How to Verify the Uncertainty Principle for a Quantum Harmonic Oscillator

Last Updated: June 2, 2021

The quantum harmonic oscillator is the quantum analogue to the classical simple harmonic oscillator. Using the ground state solution, we take the position and momentum expectation values and verify the uncertainty principle using them.

Part 1

Part 1 of 3:

Ground State Solution

1
Recall the Schrödinger equation. This partial differential equation is the fundamental equation of motion in quantum mechanics that describes how a quantum state $\psi$ evolves in time. ${\hat {H}}$ denotes the Hamiltonian, the energy operator that describes the total energy of the system.
- $i\hbar {\frac {\partial \psi }{\partial t}}={\hat {H}}\psi$
2
Write out the Hamiltonian for the harmonic oscillator. While the position and momentum variables have been replaced with their corresponding operators, the expression still resembles the kinetic and potential energies of a classical harmonic oscillator. Since we are working in physical space, the position operator is given by ${\hat {x}}=x,$ while the momentum operator is given by ${\hat {p}}=-i\hbar {\frac {\partial }{\partial x}}.$
- ${\hat {H}}={\frac {{\hat {p}}^{{2}}}{2m}}+{\frac {1}{2}}m\omega ^{{2}}{\hat {x}}^{{2}}$
Advertisement
3
Write out the time-independent Schrödinger equation. We see that the Hamiltonian does not depend explicitly on time, so the solutions to the equation will be stationary states. The time-independent Schrödinger equation is an eigenvalue equation, so solving it means that we are finding the energy eigenvalues and their corresponding eigenfunctions - the wavefunctions.
- $-{\frac {\hbar ^{{2}}}{2m}}{\frac {{\mathrm {d}}^{{2}}\psi }{{\mathrm {d}}x^{{2}}}}+{\frac {1}{2}}m\omega ^{{2}}x^{{2}}\psi =E\psi$
4
Solve the differential equation. This differential equation has variable coefficients and cannot easily be solved by elementary methods. However, after normalizing, the solution for the ground state can be written like so. Remember that this solution only describes a one-dimensional oscillator.
- $\psi (x)=\left({\frac {m\omega }{\pi \hbar }}\right)^{{1/4}}\exp \left(-{\frac {m\omega }{2\hbar }}x^{{2}}\right)$
- This is a Gaussian centered at $x=0.$ We will use the fact that this function is even to simplify our calculations in the next part.
Advertisement

Part 2

Part 2 of 3:

Expectation Values

1
Recall the formula for the uncertainty. The uncertainty of an observable such as position is mathematically the standard deviation. That is, we find the average value, take each value and subtract from the average, square those values and average, and then take the square root.
- $\sigma _{{x}}={\sqrt {\langle x^{{2}}\rangle -\langle x\rangle ^{{2}}}}$
2
Find $\langle x\rangle$ . Since the function is even, we can deduce from symmetry that $\langle x\rangle =0.$
- If you set up the integral needed to evaluate, you will find that the integrand is an odd function, because an odd function times an even function is odd.
  - $\langle x\rangle =\int _{{-\infty }}^{{\infty }}x|\psi (x)|^{{2}}{\mathrm {d}}x$
- One property of an odd function is that for every positive value of the function, there exists a doppelgänger - a corresponding negative value - that cancels them out. Since we are evaluating over all $x$ values, we know the integral evaluates to 0 without having to actually do the calculations.
3
Calculate $\langle x^{{2}}\rangle$ . Since our solution is written as a continuous wavefunction, we must employ the integral below. The integral describes the expectation value of $x^{{2}}$ integrated over all space.
- $\langle x^{{2}}\rangle =\int _{{-\infty }}^{{\infty }}x^{{2}}|\psi (x)|^{{2}}{\mathrm {d}}x$
4
Substitute the wavefunction into the integral and simplify. We know that the wavefunction is even. The square of an even function is even as well, so we can pull out a factor of 2 and change the lower bound to 0.
- $\langle x^{{2}}\rangle =2\left({\frac {m\omega }{\pi \hbar }}\right)^{{1/2}}\int _{{0}}^{{\infty }}x^{{2}}\exp \left(-{\frac {m\omega }{\hbar }}x^{{2}}\right){\mathrm {d}}x$
5
Evaluate. First, let $\alpha ={\frac {m\omega }{\hbar }}.$ Next, instead of integrating by parts, we will use the gamma function.
- ${\begin{aligned}\langle x^{{2}}\rangle &=2\left({\frac {m\omega }{\pi \hbar }}\right)^{{1/2}}\int _{{0}}^{{\infty }}x^{{2}}e^{{-\alpha x^{{2}}}}{\mathrm {d}}x,\ \ u=\alpha x^{{2}}\\&=2\left({\frac {m\omega }{\pi \hbar }}\right)^{{1/2}}\int _{{0}}^{{\infty }}{\frac {u}{\alpha }}e^{{-u}}{\mathrm {d}}u{\frac {1}{2\alpha x}},\ \ x={\sqrt {{\frac {u}{\alpha }}}}\\&=\left({\frac {m\omega }{\pi \hbar }}\right)^{{1/2}}\alpha ^{{-3/2}}\int _{{0}}^{{\infty }}u^{{1/2}}e^{{-u}}{\mathrm {d}}u\\&=\left({\frac {m\omega }{\pi \hbar }}\right)^{{1/2}}\left({\frac {m\omega }{\hbar }}\right)^{{-3/2}}\Gamma \left({\frac {3}{2}}\right),\ \ \Gamma \left({\frac {3}{2}}\right)={\frac {{\sqrt {\pi }}}{2}}\\&={\frac {\hbar }{m\omega }}{\frac {1}{{\sqrt {\pi }}}}{\frac {{\sqrt {\pi }}}{2}}\\&={\frac {\hbar }{2m\omega }}\end{aligned}}$
6
Arrive at the uncertainty in position. Using the relation we wrote in step 1 of this part, $\sigma _{{x}}$ immediately follows from our results.
- $\sigma _{{x}}={\sqrt {{\frac {\hbar }{2m\omega }}}}$
7

