momentum space

Position operator in momentum space representation

November 8, 2015 phy1520 No comments , ,

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A derivation of the position space representation of the momentum operator \( -i \Hbar \partial_x \) is made in [1], starting with the position-momentum commutator. Here I’ll repeat that argument for the momentum space representation of the position operator.

What we want to do is expand the matrix element of the commutator. First using the definition of the commutator

\begin{equation}\label{eqn:positionOperatorInMomentumSpace:20}
\bra{p’} X P – P X \ket{p”}
=
i \Hbar \braket{p’}{p”}
=
i \Hbar \delta{p’ – p”},
\end{equation}

and then by inserting an identity operation in a momentum space basis

\begin{equation}\label{eqn:positionOperatorInMomentumSpace:40}
\begin{aligned}
\bra{p’} X P – P X \ket{p”}
&=
\int dp
\bra{p’} X \ket{p}\bra{p} P \ket{p”}
-\int dp
\bra{p’} P \ket{p}\bra{p} X \ket{p”} \\
&=
\int dp
\bra{p’} X \ket{p} p \delta(p – p”)
-\int dp
p \delta(p’ – p)
\bra{p} X \ket{p”} \\
&=
\bra{p’} X \ket{p”} p”

p’ \bra{p’} X \ket{p”}.
\end{aligned}
\end{equation}

So we have

\begin{equation}\label{eqn:positionOperatorInMomentumSpace:60}
\bra{p’} X \ket{p”} p”

p’ \bra{p’} X \ket{p”}
=
i \Hbar \delta{p’ – p”}.
\end{equation}

Because the RHS is zero whenever \( p’ \ne p” \), the matrix element \( \bra{p’} X \ket{p”} \) must also include a delta function. Let

\begin{equation}\label{eqn:positionOperatorInMomentumSpace:80}
\bra{p’} X \ket{p”} = \delta(p’ – p”) X(p”).
\end{equation}

Because \ref{eqn:positionOperatorInMomentumSpace:60} is an operator equation that really only takes on meaning when applied to a wave function and integrated, we do that

\begin{equation}\label{eqn:positionOperatorInMomentumSpace:100}
\int dp” \delta(p’ – p”) X(p”) p” \psi(p”)

\int dp” p’ \delta(p’ – p”) X(p”) \psi(p”)
=
\int dp” i \Hbar \delta{p’ – p”} \psi(p”),
\end{equation}

or
\begin{equation}\label{eqn:positionOperatorInMomentumSpace:120}
i \Hbar \psi(p’)
=
X(p’) p’ \psi(p’)

p’
X(p’) \psi(p’).
\end{equation}

Provided \( X(p’) \) operates on everything to its right, this equation is solved by setting

\begin{equation}\label{eqn:positionOperatorInMomentumSpace:140}
\boxed{
X(p’) = i \Hbar \PD{p’}{}.
}
\end{equation}

References

[1] BR Desai. Quantum mechanics with basic field theory. Cambridge University Press, 2009.

Free particle propagator

September 7, 2015 phy1520 6 comments , , , , , ,

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Question: Free particle propagator ([1] pr. 2.31)

Derive the free particle propagator in one and three dimensions.

Answer

I found the description in the text confusing, so let’s start from scratch with the definition of the propagator. This is the kernel of the spatial convolution integral that encodes time evolution, and can be expressed by expanding a general time state with two sets of identity operators. Let the position relative state at time \( t \), relative to an initial time \( t_0 \) be given by \( \braket{\Bx}{\alpha, t ; t_0 } \), and expand this in terms of a complete basis of energy eigenstates \( | a’ > \) and the time evolution operator

\begin{equation}\label{eqn:freeParticlePropagator:20}
\begin{aligned}
\braket{\Bx”}{\alpha, t ; t_0 }
&= \bra{\Bx”} U \ket{\alpha, t_0 } \\
&= \bra{\Bx”} e^{-i H (t -t_0)/\Hbar} \ket{\alpha, t_0 } \\
&= \bra{\Bx”} e^{-i H (t -t_0)/\Hbar} \lr{ \sum_{a’} \ket{a’} \bra{a’ }} \ket{\alpha, t_0 } \\
&= \bra{\Bx”} \sum_{a’} e^{-i E_{a’} (t -t_0)/\Hbar} \ket{a’} \braket{a’ }{\alpha, t_0 } \\
&=
\bra{\Bx”} \sum_{a’} e^{-i E_{a’} (t -t_0)/\Hbar} \ket{a’} \bra{a’ }
\lr{ \int d^3 \Bx’
\ket{\Bx’}\bra{\Bx’}
}
\ket{\alpha, t_0 } \\
&=
\int d^3 \Bx’
\lr{
\bra{\Bx”} \sum_{a’} e^{-i E_{a’} (t -t_0)/\Hbar} \ket{a’} \braket{a’ }{\Bx’}
}
\braket{\Bx’}{\alpha, t_0 } \\
&=
\int d^3 \Bx’ K(\Bx”, t ; \Bx’, t_0) \braket{\Bx’}{\alpha, t_0 },
\end{aligned}
\end{equation}

