Heisenberg Uncertainty and the Wavepacket¶
These notes were made in preparation for a guest lecture/discussion I gave for my brother's grade 12 physics class. However, as I spoke with the class, it became clear that perhaps I'd been too ambitious in my goals for that day. Nevertheless, these notes may still serve as a helpful resource for those who have heard about some aspects of quantum mechanics before, and are looking for a deeper understanding of the Uncertainty Principle.
Pre-requisites¶
De Broglie Wavelength¶
We'll assume later on that the reader is familiar with the idea of the De Broglie wavlength, namely that the momentum and wavelength of a quantum particle are related by
$$ \lambda_p = \frac hp \,, $$
in a way similar to the Planck energy-frequency relation $E_\nu=h\nu$.
Euler's Formula¶
One essential ingredient for understanding the wavefunction of a free particle is a complex representation of the trigonometric functions, namely the expression
$$ e^{i\theta} = \cos\theta + i\,\sin\theta \,. $$
Bell Curve, or Gaussian Distribution¶
Another ingredient is the Gaussian distribution, or the "Bell curve":
$$ P(x) = N \, e^{-\left(\frac{x-y}t\right)^2} \,. $$
The variable $t$ is known as the standard deviation of the distribution, and $y$ is the mean value, as shown in the figure below:
import matplotlib.pyplot as plt
import numpy as np
y = 50
t = 10
def gaussian(x):
return 1/t/np.sqrt(np.pi) * np.exp(-(x - y)**2/t**2)
x = np.linspace(0, 100, 1000)
P = gaussian(x)
plt.plot(x, P)
plt.xlabel("Position (nm)")
plt.ylabel("P(x)")
plt.vlines(x = 50, ymin=0, ymax=gaussian(y), color = 'black', linestyle="dashed")
plt.annotate(f"y = {50}", xy=(55, 0.05))
plt.hlines(y=gaussian(60), xmin=y, xmax=y+t, color='r', linestyles="dashed")
plt.annotate(f"t = {t}", xy=(63, 0.02), color="red")
plt.ylim(bottom=0)
plt.xlim(left=0, right=100)
plt.show()
For the purposes of what we're doing here, it will be sufficient to know that the area under a Bell curve with $t=1/\sqrt{\pi}$ is exactly 1:
$$ \int_{-\infty}^\infty\!\textrm{d}u\,e^{-\pi u^2} = 1 \,. $$
(To show this, it's easy to instead consider the square of this expression, noticing that the integral
$$ \int_{-\infty}^\infty\int_{-\infty}^\infty\!\textrm{d}u\textrm{d}v\,e^{-\pi(u^2+v^2)} $$
can be thought of as the exponential sum of the areas $A$ of concentric circles with radius $r = \sqrt{u^2+v^2}$. This means that the integral above can be rewritten as
$$ \int_0^\infty\!\textrm{d}A\,e^{-A} = \left(-e^{-A}\right)\Big|_{0}^\infty = 0 - (-1) = 1\,. $$
Taking the square-root, we recover the expression above.)
