Shrodinger's equation step: -h^2/2m d^2psi/dx^2 + 1/2mw^2x^2psi = Epsi

Vol

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In schrodinger's equation it says

-h^2/2m d^2psi/dx^2 + 1/2mw^2x^2psi = Epsi which is equal to
d^2psi/dx^2 + (2mE/h^2 - m^2w^2/h^2 times x^2)psi = 0. I am completely lost. How did step 2 come out of step 1? Did we divide by E?

V(x) = 1/2mw^2x^2 by the way. So, it was inserted into Vpsi.

Help! :(
 
In schrodinger's equation it says

-h^2/2m d^2psi/dx^2 + 1/2mw^2x^2psi = Epsi which is equal to
d^2psi/dx^2 + (2mE/h^2 - m^2w^2/h^2 times x^2)psi = 0. I am completely lost. How did step 2 come out of step 1? Did we divide by E?

V(x) = 1/2mw^2x^2 by the way. So, it was inserted into Vpsi.

Help! :(

I think, if I'm interpreting your equations correctly, that you are asking how to get from

\(\displaystyle \displaystyle -\frac{h^2}{2m} \frac{d^2\psi}{dx^2} + \frac{1}{2} mw^2x^2\psi = E\psi\)

to

\(\displaystyle \displaystyle\frac{d^2\psi}{dx^2} +\left (\frac{2mE}{h^2} - \frac{m^2w^2}{h^2} \cdot x^2\right)\psi = 0\)

The answer is to multiply everything by \(\displaystyle \displaystyle-\frac{2m}{h^2}\) , and then move terms around a little. Do you see that?
 
Minor correction: Dr. Peterson is correct but there is a bit of a typo in the OP: We are using \(\displaystyle \hbar \equiv \dfrac{h}{2 \pi}\) and not Planck's constant, h, in Schrodinger's equation. (Perhaps the typo is an effort to make the typing easier?)

-Dan
 
Reply

I think, if I'm interpreting your equations correctly, that you are asking how to get from

\(\displaystyle \displaystyle -\frac{h^2}{2m} \frac{d^2\psi}{dx^2} + \frac{1}{2} mw^2x^2\psi = E\psi\)

to

\(\displaystyle \displaystyle\frac{d^2\psi}{dx^2} +\left (\frac{2mE}{h^2} - \frac{m^2w^2}{h^2} \cdot x^2\right)\psi = 0\)

The answer is to multiply everything by \(\displaystyle \displaystyle-\frac{2m}{h^2}\) , and then move terms around a little. Do you see that?

Oh, I see now. But why are we doing that? What is -2m/h^2?
 
Oh, I see now. But why are we doing that? What is -2m/h^2?

You'll have to look at what their goal is; I'm totally ignoring context (such as what the equation means, as has been pointed out) and just answering your basic algebra question (what they did). But doesn't it make sense to get the second derivative by itself, just to clean things up?
 
Oh, I see now. But why are we doing that? What is -2m/h^2?
I've never seen an actual derivation of the S-equation, but there are a few things we can put together.

There is a rule of thumb to translate Classical formulas to Quantum ones. One of these is how to get momentum: We take p and replace it with \(\displaystyle -i \hbar \dfrac{d}{dx}\). (This is an operator expression. We need this to act on the wavefunction \(\displaystyle \psi\).)

Now, the S-equation is basically Kinetic energy + potential energy = total energy. Classically speaking kinetic energy is \(\displaystyle \dfrac{p^2}{2m}\) so translating this into QM:
\(\displaystyle \dfrac{p^2}{2m} \to \dfrac{ \left ( -i \hbar \dfrac{d}{dx} \right ) \left ( -i \hbar \dfrac{d}{dx} \right ) }{2m} = - \dfrac{\hbar ^2}{2m} \dfrac{ d^2 }{ dx^2}\).

then apply it to \(\displaystyle \psi\).

Why does this work? Notice that \(\displaystyle -i \hbar \dfrac{d}{dx}\) has units of momentum. Otherwise, well..... It works, let's just leave it at that. Unless you know how to calculate expectation values?

-Dan
 
still lost

I am still lost. What allows us to multiply all the terms by -2m/h^2? That would change the equation would it not?
 
I am still lost. What allows us to multiply all the terms by -2m/h^2? That would change the equation would it not?

This is basic algebra. If you have an equation, a = b, and multiply all terms by any nonzero number c, the resulting equation (ac = bc) is equivalent to the original -- that is, one is true whenever the other is true.

All we are doing here is multiplying both sides of an equation by a number, -2m/h^2, so the new equation means the same thing as the original.

What changes an equation is when you do different things to each side.
 
Oh

This is basic algebra. If you have an equation, a = b, and multiply all terms by any nonzero number c, the resulting equation (ac = bc) is equivalent to the original -- that is, one is true whenever the other is true.

All we are doing here is multiplying both sides of an equation by a number, -2m/h^2, so the new equation means the same thing as the original.

What changes an equation is when you do different things to each side.

Oh, that's right. If you do the same thing to both sides they are still equal. I forgot to focus on the equal sign. Thanks.
 
Oh, that's right. If you do the same thing to both sides they are still equal. I forgot to focus on the equal sign. Thanks.

Yes, if it were just an expression, it would be wrong to multiply by something, which would change its value.
 
Next step

Now y = sqrt mw/h x. When you apply it to the above you get
d^2psi(y)/dy^2 + (2E/hw - y^2)psi(y) = 0
How did we get 2E/hw?
 
