In the previous article, we demonstrated that the Heisenberg uncertainty principle is satisfied by a simple wavepacket with average position $y$ and average momentum $\hbar k$. In this article, we're going to generalize these results so that they apply to any state $|\alpha\rangle$.
Preliminaries¶
We begin by introducing some convenient notation: we'll denote the state $(x-\langle x\rangle_\alpha)|\alpha\rangle$ by $|\Delta x_\alpha\rangle$, and the state $(p-\langle p\rangle_\alpha)|\alpha\rangle$ by $|\Delta p_\alpha\rangle$. This allows us to cleanly write
$$ \begin{align*} \left(\Delta x_{\textrm{RMS}}^{(\alpha)}\right)^2 &= \langle \alpha| (x - \langle x \rangle_\alpha)^2|\alpha\rangle = \langle \Delta x_\alpha|\Delta x_\alpha\rangle \\ \left(\Delta p_{\textrm{RMS}}^{(\alpha)}\right)^2 &= \langle \alpha| (p - \langle p \rangle_\alpha)^2|\alpha\rangle = \langle \Delta p_\alpha|\Delta p_\alpha\rangle \,. \end{align*} $$
We do this, in part, to draw an analogy with the case of vectors: given two vectors ${\bf A}$ and ${\bf B}$, the Cauchy inequality states that
$$ |{\bf A}|^2|{\bf B}|^2 \geq \left({\bf A}\cdot {\bf B}\right)^2\,, $$
i.e. the product of the squares of two vectors is always greater than the square of their dot product.
The same is true for the complex vectors $|\Delta x_\alpha\rangle$ and $|\Delta p_\alpha\rangle$, although in this case the dot product is given by $\langle \Delta x_\alpha|\Delta p_\alpha\rangle$. Therefore,
$$ \langle \Delta x_\alpha|\Delta x_\alpha\rangle \langle \Delta p_\alpha|\Delta p_\alpha\rangle \geq |\langle \Delta x_\alpha|\Delta p_\alpha\rangle|^2 \,. $$
This means that, if we can find a way to put a bound on $|\langle \Delta x_\alpha|\Delta p_\alpha\rangle|^2$, then we'll have an essential ingredient to showing the validity of the Heisenberg uncertainty principle.
Plane Waves Revisited¶
We begin by first writing
$$ \begin{align*} \langle \Delta x_\alpha|\Delta p_\alpha\rangle &= \langle \alpha | \left(x - \langle x \rangle_\alpha\right)\left(p-\langle p \rangle_\alpha\right)|\alpha\rangle \\ &=\langle \alpha | xp|\alpha\rangle - \langle x \rangle_\alpha \langle p \rangle_\alpha \\ &=\langle \alpha | x \left(\int_{-\infty}^\infty\!\textrm{d}x'\,|x'\rangle\langle x'|\right)p|\alpha\rangle - \langle x \rangle_\alpha \langle p \rangle_\alpha \end{align*} $$
and focus in on the contribution from $\langle x'|p|\alpha\rangle$:
$$ \begin{align*} \langle x'|p|\alpha\rangle &= \langle x'|p\left(\int_{-\infty}^\infty\!\textrm{d}p'\,|p'\rangle\langle p'|\right)|\alpha\rangle \\ &= \int_{-\infty}^\infty\!\textrm{d}p'\,p'\langle x'|p'\rangle \langle p'|\alpha\rangle \,. \end{align*} $$
From here, it's helpful to recall that the momentum state $|p\rangle$ in the position basis is given by
$$ \langle x|p\rangle = N \,e^{2\pi i x/\lambda_p} = N\,e^{ipx/\hbar} \,, $$
where $\lambda_p = h/p$ is the De Broglie wavelength and $\hbar = h/2\pi$. Therefore, we can write the expectation value $\langle x'|p|\alpha\rangle$ as
