Another Way to do the Gaussian Problem

[ehrenfest]

Shankar page 155

Homework Statement

Use the propagator equation for a free particle

U(t) = \exp\left(\frac{i}{\hbar}\left(\frac{\hbar^2t}{2m}\frac{d^2}{dx^2}\right)\right) = \sum_{n=0}^{\infty}\frac{1}{n!}\rleft(\frac{i\hbar t}{2m}\right)\frac{2^{2n}}{dx^{2n}}

The initial state of the wave packet is
\psi(x',0)= \frac{exp(-x^2/2)}{(\pi)^{1/4}}

Find psi(x,t).

Homework Equations

Hint 1: Express the initial wave function as a power series:

\psi(x',0) = (\pi)^{-1/4} \sum_{n = 0}^{\infty}{\frac{(-1)^nx^{2n}}{n!(2)^n}}

Hint 2: Find the action of a few terms

1, \left( \frac{i\hbart}{2m}\right) \frac{d^2}{dx^2}, \frac{1}{2!}\left( \frac{i\hbar t}{2m} \frac{d^2}{dx^2}\right)^2

The Attempt at a Solution

I am stuck on the second hint. How do you find the action when you do not have a Lagrangian?

Please do not solve the entire thing--just help me with this hint. Thanks.

 

[mjsd]

if I have understood the question correctly,
looks like "action" means the action of the operator U(t) rather than "action" in the variational calculus sense. since psi(x,t) = U(t) psi(x,0), i think you are asked to look at the first few terms to see whether there is a pattern that you can conclude something about what the form must psi(x,t) take.

 

[ehrenfest]

Yes. You're definitely right. The rest of the hint even says "on this power series [meaning the one from hint 1]". So the action of the ith term of the power series expression for the propagator on the wave packet is:

\frac{1}{i!}\left( \frac{i\hbar t}{2m}\frac{d^2}{dx^2}\right)^i * (\pi)^{-1/4} \sum_{n = 0}^{\infty}{\frac{(-1)^n x^{2n}}{n!(2)^n}}

=\frac{1}{i!}\left( \frac{i\hbar t}{2m} \right)^i * (\pi)^{-1/4} \sum_{n = 0}^{\infty}{\frac{(-1)^n C(2n, 2i)x^{2(n-i)}}{n!(2)^n}}

where C is a combination of 2i elements from a set of 2i. Is that in the right direction?

 

[mjsd]

probably. but your coefficient for x^(2n-2i) seems wrong. looks more like a nPr than a nCr.

 

[ehrenfest]

Yes. It should be

=\frac{1}{i!}\left( \frac{i\hbar t}{2m} \right)^i * (\pi)^{-1/4} \sum_{n = 0}^{\infty}{\frac{(-1)^n (2n)! x^{2(n-i)}}{n!(2n - 2i)!(2)^n}}

The next hint is to collect terms with the same power of x and then to notice the series expansion of the coefficient of the power x^(2n):

\left( 1 + \frac{i t\hbar}{m} \right)^{-n - 1/2}

I am only famililar with expansions of binomials with positive integer exponents. Is this an extension of that? Can someone link me to a source on binomial expansions of of non-positive non-integers...

 

[mjsd]

generalised binomial theorem:
(1+z)^{a}=1+az+\frac{a(a-1)}{2!}z^2+\frac{a(a-1)(a-2)}{3!}z^3+\cdots
for |z|<1. define binomial coeff as
\begin{pmatrix}a \\ n\end{pmatrix}=\frac{a(a-1)(a-2)\ldots(a-n+1)}{n!} for n\geq 1 and
\begin{pmatrix}a \\ 0\end{pmatrix}=1.

then for positive integer, a, it reduces to the usual nCr formula:
\begin{pmatrix}a \\ n\end{pmatrix}=\frac{a!}{n!(a-n)!}, a \in \mathbb{Z}

in some text the Pochhammer's symbol (a)_n is also used, it means simply:
(a)_n = a(a+1)(a+2)\ldots(a+n-1) for n\geq 1.

this is neat because (a)_n relates to gamma functions via:
(a)_n = \frac{\Gamma(n+a)}{\Gamma(a)}
and as you probably know
z! = \Gamma(z+1) and
\Gamma(s) = \int_0^\infty\, e^{-t}\, t^{s-1}\, dt,\;\; \text{Re}(s) &gt;0

one observation from all these: (binomial trick)
(1-x)^{-a} = \sum_{n=0}^\infty \frac{(a)_n x^n}{n!}
if |x|<1.

 

[ehrenfest]

Thanks mjsd. I understand generalized binomial coefficients now. I am still having trouble extract that generalized binomial series from

=\frac{1}{i!}\left( \frac{i\hbar t}{2m} \right)^i * (\pi)^{-1/4} \sum_{n = 0}^{\infty}{\frac{(-1)^n (2n)! x^{2(n-i)}}{n!(2n - 2i)!(2)^n}}

Any tips? How do you get rid of pi?

 

[mjsd]

i haven't been thinking too hard lately, but it may help me to understand your problem better if you explain why you are having trouble with the pi? aren't your concern is x? it is double sum I believe summing over n and i. did the question specifically say to use generalised binomial series to do this?