Suppose, for example, that we want to find the minimum energy of a particle undergoing simple harmonic motion. In classical mechanics, the energy is:

$$E = \frac{p^2_x}{2m} + \frac{1}{2} m \omega_0^2x^2$$

where $m$ is the mass of the particle and $x$ the position. I was taught that one can use Heisenberg's uncertainty principle to find an estimate (which actually turns out to be an exact answer) for the ground state energy of this system. Since the potential energy well is symmetric, one can deduce that the expectation value of $x$ and $p_x$ will be zero. The uncertainty in the momentum will be given by: $$\Delta p_x \ge \frac{\hbar}{2\Delta x}.$$

Now then I was told that the energy of the system must obey the following: $$E \ge \frac{1}{2m} \left(\frac{\hbar}{2\Delta x}\right)^2 + \frac{1}{2}m\omega_0^2(\Delta x)^2.$$

I can see that this inequality is obtained plugging in the value $p_x = \frac{\hbar}{2 \Delta x}$ and $x=\Delta x$, but I really don't get how this is "mathematically" possible. The first equation contains on both sides values of energy, momentum and position but the new inequality contains on the LHS a value for the energy and on the RHS uncertainties, how is this possible? Surely the equation must be the following: $$\Delta E \ge \frac{1}{2m} \left(\frac{\hbar}{2\Delta x}\right)^2 + \frac{1}{2}m\omega_0^2(\Delta x)^2$$

  • 1
    $\begingroup$ You forgot the mass in the kinetic energy term. $\endgroup$
    – Semoi
    Jan 26, 2020 at 14:34

2 Answers 2


Building on the last equation in Semoi's answer: \begin{equation} \left\langle E \right\rangle = \frac{(\Delta p)^2}{2m} + \frac{m\omega^2 (\Delta x)^2}{2}, \tag{1} \end{equation} if you want to arrive at the minimum energy, all you need to do is substitute the uncertainty principle, $\Delta p = \hbar / (2 \Delta x)$, into Eq. (1), and minimize the energy with respect to $\Delta x$, i.e., setting the derivative equal to zero: \begin{equation} \frac{d\left \langle E \right\rangle}{d(\Delta x)} = 0. \tag{2} \end{equation}

You will arrive at the result: \begin{equation} (\Delta x)^2 = \frac{\hbar}{2m\omega}. \tag{3} \end{equation}

Substituting Eq. (3) in the original expression you will arrive at $\left \langle E \right\rangle=\frac{1}{2}\hbar\omega$, which is precisely the ground-state energy for the harmonic oscillator.


I totally agree: This "derivation" is mathematically improper.

The energy of the system under consideration is given by $E = \frac{p^2}{2m} + \frac{m\omega^2 x^2}{2}$. However, as you know this is not true, because the energy levels would not posses any uncertainty. Thus, we should interpret this relationship by assuming that $p$ and $x$ are random variables. For example, we could assume that both are independent and normally distributed with standard deviation $\sigma_p$ and $\sigma_x$. However, as you know this is also not true, because the two uncertainties are correlated. So, in order to obtain a proper mathematical derivation, we could use a Taylor expansion. By expressing the energy as a function of two variables, $E = E(p, x) = f(q_1, q_2)$, the uncertainty (= standard deviation, $\sigma_E$) becomes (here we use the variance instead of the standard deviation) \begin{align} \sigma^2_E &\approx \sum_{i=1}^{k} \sum_{j=1}^k \frac{\partial f}{\partial q_i} \frac{\partial f}{\partial q_j} \mathrm{cov}[q_i, q_j] \\ %%% &=\sum_{i=1}^k \left(% \frac{\partial f}{\partial q_i} \right)^2 \sigma^2_{q_i} + 2 \sum_{i=1}^{k-1} \sum_{j=i+1}^k \frac{\partial f}{\partial q_i} \frac{\partial f}{\partial q_j} \mathrm{cov}[q_i, q_j] %+ \mathcal{O}(\sigma^4) \end{align} where "cov" is the covariance of the two random variables. However, this get's messy.

So, if we like to approximate the uncertainty in energy the simple phenomenological model $$ E + \Delta E \approx \frac{(p + \Delta p)^2}{2m} + \frac{m\omega^2 (x + \Delta x)^2}{2} $$ seams to be reasonable. Now, if we evaluate this relationship for the point $(p, x) = (0,0)$ in phase-space, we obtain $E=0$ and $$ \Delta E \approx \frac{(\Delta p)^2}{2m} + \frac{m\omega^2 (\Delta x)^2}{2} $$

The argument used in your notes reads "the uncertainty in position and momentum provide typical values for position and momentum". Thus, your professor did not try to estimate the uncertainty $\Delta E$, but they estimated the averaged energy $\left\langle E \right\rangle$. In order to do so, they used "typical values". In this case the LHS reads $\left\langle E \right\rangle$.

There are several ways to calculate this average energy. Your professor choose to plug in "typical values". Thus, he/she just replaced $p \to \Delta p$ and $x \to \Delta x$. This is valid, because if the expectation value of a random variable is zero, $\left\langle Q \right\rangle = 0$, then it's variance is equal to it's expectation value squared, $ (\Delta Q)^2 = \sigma_{Q}^2 = \left\langle Q^2 \right\rangle - \left\langle Q \right\rangle^2 = \left\langle Q^2 \right\rangle$. Hence,

\begin{align} \left\langle E \right\rangle &= \left\langle \frac{p^2}{2m} \right\rangle + \left\langle \frac{m\omega^2x^2}{2} \right\rangle \\ &= \frac{\left\langle p^2 \right\rangle}{2m} + \frac{m\omega^2 \left\langle x^2 \right\rangle}{2} \\ &= \frac{\sigma_p^2}{2m} + \frac{m\omega^2 \; \sigma_x^2}{2} \end{align}

  • $\begingroup$ I see. However, my doubt is around the last equation. You (correctly) inferred that that is the uncertainty in the energy, however the claim in my notes is that the that uncertainty in the energy is the minimum value for the energy itself. For more details, here is the full relevant snippet from my notes imgur.com/ZZl6L8w $\endgroup$
    – user63248
    Jan 26, 2020 at 15:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy