Dimensionless form of the time-independent Schrodinger equation

[ehrenfest]

For a free particle, show that the time-independent Schrödinger equation can be written in dimensionless form as

d^2\psi(z)/dz^2 = -\psi(z).

I do not see how you would get rid of the m (with units mass) in front of the del in the SE (or the other constants for that matter)...

 

[Meir Achuz]

Substitute z=[hbar/sqrt{2m}]x

 

[ehrenfest]

Sorry, this is probably a really basic question but if I have psi(x(z)), how do I get psi(z)?

 

[Dick]

psi(z(x))=psi(z). They are the same thing. Meir is suggesting you change the variable you differentiate wrt, not the psi. Use the chain rule.

 

[ehrenfest]

Just so we are on the same page, the equation I am starting with is:

-\frac{\hbar^2}{2m}\frac{d^2}{dx^2}\psi_E(x) = E \psi_E(x)

where (I thought) E was a constant not an operator.

So, first, I do not see how that substitution would get rid of the constant E.

Second, if I change variables I get

-\frac{\hbar^2}{2m}\frac{d^2}{dx^2}\psi_E(v(x)) = E \psi_E(v(x))

I cannot change the dx in the second derivative because that is part of an operator not a variable, right?

Then if I use the chain rule one time, I get:

d/dx psi(z(x)) = d/dx psi(z) * d/dx z(x) from which I do not see how you get a d/dz operator into the equation

 

[Meir Achuz]

Sorry, it should have been sqrt{mE}.
If z=ax, d/dx=ad/dz.
Look at your calculus text.

 

[ehrenfest]

If z=ax, d/dx=ad/dz.

Sorry, I cannot find that rule in my calculus text. Does it have a name? I know how to substitute variables with integrals, but I have not seen it with derivatives. If you were going to do it with an integral it would be

dz = a*x*dx but how do you get the derivative operator form? Sorry again, I should probably know this!

 

[nrqed]

Sorry, I cannot find that rule in my calculus text. Does it have a name? I know how to substitute variables with integrals, but I have not seen it with derivatives. If you were going to do it with an integral it would be

dz = a*x*dx but how do you get the derivative operator form? Sorry again, I should probably know this!

z = a x

Now, \frac{d}{dx} = \frac{dz}{dx} \frac{d}{dz} = a \frac{d}{dz}

It's just the chain rule applied to a very simple case.

 

[ehrenfest]

z = a x

Now, \frac{d}{dx} = \frac{dz}{dx} \frac{d}{dz} = a \frac{d}{dz}

It's just the chain rule applied to a very simple case.

The chain rule states that if a = f(b) and if b = f(c), then da/dc = da/db * db/dc.

Your comment obviously makes sense with the Liebnex notation when you treat differentials like fractions, but I do not see how you get that directly from the chain rule. Apply, d/dx to both sides of the equation doesn't work...

 

[ehrenfest]

z = a x

Now, \frac{d}{dx} = \frac{dz}{dx} \frac{d}{dz} = a \frac{d}{dz}

It's just the chain rule applied to a very simple case.

I think I see now. You can just replace d/dx with any dy/dx for any function y and you get the chain rule. Just a little too abstract for me!

 

[nrqed]

I think I see now. You can just replace d/dx with any dy/dx for any function y and you get the chain rule. Just a little too abstract for me!

To show how it comes from the chain rule, consider a function y(z) and z is a function of x itself z(x).
Then the chain rule says

dy/dx = dy/dz dz/dx

Now let's say that z= ax. Then we have dz/dx = a (where a is a constant)

So

dy/dx = a dy/dz

In this thread, you must think of the wavefunction psi as playing the role of y.

Psi is initially a function of x. But then you imagine replacing all the x interms fo z in order to get a function psi(z). Plugging back in the Schrödinger equation, you then have to deal with calculating

\frac{d \psi(z)}{dx}. Using the rule above, you see that this is equal to a \frac{d \psi(z)}{dz}.