where

\begin{equation}\label{eqn:freeParticlePropagator:40}
K(\Bx”, t ; \Bx’, t_0) =
\sum_{a’}
\braket{\Bx”}{a’}\braket{a’ }{\Bx’}
e^{-i E_{a’} (t -t_0)/\Hbar},
\end{equation}

the propagator, is the kernel of the convolution integral that takes the state \( \ket{\alpha, t_0} \) to state \( \ket{\alpha, t ; t_0} \). Evaluating this over the momentum states (where integration and not plain summation is required), we have

\begin{equation}\label{eqn:freeParticlePropagator:60}
\begin{aligned}
K(\Bx”, t ; \Bx’, t_0)
&=
\int d^3 \Bp’
\braket{\Bx”}{\Bp’}\braket{\Bp’ }{\Bx’}
e^{-i E_{\Bp’} (t -t_0)/\Hbar} \\
&=
\int d^3 \Bp’
\braket{\Bx”}{\Bp’}\braket{\Bp’ }{\Bx’}
\exp\lr{-i \frac{(\Bp’)^2 (t -t_0)}{2 m \Hbar}} \\
&=
\int d^3 \Bp’
\frac{e^{i \Bx” \cdot \Bp’/\Hbar}}{(\sqrt{2 \pi \Hbar})^3}
\frac{e^{-i \Bx’ \cdot \Bp’/\Hbar}}{(\sqrt{2 \pi \Hbar})^3}
\exp\lr{-i \frac{(\Bp’)^2 (t -t_0)}{2 m \Hbar}} \\
&=
\inv{(2 \pi \Hbar)^3}
\int d^3 \Bp’
e^{i (\Bx” -\Bx’) \cdot \Bp’/\Hbar}
\exp\lr{-i \frac{(\Bp’)^2 (t -t_0)}{2 m \Hbar}} \\
&=
\inv{ 2 \pi \Hbar }
\int_{-\infty}^\infty dp_1′
e^{i (x_1” -x_1′) p_1’/\Hbar}
\exp\lr{-i \frac{(p_1′)^2 (t -t_0)}{2 m \Hbar}} \times \\
&\quad \inv{ 2 \pi \Hbar }
\int_{-\infty}^\infty dp_2′
e^{i (x_2” -x_2′) p_2’/\Hbar}
\exp\lr{-i \frac{(p_2′)^2 (t -t_0)}{2 m \Hbar}} \times \\
&\quad \inv{ 2 \pi \Hbar }
\int_{-\infty}^\infty dp_3′
e^{i (x_3” -x_3′) p_3’/\Hbar}
\exp\lr{-i \frac{(p_3′)^2 (t -t_0)}{2 m \Hbar}}
\end{aligned}
\end{equation}

With \( a = \ifrac{(t -t_0)}{2 m \Hbar} \), each of these three integral factors is of the form

\begin{equation}\label{eqn:freeParticlePropagator:80}
\begin{aligned}
\inv{ 2 \pi \Hbar }
\int_{-\infty}^\infty dp
e^{i \Delta x p/\Hbar }
\exp\lr{-i a p^2}
&=
\inv{2 \pi \Hbar \sqrt{a}}
\int_{-\infty}^\infty du
e^{i \Delta x u/(\sqrt{a}\Hbar) }
\exp\lr{-i u^2} \\
&=
\inv{2 \pi \Hbar \sqrt{a}}
\int_{-\infty}^\infty du
e^{i \Delta x u/(\sqrt{a} \Hbar) }
\exp\lr{-i (u – \Delta x /(2\sqrt{a}\Hbar))^2 + i(\Delta x/(2\sqrt{a}\Hbar))^2} \\
&=
\inv{2 \pi \Hbar \sqrt{a}}
\exp\lr{ \frac{i(\Delta x)^2 2 m \Hbar}{4 (t -t_0) \Hbar^2} }
\int_{-\infty}^\infty dz
e^{-i z^2} \\
&= \sqrt{ \frac{ -i \pi 2 m \Hbar}{ 4 \pi^2 \Hbar^2 (t -t_0)} }
\exp\lr{ \frac{i(\Delta x)^2 m}{2 (t -t_0) \Hbar} } \\
&= \sqrt{ \frac{ m }{ 2 \pi i \Hbar (t -t_0)} }
\exp\lr{ \frac{i(\Delta x)^2 m}{2 (t -t_0) \Hbar} }.
\end{aligned}
\end{equation}