Now, more generally, if we encounter an integral of the form
$$ I_0(A,B,C) = \int_{-\infty}^\infty\!\textrm{d}x\,e^{-(Ax^2+Bx+C)} $$
for some constants $A$, $B$, and $C$, we can complete the square in the argument of the exponent to determine the result in general:
$$ \begin{align*} I_0(A,B,C) &= \int_{-\infty}^\infty\!\textrm{d}x\,e^{-A\left(x+\frac B{2A}\right)^2 + \frac{B^2}{4A} -C} \\ &= e^{\frac{B^2}{4A} -C}\int_{-\infty}^\infty\!\textrm{d}x'\,e^{-A\,x'^2} \qquad x' := x+\frac B{2A} \\ &= e^{\frac{B^2}{4A} -C} \sqrt{\frac\pi A} \int_{-\infty}^\infty\!\textrm{d}u\,e^{-\pi u^2} \qquad x' := \sqrt{\frac\pi A} \, u \\ &= e^{\frac{B^2}{4A} -C} \sqrt{\frac\pi A} \,. \end{align*} $$
Before continuing, we can lastly use a trick called "differentiating under the integral sign" to compute two other related integrals, $I_1(A,B,C)$ and $I_2(A,B,C)$, as follows:
$$ \begin{align*} I_1(A,B,C) &:= \int_{-\infty}^\infty\!\textrm{d}x\,x\,e^{-(Ax^2+Bx+C)} =-\frac{d}{dB} I_0(A,B,C) \\ & = -\frac B{2A}\, I_0(A,B,C) \\ I_2(A,B,C) &:= \int_{-\infty}^\infty\!\textrm{d}x\,x^2\,e^{-(Ax^2+Bx+C)} = -\frac{d}{dA} I_0(A,B,C)\\ & = \left(\frac{B^2}{4A^2} + \frac1{2A}\right) I_0(A,B,C) \\ \end{align*} $$
Bra-Ket Notation and Formalism¶
In quantum mechanics, it is helpful to consider the state of a quantum system to be independent of the basis in which we choose to decompose that state. For example, if we are discussing the position of a quantum state, it may be helpful to consider the projection of that state onto a state of definite position. However, if we're measuring momentum instead, such a projection may not be as helpful, and a basis of states with definite momentum would be more helpful.
Notationally, we'll describe the quantum state as $|\alpha\rangle$, and the basis states as either $\langle x |$ or $\langle p |$ in either position or momentum basis, respectively. The wavefunction in position space $\psi_\alpha(x)$ of the state $|\alpha\rangle$ is then given by the projection
$$ \psi_\alpha(x) = \langle x|\alpha\rangle \,. $$
In this formalism, the complex conjugate of the wavefunction is given by $\psi^*_\alpha(x) = \langle \alpha|x\rangle$, and the usual statement of probabilities being normalized to one (i.e. the state must appear in one of the positions $\langle x|$ above) can be written simply as
$$ \langle \alpha | \alpha \rangle = 1 \,. $$
This follows because the basis states $\langle x|$ satisfy a completeness relation
$$ \int_{-\infty}^\infty\!\textrm{d}x \, | x \rangle \langle x | = 1 $$
which implies that
$$ \begin{align*} \langle \alpha | \alpha \rangle &= \langle \alpha | \left(\int_{-\infty}^\infty\!\textrm{d}x \, | x \rangle \langle x |\right)| \alpha \rangle \\ &=\int_{-\infty}^\infty\!\textrm{d}x \, \langle \alpha|x\rangle \langle x|\alpha\rangle \\ &=\int_{-\infty}^\infty\!\textrm{d}x \,\psi^*_\alpha(x) \,\psi_\alpha(x) \\ &=\int_{-\infty}^\infty\!\textrm{d}x\, P_\alpha(x) = 1\,. \end{align*} $$
(Recall that the probability distribution $P_\alpha(x)$ is given by the complex square of the wavefunction.)
Lastly, if we wish to compute the expectation value of an observable (for example, expected position $\langle x \rangle_\alpha$ in state $\alpha$), we can express in bra-ket notation using a similar approach:
$$ \begin{align*} \langle x \rangle_\alpha & =\int_{-\infty}^\infty\!\textrm{d}x\, P_\alpha(x) \, x \\ &= \int_{-\infty}^\infty\!\textrm{d}x \,\psi^*_\alpha(x) \, x \,\psi_\alpha(x) \\ &=\int_{-\infty}^\infty\!\textrm{d}x \, \langle \alpha|x\rangle x \langle x|\alpha\rangle \\ &= \langle \alpha | x | \alpha \rangle \,. \end{align*} $$
Wavepackets¶
For our purposes here, the state $\alpha$ will consist of an oscillatory part (to describe the momentum of the state, as we'll see later) and Bell curve part, to represent the uncertainty in the position (and momentum) of the particle:
$$ \langle x | \alpha \rangle = \psi_\alpha(x) = \sqrt{N_x} \, e^{ik(x-y)}\,e^{-\frac12\left(\frac{x-y}t\right)^2} \,. $$
In the above, $N$ is a constant chosen such that the state $\alpha$ has unit probability:
$$ \begin{align*} 1 &= \langle \alpha | \alpha \rangle = \int_{-\infty}^\infty\!\textrm{d}x\,\psi^*_\alpha(x) \,\psi_\alpha(x) \\ &= N_x \int_{-\infty}^\infty\!\textrm{d}x\, e^{-\left(\frac{x-y}t\right)^2} = N_x \,I_0\left(\frac1{t^2}, - \frac{2y}{t^2}, \frac{y^2}{t^2}\right) \\ \implies N_x &= \frac1{t\sqrt{\pi}}\,. \end{align*} $$
Visually, the wavefunction has real and imaginary parts as shown below:
k = 2
Nx = 1/(t*np.sqrt(np.pi))
yR = np.sqrt(Nx) * np.cos(k*(x-y)) * np.exp(-(x-y)**2/t**2/2)
yI = np.sqrt(Nx) * np.sin(k*(x-y)) * np.exp(-(x-y)**2/t**2/2)
fig, axes = plt.subplots(2)
axes[0].plot(x, yR, color="blue")
axes[0].set(xlabel='Position (nm)', ylabel='ψᵣ(x)')
axes[1].plot(x, yI, color="orange")
axes[1].set(xlabel='Position (nm)', ylabel='ψᵢ(x)')
for ax in axes.flat:
ax.label_outer()
plt.show()
We eventually want to compute the RMS (root-mean-square) error for both the position and the momentum of the state $|\alpha\rangle$. In the former case, this is done by doing as the name suggests: starting with the error $(x-\langle x\rangle_\alpha)$, we square it, take the mean, and then finally the square-root:
$$ \Delta x_{\textrm{RMS}}^{(\alpha)} = \sqrt{\langle (x - \langle x \rangle_\alpha)^2\rangle_\alpha} = \sqrt{\langle x^2 \rangle_\alpha - \langle x \rangle_\alpha^2} \,. $$
To this end, let's calculate both $\langle x^2 \rangle_\alpha$ and $\langle x \rangle_\alpha$:
$$ \begin{align*} \langle x \rangle_\alpha &= \langle \alpha | x|\alpha\rangle \\ &=\int_{-\infty}^\infty\!\textrm{d}x \,\psi^*_\alpha(x) \, x \,\psi_\alpha(x) \\ &= N_x \int_{-\infty}^\infty \!\textrm{d}x \, e^{-\left(\frac{x-y}t\right)^2} \,x \\ &= N_x \,I_1\left(\frac1{t^2},-\frac{2y}{t^2},\frac{y^2}{t^2}\right) \\ % &= N_x \left[-\frac{B}{2A} I_0(A,B,C)\right]\Bigg|_{A=\frac1{t^2},\;B=-\frac{2y}{t^2},\;C=\frac{y^2}{t^2}} \\ &= y \, \end{align*} $$
This is somewhat expected, since the Bell curve is peaked at $x=y$. Similarly,
$$ \begin{align*} \langle x^2 \rangle_\alpha &= \langle \alpha | x^2|\alpha\rangle \\ &=\int_{-\infty}^\infty\!\textrm{d}x \,\psi^*_\alpha(x) \, x^2 \,\psi_\alpha(x) \\ &= N_x \int_{-\infty}^\infty \!\textrm{d}x \, e^{-\left(\frac{x-y}t\right)^2} \,x^2 \\ &= N_x \, I_2\left(\frac1{t^2},-\frac{2y}{t^2},\frac{y^2}{t^2}\right) \\ % &= N_x \left[\left(\frac{B^2}{4A^2} + \frac1{2A}\right) I_0(A,B,C)\right]\Bigg|_{A=\frac1{t^2},\;B=-\frac{2y}{t^2},\;C=\frac{y^2}{t^2}} \\ &= y^2 + \frac{t^2}2 \, \end{align*} $$
Combining these results, we find
$$ \Delta x_{\textrm{RMS}} = \frac t{\sqrt{2}} \,. $$
Switching to a Basis of Momentum States¶
One of the main advantages of the Bra-Ket formalism is that it becomes straightforward to consider changing bases from position to momentum space: it's sufficient to know how to represent a single momentum state in the position basis. The derivation of this can be quite involved (see, for example, Sakurai's Quantum Mechanics for details), but is consistent with what we expect: it's a plane wave of the form
$$ \langle x|p \rangle = \sqrt{N}\,e^{2\pi i x / \lambda_p} $$
where the wavelength of the wave is given by the de Broglie formula mentioned in the pre-requisites.
