Required math: calculus
Required physics: Schrödinger equation
Reference: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 2.22.
While analyzing the free particle, we saw that we could construct a normalizable combination of stationary states by writing
Given the initial wave function, we can find via Plancherel’s theorem:
The integral 1 cannot usually be done in closed form, but one case where it can is if the initial wave function is a Gaussian, of the form
To find , we normalize:
The integral comes out to from which we get
Finding requires finding via equation 2. So we get (using Maple to do the integral):
We can now use 1 again using Maple to do the integral:
where Maple was used for the integral.
Calculating can be done using Maple, but it seems to require a bit of help. First we write out the complex conjugate:
Then we calculate using the Maple command simplify(evalc()) assuming positive and we get
At , , so which is correct. The wave packet at is therefore a Gaussian centred at . As increases, gets smaller but the centre of the Gaussian does not move from so the packet spreads out. A couple of plots show this behaviour. We’ve set in both plots. In the red plot so and in the green plot is larger, at a value such that .
We can get the mean values of position and momentum by integration, although it takes a bit of work. By symmetry, . To get the other two average values, we use integration with Maple.
This shows that the wave function spreads out with time. At , but it then increases quadratically with .
Calculating starts with:
This can be evaluated with the Maple command simplify(evalc(int(-h^2*simplify(evalc(psixtconj*(diff(psixt, x$2)))), x = -infinity .. infinity))) assuming positive where psixtconj and psixt are the Maple expressions for and respectively. The result is:
The uncertainty principle thus becomes
The system has the least uncertainty at . Uncertainty increases with time as the wave packet spreads out.