# Do stationary states with higher energy necessarily have higher position-momentum uncertainty?

For simple potentials like square wells and harmonic oscillators, one can explicitly calculate the product $\Delta x \Delta p$ for stationary states. When you do this, it turns out that higher energy levels have higher values of

$\Delta x \Delta p$.

Is this true for all time-independent potentials?

Certainly, it is possible to find two states $\mid \Psi_1 \rangle$ and $\mid \Psi_2 \rangle$ with $\langle \Psi_1 \mid H \mid \Psi_1 \rangle > \langle \Psi_2 \mid H \mid \Psi_2 \rangle$ and also $\Delta x_1 \Delta p_1 < \Delta x_2 \Delta p_2$. For example, choose a quadratic potential, let $\mid \Psi_2 \rangle$ be the first state and let $\mid \Psi_1 \rangle$ be a Gaussian coherent state (thus with minimum uncertainty) and fairly high energy. So I'm asking here just about the stationary states.

As Ron pointed out in the comments, this question is most interesting if we consider potentials with only a single local minimum, and increasing potential to the right of it and decreasing to the left.

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There are trivial counterexamples, where you have a double-dip well, and the approximate ground state of the higher dip is lower uncertainty than the excited states of the lower dip. But your question is probably for monotonically increasing potentials. –  Ron Maimon Aug 31 '11 at 15:13
@Ron Good point - I didn't think of that. I've updated the question. –  Mark Eichenlaub Aug 31 '11 at 15:17
Could you clarify what you mean by "consider potentials with only a single local minimum, and increasing potential to the right of it and decreasing to the left."? I think that's a different situation than what Ron meant by "monotonically increasing". Because if the minimum considered is truly only a local minimum, and the potential decreases continuously to a lower V on the left, you don't have discrete energy levels. –  Anonymous Coward Aug 31 '11 at 17:05
@Anonymous If the potential is decreasing to the left of the local minimum and increasing to the right of it, then the local minimum is necessarily a global minimum. Perhaps the confusion is in the word "decreasing". I didn't mean that as you go further to the left, the value of the potential goes down. I meant that if you look to the left of the minimum, you see a decreasing potential there. –  Mark Eichenlaub Aug 31 '11 at 17:20
Could it actually be that for this class of potentials, the uncertainty of both coordinate and momentum grows separately as one goes to higher excited states? Intuitively, the uncertainty of coordinate will grow since the potential gets wider and the particle thus less localized, while the uncertainty of momentum will grow since the wave function becomes more and more oscillating thanks to the well-known property about the number of nodes of the stationary states. It is certainly an amusing problem! –  Tomáš Brauner Aug 31 '11 at 17:35

The answer is no, and a counterexample is the following plateau potential:

$V(x) = x^2 \ \ \ \ \; \mathrm{for}\ \ \ \ x\ge -A$

$V(x) = A^2 \ \ \ \ \mathrm{for}\ \ \ \ -A-k \le x < -A$

$V(x) = \infty\ \ \; \ \ \mathrm{for}\ \ \ \ x <-A-k$

A is imagined to be a huge constant, and k is a large constant, but not anywhere near as huge as A. The potential has a plateau between -A-k and -A, but is continuous and increasing on either side of the origin. It's loss of uncertainty happens when the energy reaches the Plateau value of $A^2$, and it happens semiclassically, so it happens for large quantum numbers.

Semiclassically, in the Bohr-Sommerfeld (WKB) approximation, the particle has the same eigenfunctions as the harmonic oscillator, until the energy equals $A^2$. At this point, the next eigenfunction oscillates around the minimum, then crawls at a very very slow speed along the plateau, reflects off the wall, and comes back very very slowly to the oscillator.

The time spent on the plateau is much longer than the time spent oscillating (for appropriate choice of A and k) because the classical velocity on the plateau is so close to zero. This means that the position and momentum uncertainty is dominated by the uncertainty on the plateau, and the value of the position uncertainty is much less than the uncertainty for the oscillation if k is much smaller than A, and the value of the momentum uncertainty is nearly zero, because the momentum on the plateau is next to zero.

### WKB expectation values are classical orbit averages

This argument uses the WKB expression for the expectation values of functions of x, which, from the WKB wavefunction,

$$\psi(x) = {1\over \sqrt{2T}} {1\over \sqrt{v}} e^{i\int^x p dx}$$,

Where v(x) is the classical velocity a particle would have at position x, and T is just a constant, a perverse way to parametrize the normalization constant of the WKB wavefunction. The expected value of any function of the X operator is equal to

$$\langle f(x)\rangle = \int |\psi(x)|^2 f(x) = {1\over 2 T} \int {1\over v(x)} f(x) dx = {1\over T}\oint f(x(t)) dt$$

Where the last integral is the integral around the full classical orbit. The last expression obviously works for functions of P (it works for any operator using the corresponding classical function on phase space). So the expectation value is just the average value of the quantity along the orbit, the factor of 2 disappears because you go over every x value twice along the orbit, and the strangely named normalization factor "T" is revealed to be the period of the classical orbit, because the average value of the unit operator is 1.

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Very interesting! I totally agree with your argument that the position uncertainty is dominated by the plateau (and thus of the order of k). However, in case of momentum, according to your last formula, the average of p^2 is proportional to the integral of v(x)dx (the average of p itself is always zero for a stationary state), so it is dominated by the oscillator region since it is much larger than the plateau and also has much larger velocity. I presume one can still tune the parameters so that ∆x∆p drops around the plateau though. That answers the original question of Mark. –  Tomáš Brauner Sep 3 '11 at 9:18
In fact, as long as the WKB approximation holds, one can conclude using your argument that ∆p grows when going to higher excited states. One integrates v(x)dx over the classical trajectory, v(x) grows with energy and one moreover integrates over a larger range of x as energy increases. –  Tomáš Brauner Sep 3 '11 at 9:22
While $p^2$ is dominated by the oscillator region, you must remember that it is a time average, so that the time spent on the plateau reduces the total integral through the normalization constant. The decrease in $p^2$ is almost exactly proportional to the increase in the period T at this level. –  Ron Maimon Sep 3 '11 at 19:05
Ah, you are right, stupid me! –  Tomáš Brauner Sep 3 '11 at 22:09
I made the same mistake at first. –  Ron Maimon Sep 3 '11 at 22:23

I will give a counter-example. Consider the potential below, $V$ is considered to be very large but not infinity. The ground state of the system should be approximate as the ground state of an infinite potential well. The uncertainty is greater than $\hbar/2$.

While when the particle is confined in the parabolic potential well, the local "ground state" corresponding to a Gaussian wave, the uncertainty is just $\hbar/2$.

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But for any finite $V$ the actual ground state is a superposition of the states localized to the square and to the parabola. –  DanielSank Sep 26 '14 at 5:48
@DanielSank If the barrier is high enough, the tunneling coefficient would be so small that the modification of the energy levels and the wave function is negligible. –  buzhidao Sep 26 '14 at 6:04
Right, so the ground state is the state localized to the square and some excited state is the one localized to the parabola. Ok, cool. –  DanielSank Sep 26 '14 at 6:18