How do we normalize an odd function in the Schrodinger equation?

[CollectiveRocker]

I realize that when we normalize a solution to the Schrödinger equation, that we are setting it equal to 1, in order to maximize our chances of finding it. The question which I have is how do you normalize : lψl^2 = Axe^((-x^2)/2). The problem which I see is the e^(-x^2)/2). What would you fellas suggest as the best way to go about this?

 

[da_willem]

Depending on your limits you can maybe use:

$$\int _0^\infty x e^{-ax^2}dx=\frac{1}{2a}$$

Note that you can also use this if your integration limits are from minus infinity to infinity. Just multiply the above by two, because your function is symmetric. If you have other limits you can only approximate the integral.

btw; you're not normalizing a wavefunction to maximize your chances of finding it, but just to make sure you get a chance of 1 for finding it somewhere.

PS: here you can find some more integrals: http://www.sosmath.com/tables/integral/integ38/integ38.html

 

[quantumdude]

The question which I have is how do you normalize : lψl^2 = Axe^((-x^2)/2).

u-substitution would work like butta'.

u=-x²/2
du=-xdx

etc
.
.
.

 

[ZapperZ]

Depending on your limits you can maybe use:

$$\int _0^\infty x e^{-ax^2}dx=\frac{1}{2a}$$

Note that you can also use this if your integration limits are from minus infinity to infinity. Just multiply the above by two, because your function is symmetric. If you have other limits you can only approximate the integral.

btw; you're not normalizing a wavefunction to maximize your chances of finding it, but just to make sure you get a chance of 1 for finding it somewhere.

PS: here you can find some more integrals: http://www.sosmath.com/tables/integral/integ38/integ38.html

Actually, the function isn't symmetry, since x is an odd function while the argument of the exponential is an even function. So you end up with an odd function. So in this case, only the limits from 0 to infinity is the one that would give a non-zero value, since -inf to +inf will add up to zero.

Zz.

 

[Gokul43201]

Actually, the function isn't symmetry, since x is an odd function while the argument of the exponential is an even function. So you end up with an odd function. So in this case, only the limits from 0 to infinity is the one that would give a non-zero value, since -inf to +inf will add up to zero.

Zz.

Which makes me suspect that the given function, $xe^{-x^2/2}=\psi (x)~not~ \psi ^2 (x)$, or it may be $\psi (r)$...

PS : Normalization is not about maximizing. It is simply ensuring existence.

 

[ZapperZ]

Which makes me suspect that the given function, $xe^{-x^2/2}=\psi (x)~not~ \psi ^2 (x)$, or it may be $\psi (r)$...

PS : Normalization is not about maximizing. It is simply ensuring existence.

Well, there's nothing wrong with that function being the wave function itself and not the the squared cousin. For example, this wavefunction can easily be a "trial" ground state wavefunction in spherical coordinates when we do the variational method. So here, x=r. Thus only the region for r = 0 to +infinity is physical. That's the region where it has to be normalized.

Zz.

Edit: Oh, I see what you mean. I reread the original posting and it does appear that the wavefunction itself had a sqrt(x) factor. I suppose there's really nothing wrong with that... it makes it consistent that the region of interest is only from 0 to +infinity, or else one would have an imaginary function in there...

 

[da_willem]

Actually, the function isn't symmetry, since x is an odd function while the argument of the exponential is an even function. So you end up with an odd function. So in this case, only the limits from 0 to infinity is the one that would give a non-zero value, since -inf to +inf will add up to zero.

Zz.

Ofcourse you are absolutely right.