Note that the integral above has value \( \sqrt{-i\pi} \) which can be found by integrating over the contour of fig. 1, letting \( R \rightarrow \infty \).

contourFig1

fig. 1. Integration contour for \( \int e^{-i z^2} \)

Multiplying out each of the spatial direction factors gives the propagator in its closed form
\begin{equation}\label{eqn:freeParticlePropagator:120}
\boxed{
K(\Bx”, t ; \Bx’, t_0)
= \lr{ \sqrt{ \frac{ m }{ 2 \pi i \Hbar (t -t_0)} } }^3
\exp\lr{ \frac{i(\Bx” – \Bx’)^2 m}{2 (t -t_0) \Hbar} }.
}
\end{equation}

In one or two dimensions the exponential power \( 3 \) need only be adjusted appropriately.

Question: Momentum space free particle propagator ([1] pr. 2.33)

Derive the free particle propagator in momentum space.

Answer

The momentum space propagator follows in the same fashion as the spatial propagator

\begin{equation}\label{eqn:freeParticlePropagator:140}
\begin{aligned}
\braket{\Bp”}{\alpha, t ; t_0 }
&= \bra{\Bp”} U \ket{\alpha, t_0 } \\
&= \bra{\Bp”} e^{-i H (t -t_0)/\Hbar} \ket{\alpha, t_0 } \\
&= \bra{\Bp”} e^{-i H (t -t_0)/\Hbar} \lr{ \sum_{a’} \ket{a’} \bra{a’ }} \ket{\alpha, t_0 } \\
&= \bra{\Bp”} \sum_{a’} e^{-i E_{a’} (t -t_0)/\Hbar} \ket{a’} \braket{a’ }{\alpha, t_0 } \\
&=
\bra{\Bp”} \sum_{a’} e^{-i E_{a’} (t -t_0)/\Hbar} \ket{a’} \bra{a’ }
\lr{ \int d^3 \Bp’
\ket{\Bp’}\bra{\Bp’}
}
\ket{\alpha, t_0 } \\
&=
\int d^3 \Bp’
\lr{
\bra{\Bp”} \sum_{a’} e^{-i E_{a’} (t -t_0)/\Hbar} \ket{a’} \braket{a’ }{\Bp’}
}
\braket{\Bp’}{\alpha, t_0 } \\
&=
\int d^3 \Bp’ K(\Bp”, t ; \Bp’, t_0) \braket{\Bp’}{\alpha, t_0 },
\end{aligned}
\end{equation}

so

\begin{equation}\label{eqn:freeParticlePropagator:160}
K(\Bp”, t ; \Bp’, t_0)
=
\sum_{a’}
\braket{\Bp”}{a’}
\braket{a’ }{\Bp’}
e^{-i E_{a’} (t -t_0)/\Hbar}.
\end{equation}

For the free particle Hamiltonian, this can be evaluated over a momentum space basis

\begin{equation}\label{eqn:freeParticlePropagator:170}
\begin{aligned}
K(\Bp”, t ; \Bp’, t_0)
&=
\int d^3 \Bp”’
\braket{\Bp”}{\Bp”’}
\braket{\Bp”’ }{\Bp’}
e^{-i E_{\Bp”’} (t -t_0)/\Hbar} \\
&=
\int d^3 \Bp”’
\braket{\Bp”}{\Bp”’}
\delta(\Bp”’ – \Bp’)
\exp\lr{ -i \frac{(\Bp”’)^2 (t -t_0)}{2 m \Hbar}} \\
&=
\braket{\Bp”}{\Bp’}
\exp\lr{ -i \frac{(\Bp’)^2 (t -t_0)}{2 m \Hbar}}
\end{aligned}
\end{equation}

or

\begin{equation}\label{eqn:freeParticlePropagator:200}
\boxed{
K(\Bp”, t ; \Bp’, t_0)
=
\delta( \Bp” – \Bp’ )
\exp\lr{ -i \frac{(\Bp’)^2 (t -t_0)}{2 m \Hbar}}.
}
\end{equation}

This is what we expect since the time evolution is given by just this exponential factor

\begin{equation}\label{eqn:freeParticlePropagator:220}
\begin{aligned}
\braket{\Bp’}{\alpha, t_0 ; t}
&= \bra{\Bp’} \exp\lr{ -i \frac{(\Bp’)^2 (t -t_0)}{2 m \Hbar}} \ket{\alpha, t_0} \\
&=
\exp\lr{ -i \frac{(\Bp’)^2 (t -t_0)}{2 m \Hbar}}
\braket{\Bp’}
{\alpha, t_0}.
\end{aligned}
\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 No comments , , , , ,

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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.