solutions, the solution to the z-component 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 Z(z) = zR_{n0}(z) }

where

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 R_{10} = \left( \frac{1}{a_0} \right) ^{3/2} z e^{ -\frac{z}{a_0} } }

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 R_{20} = \left( \frac{1}{2a_0} \right) ^{3/2} \left( 2-\frac{z}{a_0} \right) e^{ -\frac{z}{2a_0} } }


The total wave function and energies are:

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 = Ae^{i(k_x x + k_y y)}zR_{n0}(z) }

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_{k_xK_yK_z} = \frac{\hbar^2}{2m} \left( k_x^2 + k_y^2 \right) -\frac{Q^4e^4m}{2\hbar^2n^2} }

where n = 1,2,... is the quantum number for the z-direction and the bohr radius has become

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 a_0 = \frac{hbar^2}{mQ^2e^2} }


(b) Turn on electric field at t=0.

The electric field introduces a perturbation to 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 V_t = eE_0ze^{t/\tau} }

From expression 2.1.10 in Time Dependent Perturbation Section of the PHY5646 page:

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 \begin{align} P_{1 \rightarrow 2}(t \rightarrow \infty) &= |\langle n|\psi(t)\rangle|^2 \\ &= \left|\frac{1}{i\hbar}\int_{0}^{\infty}dt' e^{\frac{i}{\hbar}(E_2 - E_1)t'} \langle \psi_{n=2}|V_{t'}| \psi_{n=1}\rangle\right|^2 \\ &= \frac{e^2E_0^2}{\hbar^2} \frac{1}{\omega_{21}^2 + \frac{1}{\tau}} \left| \langle \psi_{n=2}(z) | z | \psi_{n=1}(z) \rangle \right|^2 \end{align}}


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 \begin{align} \langle \psi_{n=2}(z) | z | \psi_{n=1}(z) \rangle &= \langle zR{20} | z | zR_{10} \rangle \\ &= \int_{0}^{\infty} dz z^3 e^{ frac{3}{2} frac{z}{a_0} } \left( 2 - \frac{z}{a_0} \right) \left[ \frac{1}{ \left( \int_{0}^{\infty} dz z^2 e^{\frac{-2z}{a_0}} \right)^{1/2} \left( \int_{0}^{\infty} dz z^2 \left(2 - \frac{z}{a_0} \right)^2 e^{\frac{-z}{a_0}} \right)^{1/2} } \right] \end{align}}

where the quantity inside the brackets is the normalization for 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_1 } 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 \psi_2 } , respectively.

let 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 x = \frac{z}{a_0} }

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 \begin{align} \langle \psi_{n=2}(z) | z | \psi_{n=1}(z) \rangle &\rightarrow a_0 \int_{0}^{\infty} dx x^3 e^{ /frac{3}{2} x } \left( 2 - x \right) \left[ \frac{1}{ \left( \int_{0}^{\infty} dx x^2e^{-2x} \right)^{1/2} \left( \int_{0}^{\infty} dx x^2 \left( 2 - x \right)^2 e^{ -x } \right)^{1/2} } \right] \\ &= a_0 \frac{32 \sqrt{2}}{81} \end{align}}

So the probablity of a transition from ground state to the first excited state 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 P_{1 \rightarrow 2} = \frac{e^2E_0^2}{\hbar^2} \frac{1}{ \omega_{21}^2 + \frac{1}{\tau^2} } \frac{2^{11}}{3^{8}} }