Now y = sqrt mw/h x. When you apply it to the above you get
d^2psi(y)/dy^2 + (2E/hw - y^2)psi(y) = 0
How did we get 2E/hw?
Can you get to the point where the equation is
\(\displaystyle \dfrac{m \omega}{\hbar} \dfrac{d^2 \psi}{dy^2} + \left ( \dfrac{2mE}{\hbar ^2} - \dfrac{m \omega }{\hbar } y^2 \right ) \psi = 0\)

We want to get rid of the coefficient of the second derivative term. So multiply both sides by \(\displaystyle \dfrac{ \hbar }{m \omega }\) :
\(\displaystyle \left ( \dfrac{\hbar }{m \omega } \right ) \cdot \left ( \dfrac{m \omega}{\hbar} \dfrac{d^2 \psi}{dy^2} + \left ( \dfrac{2mE}{\hbar ^2} - \dfrac{m \omega }{\hbar } y^2 \right ) \psi \right ) = 0\)

\(\displaystyle \left ( \dfrac{\hbar }{m \omega } \cdot \dfrac{m \omega}{\hbar} \dfrac{d^2 \psi}{dy^2} \right ) + \left ( \dfrac{ \hbar }{m \omega } \right ) \cdot \left ( \dfrac{2mE}{\hbar ^2} - \dfrac{m \omega }{\hbar } y^2 \right ) \psi = 0\)

\(\displaystyle \dfrac{d^2 \psi}{dy^2} + \left ( \dfrac{ \hbar }{m \omega } \cdot \dfrac{2mE}{\hbar ^2} - \dfrac{ \hbar }{m \omega } \cdot \dfrac{m \omega }{\hbar } y^2 \right ) \psi = 0\)

\(\displaystyle \dfrac{d^2 \psi}{dy^2} + \left ( \dfrac{2E}{\hbar \omega } - y^2 \right ) \psi = 0\)

Let us know if you have an difficulties with this.

-Dan
 
Thanks.

OK. Thanks. I'll be back. Fortunately, I have a federal job and plenty of goof off time to study this. :)
 
I'm back

quantfinal4x.gif
is as far as I got.

So, you are supposed to apply

quantfinal5x.gif
into it. This is where I get lost. How do you get the following when you insert it into it? Thanks.

quantfinal6x.gif
 
quantfinal4x.gif
is as far as I got.

So, you are supposed to apply

quantfinal5x.gif
into it. This is where I get lost. How do you get the following when you insert it into it? Thanks.

quantfinal6x.gif
Vol, take a pencil and paper (instead of staring at the screen) - write the equations down and apply some algebra to those equations.

You know where you start from - you know where you end up. Now you we need to figure out the operation to get there (by the way that was explicitly stated in some posts!).
 
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Missing something

This is like Chinese water torture! OK, here is what I got but I am missing something:

From
quantfinal5x.gif
I get:

m = hy^2/wx^2 and
x^2 = hy^2/wx^2 and
y^2/x^2 = mw/h

When I plug these into

quantfinal4x.gif


I get

[FONT=&quot] [/FONT]
quantfinal6x.gif
but with the second term multiplied by mw/h

So, then I multiply all the terms by h/mw to get rid of it?
But then for the first term I end up with

h/mw d^2Y(y)/dy^2 ...what does this mean? The second derivative of psi has to be multiplied by h/mw? But that is not in the next step which should be

quantfinal6x.gif


There's no h/mw in the first term. What am I not getting?
 
I keep getting:

psi" + mw/h(2E/wh - y^2)psi(y) = 0

If I multiply this with h/mw to get rid of the mw/h I get h/mw psi". Seems like I am going around in circles.
 
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How did you get this?

Can you get to the point where the equation is
\(\displaystyle \dfrac{m \omega}{\hbar} \dfrac{d^2 \psi}{dy^2} + \left ( \dfrac{2mE}{\hbar ^2} - \dfrac{m \omega }{\hbar } y^2 \right ) \psi = 0\)

We want to get rid of the coefficient of the second derivative term. So multiply both sides by \(\displaystyle \dfrac{ \hbar }{m \omega }\) :
\(\displaystyle \left ( \dfrac{\hbar }{m \omega } \right ) \cdot \left ( \dfrac{m \omega}{\hbar} \dfrac{d^2 \psi}{dy^2} + \left ( \dfrac{2mE}{\hbar ^2} - \dfrac{m \omega }{\hbar } y^2 \right ) \psi \right ) = 0\)

\(\displaystyle \left ( \dfrac{\hbar }{m \omega } \cdot \dfrac{m \omega}{\hbar} \dfrac{d^2 \psi}{dy^2} \right ) + \left ( \dfrac{ \hbar }{m \omega } \right ) \cdot \left ( \dfrac{2mE}{\hbar ^2} - \dfrac{m \omega }{\hbar } y^2 \right ) \psi = 0\)

\(\displaystyle \dfrac{d^2 \psi}{dy^2} + \left ( \dfrac{ \hbar }{m \omega } \cdot \dfrac{2mE}{\hbar ^2} - \dfrac{ \hbar }{m \omega } \cdot \dfrac{m \omega }{\hbar } y^2 \right ) \psi = 0\)

\(\displaystyle \dfrac{d^2 \psi}{dy^2} + \left ( \dfrac{2E}{\hbar \omega } - y^2 \right ) \psi = 0\)

Let us know if you have an difficulties with this.

-Dan

How did you get the equation you started with? Seems like we have different versions of the same thing?
 
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