The normalization $N$ of the plane wave can be determined in a moment by requiring once again that the state $|\alpha\rangle$ is unit-normalized in this basis.
Given our completeness relation from before, we can insert this and find
$$ \begin{align*} \langle p | \alpha \rangle &= \langle p | \left(\int_{-\infty}^\infty\!\textrm{d}x \, | x \rangle \langle x |\right)| \alpha \rangle \\ &=\int_{-\infty}^\infty\!\textrm{d}x \, \langle p|x\rangle \langle x|\alpha\rangle \\ &=\sqrt{N_xN}\int_{-\infty}^\infty\!\textrm{d}x \, e^{-ipx/\hbar} \,e^{ik(x-y)}\,e^{-\frac12\left(\frac{x-y}t\right)^2} \\ &=\sqrt{N_xN} \,I_0\left(\frac1{2t^2}, -\frac y{t^2}+ i(p/\hbar -k), \frac{y^2}{2t^2} - iky\right) \\ &=\sqrt{N_p}\,e^{-ipy/\hbar}\,e^{-\frac{t^2}2\left(p/\hbar -k\right)^2} \,,\qquad N_p := 2\pi t^2 N_xN\,. \end{align*} $$
(Although the convergence of the integral $I_0(A,B,C)$ does depend on its arguments being real, the result above still holds via analytic continuation; we won't discuss the details of this here.) A couple things to note about the state $|\alpha\rangle$ in the momentum basis:
- the Bell curve (or Gaussian) is now peaked a $p = \hbar k$;
- the width of the Gaussian now decreases with $t$, instead of increasing as before;
- there is an oscillatory part to the wavefunction $\psi_\alpha(p)$ whenever $y\neq 0$.
Lastly, requiring that the state $|\alpha\rangle$ be unit-normalized gives $$ \begin{align*} 1 = \langle \alpha | \alpha \rangle &= \int_{-\infty}^\infty\!\textrm{d}p\, \langle \alpha| p\rangle\langle p|\alpha\rangle \\ &= N_p\, \int_{-\infty}^\infty\!\textrm{d}p\, e^{-t^2\left(p/\hbar -k\right)^2} \\ &= N_p\, I_0 \left(\frac{t^2}{\hbar^2}, -\frac{2\,t^2 k}{\hbar}, t^2k^2\right) \\ &= N_p \,\sqrt{\frac{\pi\hbar^2}{t^2}} \\ \implies N &= \frac1{2\pi \hbar} \qquad \left(\textrm{so}\;\; N_p = \frac t{\sqrt{\pi} \hbar} \right)\,. \end{align*} $$
(Since the normalization of the plane wave shouldn't depend on $t$ — or the state $|\alpha\rangle$ at all — this result is sensible.)
