D.13 The harmonic oscillator and uncertainty

The given qualitative explanation of the ground state of the harmonic oscillator in terms of the uncertainty principle is questionable. In particular, position, linear momentum, potential energy, and kinetic energy are uncertain for the ground state. This note gives a solid argument, but it uses some advanced ideas discussed in chapter 4.4 and 4.5.3.

As explained more fully in chapter 4.4, the expectation value of the kinetic energy is defined as the average value expected for kinetic energy measurements. Similarly, the expectation value of the potential energy is defined as the average value expected for potential energy measurements.

From the precise form of expectation values in quantum mechanics, it follows that total energy must be the sum of the kinetic and potential energy expectation values. For the harmonic oscillator ground state, that gives

\begin{displaymath}
E_{x0} = {\textstyle\frac{1}{2}}\hbar\omega
= \frac{1}{2...
...big\rangle + \frac{m}{2} \omega^2 \big\langle x^2\big\rangle
\end{displaymath}

Here $\big\langle.\big\rangle $ stands for the average of the enclosed quantity. Only the motion in the $x$-​direction will be considered here. The $y$ and $z$ directions go exactly the same way.

Now any value of $p_x$ can be written as equal to the average value $\big\langle p_x\big\rangle $ plus a deviation from that average $\Delta{p}_x$. Then

\begin{displaymath}
\big\langle p_x^2\big\rangle = \Big\langle(\big\langle p_x...
...\Delta p_x\big\rangle + \big\langle(\Delta p_x)^2\big\rangle
\end{displaymath}

Note that an average is a constant that is not affected by further averaging. Next note that the average of $\Delta{p}_x$ is zero, otherwise the average of $p_x+\Delta{p}_x$ would not be $\big\langle p_x\big\rangle $. So:

\begin{displaymath}
\big\langle p_x^2\big\rangle = \big\langle p_x\big\rangle ^2 + \big\langle(\Delta p_x)^2\big\rangle
\end{displaymath}

Of course, a similar expression holds for $\big\langle x^2\big\rangle $, so the ground state energy is

$\parbox{400pt}{\hspace{11pt}\hfill$\displaystyle
{\textstyle\frac{1}{2}}\hbar...
...gle
+ \frac{m}{2} \omega^2 \big\langle(\Delta x)^2\big\rangle
$\hfill(1)}$
Consider the last two terms. Call them $a^2$ and $b^2$ for now. Note that

\begin{displaymath}
(a-b)^2 \mathrel{\raisebox{-1pt}{$\geqslant$}}0 \quad\Long...
..._x)^2\big\rangle } \sqrt{\big\langle(\Delta x)^2\big\rangle }
\end{displaymath}

as follows from multiplying out the square. The $\raisebox{-.5pt}{$\geqslant$}$ becomes $\vphantom0\raisebox{1.5pt}{$=$}$ when $a$ and $b$ are equal.

Now the first square root above is a measure of the uncertainty in $p_x$. If $\Delta{p}_x$ is always zero, then $p_x$ is always its average value, without any uncertainty. Similarly, the second square root above is a measure of the uncertainty in $x$. The Heisenberg uncertainty principle can be made quantitative as, chapter 4.5.3,

\begin{displaymath}
\sqrt{\big\langle(\Delta p_x)^2\big\rangle } \sqrt{\big\la...
...el{\raisebox{-1pt}{$\geqslant$}}{\textstyle\frac{1}{2}} \hbar
\end{displaymath}

Therefore

\begin{displaymath}
a^2 + b^2 \mathrel{\raisebox{-1pt}{$\geqslant$}}{\textstyle\frac{1}{2}} \hbar \omega
\end{displaymath}

So the minimum value of the final two terms in the expression (1) for the ground state energy is the complete ground state energy. Therefore, in order that the right hand side in (1) does not exceed the left hand side, the first two terms must be zero. So the average particle momentum and position are both zero. In addition, for the estimates of the final two terms, equalities are needed, not inequalities. That means that $a$ must be $b$. That then means that the expectation kinetic energy must be the expectation potential energy. And the two must be the very minimum allowed by the Heisenberg relation; otherwise there is still that inequality.