Schrodinger equation

Dirac delta function potential

November 19, 2015 phy1520 , ,

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Note: there’s an error below (and in the associated PDF).  10 points to anybody that finds it (I’ve fixed it in my working version of phy1520.pdf)

Q:Dirac delta function potential

Problem 2.24/2.25 [1] introduces a Dirac delta function potential

\begin{equation}\label{eqn:diracPotential:20}
H = \frac{p^2}{2m} – V_0 \delta(x),
\end{equation}

which vanishes after \( t = 0 \). Solve for the bound state for \( t < 0 \) and then the time evolution after that.

A:

The first part of this problem was assigned back in phy356, where we solved this for a rectangular potential that had the limiting form of a delta function potential. However, this problem can be solved directly by considering the \( \Abs{x} > 0 \) and \( x = 0 \) regions separately.

For \( \Abs{x} > 0 \) Schrodinger’s equation takes the form

\begin{equation}\label{eqn:diracPotential:40}
E \psi = -\frac{\Hbar^2}{2m} \frac{d^2 \psi}{dx^2}.
\end{equation}

With

\begin{equation}\label{eqn:diracPotential:60}
\kappa =
\frac{\sqrt{-2 m E}}{\Hbar},
\end{equation}

this has solutions

\begin{equation}\label{eqn:diracPotential:80}
\psi = e^{\pm \kappa x}.
\end{equation}

For \( x > 0 \) we must have
\begin{equation}\label{eqn:diracPotential:100}
\psi = a e^{-\kappa x},
\end{equation}

and for \( x < 0 \)
\begin{equation}\label{eqn:diracPotential:120}
\psi = b e^{\kappa x}.
\end{equation}

requiring that \( \psi \) is continuous at \( x = 0 \) means \( a = b \), or

\begin{equation}\label{eqn:diracPotential:140}
\psi = \psi(0) e^{-\kappa \Abs{x}}.
\end{equation}

For the \( x = 0 \) region, consider an interval \( [-\epsilon, \epsilon] \) region around the origin. We must have

\begin{equation}\label{eqn:diracPotential:160}
E \int_{-\epsilon}^\epsilon \psi(x) dx = \frac{-\Hbar^2}{2m} \int_{-\epsilon}^\epsilon \frac{d^2 \psi}{dx^2} dx – V_0 \int_{-\epsilon}^\epsilon \delta(x) \psi(x) dx.
\end{equation}

The RHS is zero

\begin{equation}\label{eqn:diracPotential:180}
E \int_{-\epsilon}^\epsilon \psi(x) dx
=
E \frac{ e^{-\kappa (\epsilon)} – 1}{-\kappa}
-E \frac{ 1 – e^{\kappa (-\epsilon)}}{\kappa}
\rightarrow
0.
\end{equation}

That leaves
\begin{equation}\label{eqn:diracPotential:200}
\begin{aligned}
V_0 \int_{-\epsilon}^\epsilon \delta(x) \psi(x) dx
&=
\frac{-\Hbar^2}{2m} \int_{-\epsilon}^\epsilon \frac{d^2 \psi}{dx^2} dx \\
&=
\frac{-\Hbar^2}{2m} \evalrange{\frac{d \psi}{dx}}{-\epsilon}{\epsilon} \\
&=
\frac{-\Hbar^2}{2m}
\psi(0)
\lr
{
-\kappa e^{-\kappa (\epsilon)}
–
\kappa e^{\kappa (-\epsilon)}
}.
\end{aligned}
\end{equation}

In the \( \epsilon \rightarrow 0 \) limit this gives

\begin{equation}\label{eqn:diracPotential:220}
V_0 = \frac{\Hbar^2 \kappa}{m}.
\end{equation}

Equating relations for \( \kappa \) we have

\begin{equation}\label{eqn:diracPotential:240}
\kappa = \frac{m V_0}{\Hbar^2} = \frac{\sqrt{-2 m E}}{\Hbar},
\end{equation}

or

\begin{equation}\label{eqn:diracPotential:260}
E = -\inv{2 m} \lr{ \frac{m V_0}{\Hbar} }^2,
\end{equation}

with

\begin{equation}\label{eqn:diracPotential:280}
\psi(x, t < 0) = C \exp\lr{ -i E t/\hbar – \kappa \Abs{x}}.
\end{equation}

