A Normalization of Morse potential wavefunctions

[Malamala]

Hello! I am trying to use the wavefunctions of a Morse potential as defined in the link provided. They define a parameter $z$ and the wavefunctions are in terms of z. In my particular case, given their definitions, I have $\lambda = 132.19377$, $a=1.318 A^{-1}$ and $R_e = 2.235 A$. I am not sure how to check the normalization of that function. Naively, I would expect to calculate:

$$\int_{z_i}^{z_f}{\psi_N(z)^2}dz$$

and this should be equal to one. However I am getting -263.388. For the limits of integration, I assume that $z_i$ corresponds to the case when $R=0$ i.e. $z_i = 2\lambda e^{aR}= 5029.85$ and $z_f =0$ i.e. when $R = \infty$. I am pasting below the Mathematica code I am using for the calculations for reference:

lmbda0 = 132.19377;
a0 = 1.318;
Re0 = 2.235;
xe0 = a0*Re0;
n0 = 0;
alpha0 = 2 lmbda0 - 2 n0 - 1;

f[R_?NumericQ] := (Exp[0.5*(Log[n0!*alpha0] - Log[Gamma[2 lmbda0 - n0]]) + (lmbda0 - n0 - 0.5) Log[z] - 0.5 *z + Log[LaguerreL[n0, alpha0, z]]])^2;

Plot[f[z], {z, 411.145, 163.424}]

NIntegrate[f[z], {z, 500, 100}]

However, if I instead integrate

$$\int_{z_i}^{z_f}{\psi_N(z)^2/z}dz$$

I get one, as expected. Am I doing something wrong? Is the normalization defined differently? Or are my integration bounds wrong? I checked the formula on multiple websites and it seems to be correct, so I am not sure why I need to add by hand that $1/z$ Thank you!

 

[hutchphd]

You should not expect your naive assumption to work. The normalization is defined uniformly bover all space x. When you substitute for x you need to substitute for dx (in terms of dz and z ) as well. I think it will do what you need.

 

[Malamala]

You should not expect your naive assumption to work. The normalization is defined uniformly bover all space x. When you substitute for x you need to substitute for dx (in terms of dz and z ) as well. I think it will do what you need.

I am not sure I understand. If the function is defined in terms of $z$, shouldn't the usual normalization formula in quantum mechanics work? How would I know what is the variable of integration in this case ($z$, $x$, $R$ or something else)?

 

[hutchphd]

The normalization constant is a convention chosen for convenience. We choose it so that the wavefunction (mod) squared when integrated over all space yields unity.
What is the "usual" normalization formula? What you are assuming does not yield consistent results.

 

[Malamala]

The normalization constant is a convention chosen for convenience. We choose it so that the wavefunction (mod) squared when integrated over all space yields unity.
What is the "usual" normalization formula? What you are assuming does not yield consistent results.

I thought that in QM, a given wavefunction is normalized if

$$\int_{-\infty}^{\infty}\psi(x)^*\psi(x)dx = 1$$

in our case $x = z$, the boundaries are as I mentioned above and the function is real so we get just the square of the function. This is the formula I mentioned first above. And the normalization constant is specifically chosen such that this formula is fulfilled. I don't understand why it is wrong.

 

[hutchphd]

in our case x=z,

No. Please read the wikipedia article you quoted carefully.