For the harmonic oscillator we have $\sigma_x \sigma_p = \hbar(n+1/2) $ and by the uncertainty principle $\sigma_x \sigma_p \geq \frac{\hbar}{2}$.

In one of the exercises I was doing I was asked to comment on this result. This makes me think that I am missing something special, as the most I can say is that

  • there is some connection between energy and uncertainty as $\frac{E_n}{\omega}= \sigma_x \sigma_p = \hbar(n+1/2) $

  • the groundstate is a minimum uncertainty state.

What's so special about the above result? Can anyone help me out?

Many thanks!

  • $\begingroup$ I guess one could also say "As the frequency decreases the uncertainty increases" and "High energy states have greater uncertainty". $\endgroup$
    – user
    Jun 24, 2020 at 14:13
  • 2
    $\begingroup$ I am guessing that they were looking for your second statement, that the ground state minimizes uncertainty. This is a very important fact when one goes on to study coherent states (en.wikipedia.org/wiki/Coherent_state). $\endgroup$
    – Rococo
    Jun 26, 2020 at 15:24
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    $\begingroup$ Thanks! @Rococo if you want to add as answer I will accept it. $\endgroup$
    – user
    Jun 26, 2020 at 17:39

3 Answers 3


Your professor will most likely be fine with your second point, which is correct; the ground state is a minimum uncertainty state. The reason why has to do with the form of the Hamiltonian of the quantum harmonic oscillator; it is uniquely suited to minimization, such that one can find the ground state and energy level without solving the Schrödinger equation, which is unusual, to say the least. To see why, there's nothing like getting our hands wet; let's try and minimize the Hamiltonian's expectation value! Taking a normalized arbitrary state:

$$\langle H\rangle=\frac{\langle P^2\rangle}{2m}+\frac{1}{2}m\omega\langle X^2\rangle$$

We can make this a function purely of $\Delta P$ and $\Delta X$ by using the standard equations for the standard deviation: $(\Delta A)^2=\langle A^2 \rangle-\langle A\rangle^2$, where $A$ is an arbitrary operator. Using these, we get

$$\langle H\rangle=\frac{(\Delta P)^2+\langle P\rangle^2}{2m}+\frac{1}{2}m\omega((\Delta X)^2+\langle X\rangle^2)$$

Now, $\langle X\rangle$ and $\langle P\rangle$ are independent of each other and $(\Delta X)^2$ and $(\Delta P)^2$, so to continue minimizing let's set them both to zero. Now, we have $$\langle H\rangle=\frac{(\Delta P)^2}{2m}+\frac{1}{2}m\omega(\Delta X)^2$$

Using the uncertainty relation, we have $$\langle H\rangle \geq \frac{\hbar^2}{8m(\Delta X)^2}+\frac{1}{2}m\omega(\Delta X)^2$$

Now, it is a fact that the uncertainty relation is an equality only for the case of a Gaussian. I'll direct you to a proof of this fact at the end of this answer. So, saying $\langle x | \psi\rangle=a \exp\left(-\frac{1}{4}\left(\frac{x}{\Delta X}\right)^2\right)$ such that $a$ normalizes the function, we have

$$\langle H\rangle = \frac{\hbar^2}{8m(\Delta X)^2}+\frac{1}{2}m\omega(\Delta X)^2$$

Finally, let's choose the value of $\Delta X$ such that the expectation value is minimized. We have

$$\frac{\partial\langle H\rangle}{\partial (\Delta X)^2}=0=-\frac{\hbar^2}{8m(\Delta X)^4}+\frac{1}{2}m\omega$$

Solving for $(\Delta X)^2$, we have $(\Delta X)^2=\frac{\hbar}{2m\omega}$ and $\langle H\rangle_{min}=\frac{h\omega}{2}$. Shoving this into the Gaussian, we get $$\langle x|\psi\rangle_{min}=\left(\frac{m\omega}{\pi\hbar}\right)^{\frac{1}{4}}\exp\left(-\frac{m\omega x^2}{2\hbar}\right), \; \langle H\rangle_{min}=\frac{\hbar\omega}{2}$$