The normalization requires

\begin{equation}\label{eqn:diracPotential:300}
1
= 2 \Abs{C}^2 \int_0^\infty e^{- 2 \kappa x} dx
= 2 \Abs{C}^2 \evalrange{\frac{e^{- 2 \kappa x}}{-2 \kappa}}{0}{\infty}
= \frac{\Abs{C}^2}{\kappa},
\end{equation}

so
\begin{equation}\label{eqn:diracPotential:320}
\boxed{
\psi(x, t < 0) = \inv{\sqrt{\kappa}} \exp\lr{ -i E t/\hbar – \kappa \Abs{x}}. } \end{equation} There is only one bound state for such a potential. After turning off the potential, any plane wave \begin{equation}\label{eqn:diracPotential:360} \psi(x, t) = e^{i k x – i E(k) t/\Hbar}, \end{equation} where \begin{equation}\label{eqn:diracPotential:380} k = \frac{\sqrt{2 m E}}{\Hbar}, \end{equation} is a solution. In particular, at \( t = 0 \), the wave packet \begin{equation}\label{eqn:diracPotential:400} \psi(x,0) = \inv{\sqrt{2\pi}} \int_{-\infty}^\infty e^{i k x} A(k) dk, \end{equation} is a solution. To solve for \( A(k) \), we require \begin{equation}\label{eqn:diracPotential:420} \inv{\sqrt{2\pi}} \int_{-\infty}^\infty e^{i k x} A(k) dk = \inv{\sqrt{\kappa}} e^{ – \kappa \Abs{x} }, \end{equation} or \begin{equation}\label{eqn:diracPotential:440} \boxed{ A(k) = \inv{\sqrt{2\pi \kappa}} \int_{-\infty}^\infty e^{-i k x} e^{ – m V_0 \Abs{x}/\Hbar^2 } dx. } \end{equation} The initial time state established by the delta function potential evolves as \begin{equation}\label{eqn:diracPotential:480} \boxed{ \psi(x, t > 0) = \inv{\sqrt{2\pi}} \int_{-\infty}^\infty e^{i k x – i \Hbar k^2 t/2m} A(k) dk.
}
\end{equation}

References

[1] Jun John Sakurai and Jim J Napolitano. Modern quantum mechanics. Pearson Higher Ed, 2014.

Gauge transformed probability current

September 17, 2015 math and physics play , , , ,

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Question: Gauge transformed probability current ([1] pr. 2.37 (b))

For the gauge transformed Schrodinger equation

\begin{equation}\label{eqn:gaugeTxCurrent:20}
\inv{2m} \BPi(\Bx) \cdot \BPi(\Bx) \psi(\Bx, t) + e \phi(\Bx) \psi(\Bx, t) = i \Hbar \PD{t}{}\psi(\Bx, t),
\end{equation}

where

\begin{equation}\label{eqn:gaugeTxCurrent:40}
\BPi(\Bx) = -i \Hbar \spacegrad – \frac{e}{c} \BA(\Bx),
\end{equation}

find the probability current defined by

\begin{equation}\label{eqn:gaugeTxCurrent:60}
\PD{t}{\psi} + \spacegrad \cdot \Bj.
\end{equation}

Answer

Equation \ref{eqn:gaugeTxCurrent:20} and its conjugate are

\begin{equation}\label{eqn:gaugeTxCurrent:22}
\begin{aligned}
\inv{2m} \BPi \cdot \BPi \psi + e \phi \psi &= i \Hbar \PD{t}{\psi} \\
\inv{2m} \BPi^\conj \cdot \BPi^\conj \psi^\conj + e \phi \psi^\conj &= -i \Hbar \PD{t}{\psi^\conj}
\end{aligned}
\end{equation}

which can be used immediately in a chain rule expansion of the probability time derivative

\begin{equation}\label{eqn:gaugeTxCurrent:80}
\begin{aligned}
i \Hbar \PD{t}{\rho}
&=
i \Hbar \psi^\conj \PD{t}{\psi} +
i \Hbar \psi \PD{t}{\psi^\conj} \\
&=
\psi^\conj \lr{ \inv{2m} \BPi \cdot \BPi \psi + e \phi \psi } –
\psi \lr{ \inv{2m} \BPi^\conj \cdot \BPi^\conj \psi^\conj + e \phi \psi^\conj
} \\
&=
\inv{2m} \lr{
\psi^\conj \BPi \cdot \BPi \psi
-\psi \BPi^\conj \cdot \BPi^\conj \psi^\conj
}.
\end{aligned}
\end{equation}

We have a difference of conjugates, so can get away with expanding just the first term

\begin{equation}\label{eqn:gaugeTxCurrent:100}
\begin{aligned}
\psi^\conj \BPi \cdot \BPi \psi
&=
\psi^\conj
\psi \\
&=
\psi^\conj
\lr{ -i \Hbar \spacegrad – \frac{e}{c} \BA } \cdot \lr{ -i \Hbar \spacegrad – \frac{e}{c} \BA }
\psi \\
&=
\psi^\conj
\lr{
-\Hbar^2 \spacegrad^2 + \frac{i \Hbar e}{c} \lr{ \BA \cdot \spacegrad + \spacegrad \cdot \BA }
+ \frac{e^2}{c^2} \BA^2
}
\psi.
\end{aligned}
\end{equation}

Note that in the directional derivative terms, the gradient operates on everything to its right, including \( \BA \). Also note that the last term has no imaginary component, so it will not contribute to the difference of conjugates.

This gives

\begin{equation}\label{eqn:gaugeTxCurrent:120}
\begin{aligned}
\psi^\conj \BPi \cdot \BPi \psi – \psi \BPi^\conj \cdot \BPi^\conj \psi^\conj
&=
\psi^\conj
\lr{
-\Hbar^2 \spacegrad^2 \psi + \frac{i \Hbar e}{c} \lr{ \BA \cdot \spacegrad \psi + \spacegrad \cdot (\BA \psi) }
} \\
&\quad –
\psi
\lr{
-\Hbar^2 \spacegrad^2 \psi^\conj – \frac{i \Hbar e}{c} \lr{ \BA \cdot \spacegrad \psi^\conj + \spacegrad \cdot (\BA \psi^\conj) }
} \\
&=
-\Hbar^2 \lr{ \psi^\conj \spacegrad^2 \psi – \psi \spacegrad^2 \psi^\conj } \\
&\quad +
\frac{i \Hbar e}{c}
\lr{
\psi^\conj
\BA \cdot \spacegrad \psi + \psi^\conj \spacegrad \cdot (\BA \psi)
+
\psi
\BA \cdot \spacegrad \psi^\conj + \psi \spacegrad \cdot (\BA \psi^\conj)
}
\end{aligned}
\end{equation}

The first term is recognized as a divergence

\begin{equation}\label{eqn:gaugeTxCurrent:140}
\begin{aligned}
\spacegrad \cdot \lr{ \psi^\conj \spacegrad \psi – \psi \spacegrad \psi^\conj }
&=
\psi^\conj \spacegrad \cdot \spacegrad \psi
+
\spacegrad \psi \cdot \spacegrad \psi^\conj
–
\psi \spacegrad \cdot \spacegrad \psi^\conj
–
\spacegrad \psi^\conj \cdot \spacegrad \psi \\
&= \psi^\conj \spacegrad^2 \psi – \psi \spacegrad^2 \psi^\conj.
\end{aligned}
\end{equation}

