# Stationary States: Differential Equations Issues

1. Jun 24, 2011

### Elwin.Martin

So in Griffith's (ed. 2 page 37) there's an equation that says that
**pretend the h's are h-bars...I don't know Latex very well**

ih$\frac{1}{\varphi}$$\frac{d\varphi}{dt}$=-$\frac{h^{2}}{2m}$$\frac{1}{\psi}$$\frac{d^{2}\psi}{dx^{2}}$+V

Since in this simplified case V where is a function of x alone he says that each side is equal to a constant but I'm still trying to figure out why.

I can see that the LHS is a function of t alone because he made the wave function separable and the $\varphi$ is a function of t and the $\psi$ is a function of x but I'm not sure why it's important. I see that if we set the equation equal to a constant the rest of the math works out nicely but I can't see what allows us to do this.

If we had say $\frac{dy}{dx}$=$\frac{dz}{dt}$ and we set the whole thing equal to c, how would I know that y(x) = cx and z(t) = ct (they're the same constant right?) so y(x)/x=z(t)/t ?

Any direction would be great and thank you for your time,
elwin.

2. Jun 24, 2011

### Staff: Mentor

h-bar in Latex is \hbar (surprise! :surprised).

The equation has to be true for any value of x and any value of t.

First pick a value of t and use it to evaluate the LHS. This gives you a number. No matter what value you use for x, the RHS must evaluate to that number. So the RHS is constant.

Now start over. Pick a value of x and use it to evaluate the RHS... (see if you can fill in the rest.)

3. Jun 24, 2011

### SpectraCat

Can you see another possibility? Remember that mathematically, a variable is free to take any value in its domain. So if the LHS is only a function of t, and the RHS is only a function of x, can you find any function f(x,t) that is NOT a constant, yet still satisfies both sides of the equation independently?

In other words, what you are given is:

g(x)=h(t)

and I am asking you to do is find any f(x,t) that is NOT a constant and satisfies BOTH

g(x)=f(x,t)

AND

h(t)=f(x,t)

for ALL allowable values of x and t.

4. Jun 24, 2011

### Elwin.Martin

haha thank you for that

I think I see what you mean now.

5. Jun 24, 2011

### Elwin.Martin

That makes a whole lot more sense!

thank you very much for your time,
elwin.