# Differential equations in the schrodinger equation.

1. Oct 5, 2013

### neptune12XII

i got a book on differential equations that says a shortcut to solving the general differential equation f'(x)+p(x)f(x)=g(x) is to take the antiderivative of g(x) dx times exp(-p(x) dx times x) to solve for f(x) where dx represents the functions antiderivative. (i kno its supposed to represent the infinitesml area under part of a curve and i i were proper id write it as an improper integral. some unhelpful guy on yahoo answer pointed this out.) and yes, they explained how it worked. anyway, i was watching a tutorial on the schrodinger equation and i believe solving for the T(t) component of the wavefunction came down to the equation T'(t)+iE/(h-bar)T(t)=0. in this case p would be iE/(h-bar) and g=0 which i understand are constants. they said that the solution is T(t)=e^iEt/(h-bar). but i dont understand how the process i learned applies. if somebody could explain the process used thatd be great. im pretty sure i just got lost and i hope my question is clear.

2. Oct 5, 2013

### fzero

The first method that you mention is known as the integrating factor method. The 2nd equation, for the time-dependent factor of the Schrodinger wavefunction, is actually simple and the integrating factor is not required to solve the equation.

As you mention, the antiderivative of a function $f(x)$ is the function

$$F(x) = \int^x_{x_0} f(x') dx'.$$

Here $x'$ is what's called a "dummy" variable, since once we integrate over it, the expression no longer depends on it. This formula is justified in standard calculus texts, so I won't say more about it.

Now if we consider differential equations, we can quickly realize that the simplest types of differential equation to solve are ones of the form

$$\frac{df(x)}{dx} = g(x), ~~~~~(1)$$

i.e., the expression containing the derivatives is actually a total derivative. Then we can see that the solution to the equation is just

$$f(x) = \int_{x_0}^x g(x') dx',~~~~(2)$$

where $x_0$ is a constant that supplies the constant of integration for the general solution. Equations of the form (1) are sometimes called "integrable," since they have a direct solution (even if we can't simplify the actual integral on the right-hand side of (2).)

Now an equation of the form

$$T'(t) + k T(t) = 0 ,~~~~(3)$$

for constant $t$ is also integrable, because

$$-k = \frac{T'(t)}{T(t)} = \frac{d}{dt} \left( \ln T(t) \right).$$

We will find that

$$T(t) = A e^{-k t},$$

for a constant $A$.

Now that we have these pieces in place, we can actually explain why integrating factors exist.
Consider our equation

$$\frac{df(x)}{dx} + p(x) f(x) = g(x).$$

The left-hand side is not a total derivative, but we can ask if some multiple of it is. Namely, can we find a function $m(x)$ such that

$$\frac{d}{dx} \Bigl( m(x) f(x) \Bigr) = m(x) \frac{df(x)}{dx} + m(x) p(x) f(x) = m(x) g(x).~~~~~(4)$$

If we expand the derivative on the left, we find that such an $m(x)$ must satisfy the differential equation

$$\frac{d m(x) }{dx} + p(x) m(x) = 0.$$

This equation of the same form as the time equation (3), except now instead of the constant $k$, we have the function $p(x)$. We can still use the logarithm to solve for

$$m(x) = \exp \left( \int^x p(x') dx' \right).$$

Once $m(x)$ is determined, we can also integrate (4) to find

$$f(x) = \frac{1}{m(x)} \left( \int^x m(x') g(x') dx' + C \right).$$

To get to your question, we can try to apply this to equation (3), where we didn't really need an integration factor. We simply have $p(t) =k$, so that $m(t) = e^{kt}$. Since $g(t)=0$, we simply have $f(x) = C e^{-kt}$. So the integrating factor method still works, it is just overkill. Also, we really needed to know how to solve equations of the form (3) in order to derive the method in the first place.