A plot of the real and imaginary pars of the momentum-space wavefunction are given below:
hbar = 1.05457182e-34
N = 1/2/np.pi #/hbar
Np = 2*np.pi*t**2*N*Nx
p = np.linspace(0, 4, 1000)
yR = np.sqrt(Np) * np.cos(-p*y) * np.exp(-(p-k)**2*t**2/2)
yI = np.sqrt(Np) * np.sin(-p*y) * np.exp(-(p-k)**2*t**2/2)
fig, axes = plt.subplots(2)
axes[0].plot(p, yR, color="blue")
axes[0].set(xlabel='Momentum (ℏ / nm)', ylabel='ψᵣ(p)')
axes[1].plot(p, yI, color="orange")
axes[1].set(xlabel='Momentum (ℏ / nm)', ylabel='ψᵢ(p)')
for ax in axes.flat:
ax.label_outer()
plt.show()
Putting it All Together¶
To demonstrate the Heisenberg uncertainty relation for this wavepacket, we'll need to compute the RMS error in the momentum in the same way as we did before:
$$ \Delta p_{\textrm{RMS}} = \sqrt{\langle p^2 \rangle_\alpha - \langle p \rangle_\alpha^2} \,. $$
To this end, let's compute the expectation values for $p$ and $p^2$. For the former, we indeed find that the average momentum is given by $\hbar k$, as suggested by the plot:
$$ \begin{align*} \langle p \rangle_\alpha &= \langle \alpha | p |\alpha \rangle \\ &= \int_{-\infty}^\infty \!\textrm{d}p \, \psi^*_\alpha(p) \, p \,\psi_\alpha(p) \\ &= N_p \,\int_{-\infty}^\infty \!\textrm{d}p \, e^{-t^2\left(p/\hbar -k\right)^2} \,p \\ &= N_p \, I_1\left(\frac{t^2}{\hbar^2}, -\frac{2kt^2}\hbar, k^2t^2\right) \\ &= \hbar k \,. \end{align*} $$
Similarly, the expectation value of $p^2$ is given by
$$ \begin{align*} \langle p^2 \rangle_\alpha &= \langle \alpha | p^2|\alpha\rangle \\ &=\int_{-\infty}^\infty\!\textrm{d}p \,\psi^*_\alpha(p) \, p^2 \,\psi_\alpha(p) \\ &= N_p \,\int_{-\infty}^\infty \!\textrm{d}p \, e^{-t^2\left(p/\hbar -k\right)^2} \,p^2 \\ &= N_p \, I_2\left(\frac{t^2}{\hbar^2}, -\frac{2kt^2}\hbar, k^2t^2\right) \\ &= \hbar^2 k^2 + \frac{\hbar^2}{2t^2} \,. \end{align*} $$
Putting this all together, we find that the RMS error in the momentum is given by
$$ \Delta p_{\textrm{RMS}}^{(\alpha)} = \frac{\hbar}{\sqrt{2}t} \,. $$
This result shows that – as a direct result of the De Broglie relation – there is a corresponding uncertainty in the momentum of the state $|\alpha\rangle$ which increases as we shrink the thickness of our Bell curve in position space.
Combining this result with the the RMS error in position, we find that the product of these uncertainities is fixed, for all $k$, $y$, and $t$:
$$ \Delta x_{\textrm{RMS}}^{(\alpha)} \, \Delta p_{\textrm{RMS}}^{(\alpha)} = \frac\hbar 2 \,. $$
Conclusion¶
This is an explicit demonstration of the Heisenberg uncertainty relation, as satisfied for a free particle at rest. Since the Heisenberg uncertainty relation is in fact an inequality, we see that the state $|\alpha\rangle$ saturates this inequality. For this reason, the state $|\alpha\rangle$ is referred to as a minimum uncertainty wavepacket.
In the next post, we'll derive this relationship on more general grounds, using the Cauchy inequality as applied to arbitrary complex functions $\langle x|\alpha\rangle$.