The WKB approximation to the Schrodinger Equation

**bugatti79** · December 8th 2010, 02:16 PM

Hi all,

I am examining the approximate solutions to the Schrodinger equation using the WKB method. I have a simple differentiation query as follows. I want to find A.

$2A'(x) \phi (x)+A(x) \phi^{''} (x) = 0$ or

$(A^2(x) \phi^{'} (x))' = 0$ can be solved by

$A^2(x) \phi^{'} (x) = C^2$

or $A=\frac{C}{\sqrt{\phi^{'}}}$

How do you get to the 3rd or 4th line?

Thanks

**Ackbeet** · December 8th 2010, 05:34 PM

Integrate both sides. Or are you asking why they choose $C^{2}$ instead of $C?$

**bugatti79** · December 8th 2010, 10:16 PM

Originally Posted by Ackbeet

Integrate both sides. Or are you asking why they choose $C^{2}$ instead of $C?$

Both.

1) Integrating both sides of line 2 in post # 1

On first thoughts, I considered integrating both side but then the C^2 threw me off. On second thoughts I went ahead with it but with my own way of defining C. See attached.

2) Note line 1 in post 1 should read

$2A'(x) \phi' (x)+A(x) \phi^{''} (x) = 0$ . I left out a prime on phi.

Would it also be possible to integrate the line above as an alternative using integration by parts?

Thanks

**Ackbeet** · December 9th 2010, 02:44 AM

Well, when you integrate 0, you get a constant. You can call it whatever you want. In this situation, whoever is doing the integration must have had a reason for not wanting the constant to be negative (in this case, it's probably the square root that occurs later, thus avoiding complex numbers). Although, come to think of it, in quantum mechanics, complex numbers are common enough. I would need more context to be able to say why A must be real.

I left out a prime on phi.

Yep. I was able to figure that out due to context.

Actually, in order for you to have a total differential, you should have this original expression:

$2A(x)A'(x)\phi'(x)+A^{2}(x)\phi''(x)=0.$

If you start with this expression instead of your original expression, integration by parts works just fine. You'll get a cancellation that leads you to the same equation as above. Canceling an $A$ gives you your original expression.

Cheers.

**bugatti79** · December 9th 2010, 09:57 AM

Originally Posted by Ackbeet

Well, when you integrate 0, you get a constant. You can call it whatever you want. In this situation, whoever is doing the integration must have had a reason for not wanting the constant to be negative (in this case, it's probably the square root that occurs later, thus avoiding complex numbers). Although, come to think of it, in quantum mechanics, complex numbers are common enough. I would need more context to be able to say why A must be real.

Yep. I was able to figure that out due to context.

Actually, in order for you to have a total differential, you should have this original expression:

$2A(x)A'(x)\phi'(x)+A^{2}(x)\phi''(x)=0.$

If you start with this expression instead of your original expression, integration by parts works just fine. You'll get a cancellation that leads you to the same equation as above. Canceling an $A$ gives you your original expression.

Cheers.

A is real because i believe he assumes that the total energy is greater than the potential V(x), ie we are examining the 'classical region' or permissable region of the particle. In general psi is a complex function, thats my take.

ah ok, I see where the first and third terms cancel using your above expression integrated wrt to x (only the second term!) giving
$2\int A(x) A'(x) \phi ' (x)dx + \int \phi '' (x) A^2 (x)dx=\int0 dx$

$2\int A(x) A'(x) \phi ' (x)dx + A^2(x) \phi'(x) -2\int \phi'(x)A(x) A'(x)dx=C$

What I dont understand how you just chucked in a A(x) into your expression...

**Ackbeet** · December 9th 2010, 10:19 AM

What I dont understand how you just chucked in a A(x) into your expression...

Do you mean you don't understand how I did it? The answer there is that I can multiply an entire equation by any quantity whatsoever, and not change the equation. I just multiplied by the quantity A(x).

If you meant how did I see that I needed to do that? Well, by noticing that the LHS was not, as is, a total differential. It needed the A(x) multiplied in there (I saw that just by experience).

**bugatti79** · December 9th 2010, 11:05 AM

Originally Posted by Ackbeet

Do you mean you don't understand how I did it? The answer there is that I can multiply an entire equation by any quantity whatsoever, and not change the equation. I just multiplied by the quantity A(x).

If you meant how did I see that I needed to do that? Well, by noticing that the LHS was not, as is, a total differential. It needed the A(x) multiplied in there (I saw that just by experience).

Both and both answered :-)
Intuition and experience is wha I need to build up! How do you recognise a total differential in this case BTW

**Ackbeet** · December 9th 2010, 11:27 AM

It's a bit of an art. In this case, the double derivative of phi in one term, with the single derivative of phi in the other, means that the single derivative of phi must be one of the original pieces of the product that your differentiating. Also, if the coefficients of the two terms are different, as in this case, you can guess that the ratio of the coefficients (in this case, 2) is probably an exponent somewhere. That's how I got to the conclusion that the A should be squared in the left-most term. You have to use your most important mathematical tool of all time here: imagination. I imagined what the derivative of A^2 phi' would look like, and presto! it looked very similar, minus a factor of A which was easily supplied, to what we had.

**bugatti79** · December 9th 2010, 11:35 AM

Originally Posted by Ackbeet

It's a bit of an art. In this case, the double derivative of phi in one term, with the single derivative of phi in the other, means that the single derivative of phi must be one of the original pieces of the product that your differentiating. Also, if the coefficients of the two terms are different, as in this case, you can guess that the ratio of the coefficients (in this case, 2) is probably an exponent somewhere. That's how I got to the conclusion that the A should be squared in the left-most term. You have to use your most important mathematical tool of all time here: imagination. I imagined what the derivative of A^2 phi' would look like, and presto! it looked very similar, minus a factor of A which was easily supplied, to what we had.

Wow!

Intuition, experience and imagination. I have work to do! Thanks Ackbeet. Really Appreciated.

**Ackbeet** · December 9th 2010, 11:46 AM

Yeah, a lot of people thing mathematics is about formulas and methods of solving problems. Those are certainly important. But it's mostly about recognizing patterns. That requires imagination, in my opinion the most important quality a mathematician can possess. So don't neglect reading! I mean fiction, nonfiction, poetry, etc. That's some of the best imagination workout you can get. Avoid watching much television or movies, because that deadens the imagination.

**bugatti79** · December 9th 2010, 11:58 AM

Originally Posted by Ackbeet

Yeah, a lot of people thing mathematics is about formulas and methods of solving problems. Those are certainly important. But it's mostly about recognizing patterns. That requires imagination, in my opinion the most important quality a mathematician can possess. So don't neglect reading! I mean fiction, nonfiction, poetry, etc. That's some of the best imagination workout you can get. Avoid watching much television or movies, because that deadens the imagination.

Thats really interesting. You just confirmed my beliefs, imagination is the creative side of the brain whereas logical and analytical are the other side. So the likes of Einstein had both....he came up with mad ideas about the Universebut yet he could prove them analytically. 
My creative side isn’t very good but I also believe that the brain is very plastic and you can always improve your brains weakness to some extent based on this.

**Ackbeet** · December 9th 2010, 12:46 PM

Both halves of the brain will be important, you can bet.

**Danny** · December 10th 2010, 05:41 AM

Originally Posted by bugatti79

Hi all,

I am examining the approximate solutions to the Schrodinger equation using the WKB method. I have a simple differentiation query as follows. I want to find A.

$2A'(x) \phi (x)+A(x) \phi^{''} (x) = 0$ or

$(A^2(x) \phi^{'} (x))' = 0$ can be solved by

$A^2(x) \phi^{'} (x) = C^2$

or $A=\frac{C}{\sqrt{\phi^{'}}}$

How do you get to the 3rd or 4th line?

Thanks

I might add that you can separate and integrate, i.e.

$\dfrac{A'}{A} = - \dfrac{1}{2}\dfrac{\phi ''}{\phi '}\;\; \Rightarrow\;\;\ln A = - \frac{1}{2} \ln \phi ' + \ln C\;\;\Rightarrow\;\; A = C \phi'^{-1/2}$

**bugatti79** · December 11th 2010, 11:33 AM

Originally Posted by Danny

I might add that you can separate and integrate, i.e.

$\dfrac{A'}{A} = - \dfrac{1}{2}\dfrac{\phi ''}{\phi '}\;\; \Rightarrow\;\;\ln A = - \frac{1}{2} \ln \phi ' + \ln C\;\;\Rightarrow\;\; A = C \phi'^{-1/2}$

Well spotted!! Thanks Danny.

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