Find $\langle p\rangle$ . As with average position, a symmetry argument can be made that leads to $\langle p\rangle =0.$
8
Calculate $\langle p^{{2}}\rangle$ . Instead of using the wavefunction to calculate this expectation value directly, we can use the energy of the wavefunction to simplify the calculations needed. The energy of the ground state of the harmonic oscillator is given below.
- $E_{{0}}={\frac {1}{2}}\hbar \omega$
9
Relate the ground state energy with the particle's kinetic and potential energy. We expect this relation to hold not just for any position and momentum but also for their expectation values as well.
- ${\frac {1}{2}}\hbar \omega ={\frac {\langle p^{{2}}\rangle }{2m}}+{\frac {1}{2}}m\omega ^{{2}}\langle x^{{2}}\rangle$
10
Solve for $\langle p^{{2}}\rangle$ .
- $m\hbar \omega =\langle p^{{2}}\rangle +m^{{2}}\omega ^{{2}}{\frac {\hbar }{2m\omega }}$
- $\langle p^{{2}}\rangle ={\frac {m\hbar \omega }{2}}$
11
Arrive at the uncertainty in momentum.
- $\sigma _{{p}}={\sqrt {{\frac {m\hbar \omega }{2}}}}$
Advertisement

Part 3

Part 3 of 3:

Verification of the Uncertainty Principle

1
Recall Heisenberg's uncertainty principle for position and momentum. The uncertainty principle is a fundamental limit to the precision with which we can measure certain pairs of observables, such as position and momentum. See the tips for more background on the uncertainty principle.
- $\sigma _{{x}}\sigma _{{p}}\geq {\frac {\hbar }{2}}$
2
Substitute the uncertainties of the quantum harmonic oscillator.
- ${\begin{aligned}{\sqrt {{\frac {\hbar }{2m\omega }}}}{\sqrt {{\frac {m\hbar \omega }{2}}}}&\geq {\frac {\hbar }{2}}\\{\frac {\hbar }{2}}&\geq {\frac {\hbar }{2}}\end{aligned}}$
- Our results are in agreement with the uncertainty principle. In fact, this relation only achieves equality in the ground state - if higher energy states are used, then the uncertainties in the position and momentum only grow.
Advertisement

Expert Q&A

Search

Add New Question

Ask a Question

200 characters left

Include your email address to get a message when this question is answered.

Submit

Tips

There are two backgrounds as to why the uncertainty principle exists.
- From a wave mechanics perspective, the expressions of the wavefunction in terms of position and in terms of momentum are Fourier transforms of one another. One property of the Fourier transform is that a function and its Fourier transform cannot both be sharply localized.
- A simple example is the Fourier transform of the rectangular function. As the width of the function decreases (becomes more localized), the Fourier transform (a sinc curve) becomes flatter and flatter. An extreme example is the Dirac delta function, where the width is infinitesimal (perfect locality). Its Fourier transform is a constant (infinite uncertainty).
- The other way to look at it is from matrix mechanics. The position and momentum operators have a nonzero commutation relation. If two operators commute, then their commutation relation, as signified by the brackets below, would be 0.
  - $[{\hat {x}},{\hat {p}}]={\hat {x}}{\hat {p}}-{\hat {p}}{\hat {x}}=i\hbar$
- It turns out that this commutation relation must imply a fundamental uncertainty principle. When an operator ${\hat {x}}$ acts on a state, then the wavefunction collapses into the eigenstate of ${\hat {x}}$ with a unique measurement (the eigenvalue). However, the eigenstate of ${\hat {x}}$ need not be an eigenstate of another operator ${\hat {p}}.$ If so, then there is no unique measurement for observable $p,$ which means that the state can only be written as a linear combination of momentum basis eigenstates. (When two operators do commute, then they have a simultaneous set of eigenstates in common (this is referred to as degeneracy ) and the two observables can simultaneously be measured to arbitrary precision. This is always the case in classical mechanics.)
- This is the source of the uncertainty principle. It is not due to the limitations of our instruments that we cannot measure a particle's position and momentum to arbitrary precision. Rather, it is a fundamental property of the particles themselves.
Thanks
Helpful 0 Not Helpful 0

How to Verify the Uncertainty Principle for a Quantum Harmonic Oscillator

Steps

Ground State Solution

Expectation Values

Verification of the Uncertainty Principle

Expert Q&A

Tips

You Might Also Like

About this article

Did this article help you?

About this article

You Might Also Like

Explore Categories