The second term can also be factored into a divergence operation

\begin{equation}\label{eqn:gaugeTxCurrent:160}
\begin{aligned}
\psi^\conj
\BA \cdot \spacegrad \psi &+ \psi^\conj \spacegrad \cdot (\BA \psi)
+
\psi
\BA \cdot \spacegrad \psi^\conj + \psi \spacegrad \cdot (\BA \psi^\conj) \\
&=
\lr{ \psi^\conj\BA \cdot \spacegrad \psi
+\psi \spacegrad \cdot (\BA \psi^\conj)
}
+\lr{
\psi \BA \cdot \spacegrad \psi^\conj
+\psi^\conj \spacegrad \cdot (\BA \psi)
} \\
&= 2 \spacegrad \cdot \lr{ \BA \psi \psi^\conj } \\
\end{aligned}
\end{equation}

Putting all the pieces back together we have

\begin{equation}\label{eqn:gaugeTxCurrent:180}
\begin{aligned}
\PD{t}{\rho}
&=
\inv{2m i \Hbar} \lr{
\psi^\conj \BPi \cdot \BPi \psi
-\psi \BPi^\conj \cdot \BPi^\conj \psi^\conj
} \\
&=
\spacegrad \cdot
\inv{2m i \Hbar} \lr{
-\Hbar^2
\lr{ \psi^\conj \spacegrad \psi – \psi \spacegrad \psi^\conj }
+ \frac{ i \Hbar e}{c} 2 \BA \psi \psi^\conj
} \\
&=
\spacegrad \cdot
\lr{
\frac{i \Hbar}{2 m} \lr{ \psi^\conj \spacegrad \psi – \psi \spacegrad \psi^\conj }
+ \frac{e}{m c} \BA \psi \psi^\conj
}.
\end{aligned}
\end{equation}

From \ref{eqn:gaugeTxCurrent:60}, the probability current must be

\begin{equation}\label{eqn:gaugeTxCurrent:200}
\Bj
=
\frac{\Hbar}{2 i m} \lr{ \psi^\conj \spacegrad \psi – \psi \spacegrad \psi^\conj }
– \frac{e}{m c} \BA \psi \psi^\conj,
\end{equation}

or
\begin{equation}\label{eqn:gaugeTxCurrent:220}
\boxed{
\Bj
=
\frac{\Hbar}{m} \textrm{Im} \lr{ \psi^\conj \spacegrad \psi }
– \frac{e}{m c} \BA \psi \psi^\conj.
}
\end{equation}

References

[1] Jun John Sakurai and Jim J Napolitano. Modern quantum mechanics. Pearson Higher Ed, 2014.

Momentum space representation of Schrodinger equation

September 2, 2015 phy1520 , , , , ,

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Question: momentum space representation of Schrodinger equation ([1] pr. 2.15)

Using

\begin{equation}\label{eqn:shoMomentumSpace:20}
\braket{x’}{p’} = \inv{\sqrt{2 \pi \Hbar}} e^{i p’ x’/\Hbar},
\end{equation}

show that

\begin{equation}\label{eqn:shoMomentumSpace:40}
\bra{p’} x \ket{\alpha} = i \Hbar \PD{p’}{} \braket{p’}{\alpha}.
\end{equation}

Use this to find the momentum space representation of the Schrodinger equation for the one dimensional SHO and the energy eigenfunctions in their momentum representation.

Answer

To expand the matrix element, introduce both momentum and position space identity operators

\begin{equation}\label{eqn:shoMomentumSpace:60}
\begin{aligned}
\bra{p’} x \ket{\alpha}
&=
\int dx’ dp” \braket{p’}{x’}\bra{x’}x \ket{p”}\braket{p”}{\alpha} \\
&=
\int dx’ dp” \braket{p’}{x’}x’\braket{x’}{p”}\braket{p”}{\alpha} \\
&=
\inv{2 \pi \Hbar}
\int dx’ dp” e^{-i p’ x’/\Hbar} x’ e^{i p” x’/\Hbar} \braket{p”}{\alpha} \\
&=
\inv{2 \pi \Hbar}
\int dx’ dp” x’ e^{i (p” – p’) x’/\Hbar} \braket{p”}{\alpha} \\
&=
\inv{2 \pi \Hbar}
\int dx’ dp” \frac{d}{dp”}\lr{ \frac{-i \Hbar} e^{i (p” – p’) x’/\Hbar} }
\braket{p”}{\alpha} \\
&=
i \Hbar
\int dp”
\lr{ \inv{2 \pi \Hbar}
\int dx’ e^{i (p” – p’) x’/\Hbar} } \frac{d}{dp”} \braket{p”}{\alpha} \\
&=
i \Hbar
\int dp” \delta(p”- p’)
\frac{d}{dp”} \braket{p”}{\alpha} \\
&=
i \Hbar
\frac{d}{dp’} \braket{p’}{\alpha}.
\end{aligned}
\end{equation}

Schrodinger’s equation for a time dependent state \( \ket{\alpha} = U(t) \ket{\alpha,0} \) is

\begin{equation}\label{eqn:shoMomentumSpace:80}
i \Hbar \PD{t}{} \ket{\alpha} = H \ket{\alpha},
\end{equation}

with the momentum representation

\begin{equation}\label{eqn:shoMomentumSpace:100}
i \Hbar \PD{t}{} \braket{p’}{\alpha} = \bra{p’} H \ket{\alpha}.
\end{equation}

Expansion of the Hamiltonian matrix element for a strictly spatial dependent potential \( V(x) \) gives

\begin{equation}\label{eqn:shoMomentumSpace:120}
\begin{aligned}
\bra{p’} H \ket{\alpha}
&=
\bra{p’} \lr{\frac{p^2}{2m} + V(x) } \ket{\alpha} \\
&=
\frac{(p’)^2}{2m}
+ \bra{p’} V(x) \ket{\alpha}.
\end{aligned}
\end{equation}

Assuming a Taylor representation of the potential \( V(x) = \sum c_k x^k \), we want to calculate

\begin{equation}\label{eqn:shoMomentumSpace:140}
\bra{p’} V(x) \ket{\alpha}
= \sum c_k \bra{p’} x^k \ket{\alpha}.
\end{equation}

With \( \ket{\alpha} = \ket{p”} \) \ref{eqn:shoMomentumSpace:40} provides the \( k = 1 \) term

\begin{equation}\label{eqn:shoMomentumSpace:160}
\begin{aligned}
\bra{p’} x \ket{p”}
&= i \Hbar \frac{d}{dp’} \braket{p’}{p”} \\
&= i \Hbar \frac{d}{dp’} \delta(p’ – p”),
\end{aligned}
\end{equation}

where it is implied here that the derivative is operating on not just the delta function, but on all else that follows.

Using this the higher powers of \( \bra{p’} x^k \ket{\alpha} \) can be found easily. For example for \( k = 2 \) we have

\begin{equation}\label{eqn:shoMomentumSpace:180}
\begin{aligned}
\bra{p’} x^2 \ket{\alpha}
&=
\int dp”
\bra{p’} x \ket{p”}\bra{p”} x \ket{\alpha} \\
&=
\int dp”
i \Hbar
\frac{d}{dp’} \delta(p’ – p”) i \Hbar \frac{d}{dp”} \braket{p”}{\alpha} \\
&=
\lr{ i \Hbar }^2 \frac{d^2}{d(p’)^2} \braket{p’}{\alpha}.
\end{aligned}
\end{equation}

This means that the potential matrix element is

\begin{equation}\label{eqn:shoMomentumSpace:200}
\begin{aligned}
\bra{p’} V(x) \ket{\alpha}
&=
\sum c_k \lr{ i \Hbar \frac{d}{dp’} }^k \braket{p’}{\alpha} \\
&= V\lr{ i \Hbar \frac{d}{dp’} }.
\end{aligned}
\end{equation}

Writing \( \Psi_\alpha(p’) = \braket{p’}{\alpha} \), the momentum space representation of Schrodinger’s equation for a position dependent potential is

\begin{equation}\label{eqn:shoMomentumSpace:220}
\boxed{
i \Hbar \PD{t}{} \Psi_\alpha(p’)
=
\lr{ \frac{(p’)^2}{2m} + V\lr{ i \Hbar \PDi{p’}{} } } \Psi_\alpha(p’).
}
\end{equation}