$$ \begin{align*} \langle x'|p|\alpha\rangle &= N\int_{-\infty}^\infty\!\textrm{d}p'\,p'e^{ip'x'/\hbar} \langle p'|\alpha\rangle \\ &= -i\hbar N\int_{-\infty}^\infty\!\textrm{d}p'\,\frac d{dx'}\left(e^{ip'x'/\hbar}\right) \langle p'|\alpha\rangle \\ &= -i\hbar \int_{-\infty}^\infty\!\textrm{d}p'\,\frac d{dx'}\Big(\langle x'|p'\rangle\Big) \langle p'|\alpha\rangle \\ &= -i\hbar \frac d{dx'}\langle x'|\alpha\rangle \,. \end{align*} $$
This means the dot product $\langle \Delta x_\alpha | \Delta p_\alpha\rangle$ is given by the following integral:
$$ \langle \Delta x_\alpha| \Delta p_\alpha\rangle = -i\hbar\int_{-\infty}^\infty\!\textrm{d}x'\,\psi^*_\alpha(x') x'\left(\frac d{dx'}\psi_\alpha(x')\right) - \langle x \rangle_\alpha \langle p \rangle_\alpha \,. $$
Real and Imaginary Parts¶
In computing the bound $|\langle \Delta x_\alpha|\Delta p_\alpha\rangle|^2$, it's helpful to consider separately the real and imaginary parts as the latter can be assembled to give the former as follows:
$$ |\langle \Delta x_\alpha|\Delta p_\alpha\rangle|^2 = \left[\mathfrak{Re}\langle \Delta x_\alpha|\Delta p_\alpha\rangle\right]^2 + \left[\mathfrak{Im}\langle \Delta x_\alpha|\Delta p_\alpha\rangle\right]^2 \,. $$
As it turns out, the imaginary part of $\langle \Delta x_\alpha|\Delta p_\alpha\rangle$ is particularly simple:
$$ \begin{align*} \mathfrak{Re}\langle \Delta x_\alpha|\Delta p_\alpha\rangle &= -\frac{i\hbar}2\int_{-\infty}^\infty\!\textrm{d}x\,x\left(\psi^*_\alpha\,\frac{d\psi_\alpha}{dx} - \psi_\alpha \,\frac{d\psi_\alpha^*}{dx} \right) - \langle x \rangle_\alpha \langle p \rangle_\alpha \\ \mathfrak{Im}\langle \Delta x_\alpha|\Delta p_\alpha\rangle &= -\frac{\hbar}2\int_{-\infty}^\infty\!\textrm{d}x\,x\left[\psi^*_\alpha\,\frac{d\psi_\alpha}{dx} + \psi_\alpha \,\frac{d\psi_\alpha^*}{dx} \right] \\ &=-\frac{\hbar}2\int_{-\infty}^\infty\!\textrm{d}x\,x\,\frac{dP_\alpha}{dx}\\ &= \left[xP_\alpha(x)\right]\Big|_{-\infty}^\infty+\frac{\hbar}2\int_{-\infty}^\infty\!\textrm{d}x\,P_\alpha(x) \\ &= \frac{\hbar}2 \end{align*} $$
In the second last line, we assume that the (unit-normalized) probability function goes to zero at least as quickly as $1/x^2$ as $x\to\pm\infty$.
Since the contribution to $|\langle \Delta x_\alpha|\Delta p_\alpha\rangle|^2$ from the real part of $\langle \Delta x_\alpha|\Delta p_\alpha\rangle$ is strictly positive, we can safely conclude that
$$ |\langle \Delta x_\alpha|\Delta p_\alpha \rangle|^2 \geq \frac {\hbar^2}4 \,. $$
All Together Now¶
Since the Cauchy inequality guarantees that
$$\left(\Delta x_{\textrm{RMS}}^{(\alpha)}\right)^2 \left(\Delta p_{\textrm{RMS}}^{(\alpha)}\right)^2 \geq |\langle \Delta x_\alpha|\Delta p_\alpha \rangle|^2 \,, $$
and since we've just shown that the right-hand side of the above is bounded below by $\hbar^2/4$, we conclude the the Heisenberg Uncertainty Relation holds for any unit-normalized state $|\alpha\rangle$:
$$ \Delta x_{\textrm{RMS}}^{(\alpha)} \,\Delta p_{\textrm{RMS}}^{(\alpha)} \geq \frac\hbar 2\,. $$