This is the ground state and energy. We know that $\langle H\rangle_{min}\leq \langle H\rangle$ (for arbitrary $|\psi\rangle$). Allow $|\psi\rangle=|\psi_0\rangle$ (denoting the ground state) and we get $$\langle \psi_0|H|\psi_0\rangle\leq\langle H\rangle_{min}\leq\langle \psi_0|H|\psi_0\rangle$$

This makes the ground state energy equal to $\langle H\rangle_{min}$, and since only one state has that energy $\left(|\psi_{min}\rangle\right)$, $|\psi_{min}\rangle=|\psi_{0}\rangle$.

Consider the process we went through, and how lucky we were that everything lined up just right. If we hadn't had the $P^2$ and $X^2$ terms separate and of that power, we could not have used the standard deviation relations to get it in a form where we can use the uncertainty principle (try doing this, for example, on the hydrogen atom's Hamiltonian; you'll find you can't do it exactly). This is (one of the many reasons) why the harmonic oscillator is special; the form of its Hamiltonian allows for the ground state to be a minimum state of uncertainty, which normally does not happen.

Now, you may ask, what's the deal with this? I get that it's rare, but why does it matter? Well, these are what we call coherent states, which are states that have expectation values that evolve like the classical equivalent. They have all sorts of lovely properties and form the backbone of quantum optics; check out the Wikipedia page!

This whole discussion was heavily inspired by R. Shankar's absolutely lovely book on quantum mechanics. I direct you to chapter 7 for a more elaborated version of this discussion, and to chapter 9 for the proof that a Gaussian minimizes uncertainty.

  • 1
    $\begingroup$ Amazing answer! Just what I was looking for. Will award you the bounty asap. Many thanks! $\endgroup$
    – user
    Jun 27, 2020 at 6:22

" High energy states have greater uncertainty"

This also implies that the lowest energy state or ground state of any quantum system described by a harmonic oscillator has the lowest uncertainty and the lowest energy As.. $$E_n =\omega \hbar(n+1/2)$$ For the ground state or the lowest energy state we get $$E_0 = \frac{\hbar\omega}{2}$$ This is a significant result as this implies that the energy of a system described by a harmonic oscillator can never be zero. Physical systems such as atoms in a solid lattice or in polyatomic molecules in a gas cannot have zero energy even at absolute zero temperature. The energy of the ground vibrational state is often referred to as "zero point vibration". A classical harmonic oscillator can however have zero energy.The zero point energy is sufficient to prevent liquid helium-4 from freezing at atmospheric pressure, no matter how low the temperature.

  • 1
    $\begingroup$ This seems a bit circular to me... We calculated the energy values and therefore know they can't zero. Isn't the uncertainty principle then just telling us something we already knew ( by direct calculation)? Thanks! $\endgroup$
    – user
    Jun 24, 2020 at 16:19
  • $\begingroup$ @user you can shift the energy origin and make the ground state energy zero, but the uncertainty will remain... unlike in classical physics. $\endgroup$
    – Roger V.
    Jun 26, 2020 at 14:14
  • $\begingroup$ Doesn't this contradict the above post? Also if the energy can be arbitrarly shifted, why do we care about the specific number in the first place? Thanks! $\endgroup$
    – user
    Jun 26, 2020 at 14:30
  • $\begingroup$ @user when you shift your energy the nontrivial thing is zero state energy will be above the shifted energy by an exact amount of $\hbar\omega/2$ (you can do it by just shifting the potential energy term in Schrodinger equation) so we have to live with this additional energy. $\endgroup$
    – aitfel
    Jun 26, 2020 at 15:07

I cannot read the mind of the person who asked you to comment, but my guess is that they were looking for something like "In this case, the uncertainty principle doesn't tell me anything beyond what I already knew because it's obvious that $n+1/2\ge 1/2$."

  • $\begingroup$ Thanks! I was just wondering if there was something standard/special about this result that I was unaware of. My reasoning was "If they want me to comment, most likely there is something to comment on". $\endgroup$
    – user
    Jun 26, 2020 at 14:39

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