For the SHO Hamiltonian the potential is \( V(x) = (1/2) m \omega^2 x^2 \), so the Schrodinger equation is

\begin{equation}\label{eqn:shoMomentumSpace:240}
\begin{aligned}
i \Hbar \PD{t}{} \Psi_\alpha(p’)
&=
\lr{ \frac{(p’)^2}{2m} – \inv{2} m \omega^2 \Hbar^2
\frac{\partial^2}{\partial(p’)^2} } \Psi_\alpha(p’) \\
&=
\inv{2 m} \lr{ (p’)^2 – m^2 \omega^2 \Hbar^2 \frac{\partial^2}{\partial(p’)^2} } \Psi_\alpha(p’).
\end{aligned}
\end{equation}

To determine the wave functions, let’s non-dimensionalize this and compare to the position space Schrodinger equation. Let

\begin{equation}\label{eqn:shoMomentumSpace:260}
p_0^2 = m \omega \hbar,
\end{equation}

so
\begin{equation}\label{eqn:shoMomentumSpace:280}
\begin{aligned}
i \Hbar \PD{t}{} \Psi_\alpha(p’)
&=
\frac{p_0^2}{2 m} \lr{ \lr{\frac{p’}{p_0}}^2 –
\frac{\partial^2}{\partial(p’/p_0)^2} } \Psi_\alpha(p’) \\
&=
\frac{\omega \Hbar}{2}\lr{
– \frac{\partial^2}{\partial(p’/p_0)^2} +
\lr{\frac{p’}{p_0}}^2
} \Psi_\alpha(p’).
\end{aligned}
\end{equation}

Compare this to the position space equation with \( x_0^2 = m \omega/\Hbar \),

\begin{equation}\label{eqn:shoMomentumSpace:300}
\begin{aligned}
i \Hbar \PD{t}{} \Psi_\alpha(x’)
&=
\lr{ -\frac{\Hbar^2}{2m} \frac{\partial^2}{\partial(x’)^2}
+
\inv{2} m \omega^2 (x’)^2 }
\Psi_\alpha(x’) \\
&=
\frac{\Hbar^2}{2m}
\lr{ -\frac{\partial^2}{\partial(x’)^2}
+
\frac{m^2 \omega^2}{\Hbar^2} (x’)^2 }
\Psi_\alpha(x’) \\
&=
\frac{\Hbar^2 x_0^2}{2m}
\lr{
-\frac{\partial^2}{\partial(x’/x_0)^2}
+
\lr{\frac{x’}{x_0}}^2
}
\Psi_\alpha(x’) \\
&=
\frac{\Hbar \omega}{2}
\lr{
-\frac{\partial^2}{\partial(x’/x_0)^2}
+
\lr{\frac{x’}{x_0}}^2
}
\Psi_\alpha(x’).
\end{aligned}
\end{equation}

It’s clear that there is a straightforward duality relationship between the respective wave functions. Since

\begin{equation}\label{eqn:shoMomentumSpace:320}
\braket{x’}{n} =
\inv{\pi^{1/4} \sqrt{2^n n!} x_0^{n + 1/2}} \lr{ x’ – x_0^2 \frac{d}{dx’} }^n \exp\lr{ -\inv{2} \lr{\frac{x’}{x_0}}^2 },
\end{equation}

the momentum space wave functions are

\begin{equation}\label{eqn:shoMomentumSpace:340}
\braket{p’}{n} =
\inv{\pi^{1/4} \sqrt{2^n n!} p_0^{n + 1/2}} \lr{ p’ – p_0^2 \frac{d}{dp’} }^n \exp\lr{ -\inv{2} \lr{\frac{p’}{p_0}}^2 }.
\end{equation}

References

[1] Jun John Sakurai and Jim J Napolitano. Modern quantum mechanics. Pearson Higher Ed, 2014.