Electron on Helium Surface: Difference between revisions

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(New page: 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(h...)
 
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   \begin{cases}  
   \begin{cases}  
       -\frac{Q^2e^2}{z} &\mbox{if} \qquad z>0\\
       -\frac{Q^2e^2}{z} &\mbox{if} \qquad z>0\\
       infinity &\mbox{if} \qquad else
       \infty &\mbox{if} \qquad else
   \end{cases}
   \end{cases}
</math>
</math>


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


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


(b) An electric field is turned on at t=0 which produces the perturbation:
(b) An electric field is turned on at t=0 which produces the perturbation:
Line 25: Line 27:


== Solution...==
== Solution...==
(a) Solve the Schrödinger equation.
<math> H = -\frac{\hbar^2}{2m}\nabla^2 + V(z) </math>
The Schrödinger equation for when <math>z>0</math> is:
<math>
  \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})
</math>
Using separation of variables:
<math> \psi (x,y,z) = X(x)Y(y)Z(z) </math>
For X and Y we get place waves.
<math> X(x)Y(y) = Ae^{i(k_x x + k_y y)} </math>
This corresponds to motion parallel to the helium surface.
For z-component the Schroedinger equation becomes:
<math>
  \left[
    \frac{\hbar^2}{2m}\frac{\partial ^2}{\partial z^2} -   
    \frac{Q^2e^2}{z}
  \right] Z(z) =
</math>
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:
<math> Z(z) = zR_{n0}(z) </math>
where
<math>
  R_{10} = \left( \frac{1}{a_0} \right) ^{3/2} z e^{ -\frac{z}{a_0} }
</math>
<math>
  R_{20} = \left( \frac{1}{2a_0} \right) ^{3/2}
    \left( 2-\frac{z}{a_0} \right) e^{ -\frac{z}{2a_0} }
</math>
The total wave function and energies are:
<math> \psi = Ae^{i(k_x x + k_y y)}zR_{n0}(z) </math>
<math>
  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}
</math>
where n = 1,2,... is the quantum number for the z-direction
and the bohr radius has become
<math> a_0 = \frac{\hbar^2}{mQ^2e^2} </math>
(b) Turn on electric field at t=0.
The electric field introduces a perturbation to the hamiltonian:
<math> V_t = eE_0ze^{t/\tau} </math>
From expression 2.1.10 in [http://wiki.physics.fsu.edu/wiki/index.php/Phy5646#Formalism Time Dependent Perturbation] Section of the PHY5646 page:
<math>\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}</math>
<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>
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{3}{2} \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>

Latest revision as of 19:51, 26 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{3}{2} \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}} }

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