- #1

James_1978

- 37

- 3

- Homework Statement
- I am solving for the expectation energy for the deuteron. However, I must be missing something because my answer is very complicated compared to what is expected. I am looking at problem 4.3 from Krane.

- Relevant Equations
- ## <T> = \frac{\hbar^{2}}{2m} \int_{0}^{\infty} |\frac{\partial\psi}{\partial r}|^{2} ##

## <T> = \frac{\hbar^{2}}{2m} A^{2} [\frac{1}{2} k_{1}^{2} + \frac{1}{4}\sin{2k_{1}r} + \frac{k_{2}}{2}\sin{k_{1}r}^{2} ]##

##A\sin{k_{1}r}## for ##r < R##

##Ce^{-k_{2}r}## for ##r > R##

Dear Forum,

I am solving for the expectation value of the kinetic energy for the deuteron (Krane problem 4.3). I must be missing something since this has become far more complicated than I remember.

The problem is as follows:

## <T> = \frac{\hbar^{2}}{2m} \int_{0}^{\infty} |\frac{\partial\psi}{\partial r}|^{2} ##

And show

## <T> = \frac{\hbar^{2}}{2m} A^{2} [\frac{1}{2} k_{1}^{2} + \frac{1}{4}\sin{2k_{1}r} + \frac{k_{2}}{2}\sin{k_{1}r}^{2} ]##

The correct way (I believe) is to use the Laplacian with the wave functions for the deuteron which are:

##A\sin{k_{1}r}## for ##r < R##

And

##Ce^{-k_{2}r}## for ##r > R##

My attemp is as follows. For the kinetic energy we have ##<T> = <\frac{p^{2}}{2m}>## Where ##p = i\hbar\nabla^{2}## To solve for the expectation value I do the following.

##<T> = \int\Psi^{*}\nabla^{2}\Psi d^{3}\overrightarrow{r}##

##<T> = \int\Psi^{*}\frac{1}{r^{2}}\frac{\partial}{\partial r}(r^{2}\frac{\partial}{\partial r})\Psi d^{3}\overrightarrow{r}##

##<T> =\frac{-\hbar^{2}}{2m}[ \int_{0}^{R}A^{*}\sin{k_{1}}\frac{1}{r^{2}}\frac{\partial}{\partial r}(r^{2}\frac{\partial}{\partial r})A\sin{k_{1}}r^{2}dr + \int_{R}^{\infty}C^{*}e^{-k_{2}r}\frac{1}{r^{2}}\frac{\partial}{\partial r}(r^{2}\frac{\partial}{\partial r})Ce^{-k_{2}r}r^{2}dr]##

I understand we don't need to include the Angular components since they vannish with a symmetric wave function. What I don't understand is that when I do this above the equations becomes very complicated and does not reduce down as expected. However, I think we are only asked to do the following.## <T> = \frac{\hbar^{2}}{2m} \int_{0}^{\infty} |\frac{\partial\psi}{\partial r}|^{2} = \frac{-\hbar^{2}}{2m}[ \int_{0}^{R}A^{2}\sin^{2}{k_{1}}dr + \int_{R}^{\infty}C^{2}e^{-2k_{2}r}dr]##

My question why are we not supposed to apply the Laplacian? The answers works out but I am not sure why I do not need to use the Laplacian.

I am solving for the expectation value of the kinetic energy for the deuteron (Krane problem 4.3). I must be missing something since this has become far more complicated than I remember.

The problem is as follows:

## <T> = \frac{\hbar^{2}}{2m} \int_{0}^{\infty} |\frac{\partial\psi}{\partial r}|^{2} ##

And show

## <T> = \frac{\hbar^{2}}{2m} A^{2} [\frac{1}{2} k_{1}^{2} + \frac{1}{4}\sin{2k_{1}r} + \frac{k_{2}}{2}\sin{k_{1}r}^{2} ]##

The correct way (I believe) is to use the Laplacian with the wave functions for the deuteron which are:

##A\sin{k_{1}r}## for ##r < R##

And

##Ce^{-k_{2}r}## for ##r > R##

My attemp is as follows. For the kinetic energy we have ##<T> = <\frac{p^{2}}{2m}>## Where ##p = i\hbar\nabla^{2}## To solve for the expectation value I do the following.

##<T> = \int\Psi^{*}\nabla^{2}\Psi d^{3}\overrightarrow{r}##

##<T> = \int\Psi^{*}\frac{1}{r^{2}}\frac{\partial}{\partial r}(r^{2}\frac{\partial}{\partial r})\Psi d^{3}\overrightarrow{r}##

##<T> =\frac{-\hbar^{2}}{2m}[ \int_{0}^{R}A^{*}\sin{k_{1}}\frac{1}{r^{2}}\frac{\partial}{\partial r}(r^{2}\frac{\partial}{\partial r})A\sin{k_{1}}r^{2}dr + \int_{R}^{\infty}C^{*}e^{-k_{2}r}\frac{1}{r^{2}}\frac{\partial}{\partial r}(r^{2}\frac{\partial}{\partial r})Ce^{-k_{2}r}r^{2}dr]##

I understand we don't need to include the Angular components since they vannish with a symmetric wave function. What I don't understand is that when I do this above the equations becomes very complicated and does not reduce down as expected. However, I think we are only asked to do the following.## <T> = \frac{\hbar^{2}}{2m} \int_{0}^{\infty} |\frac{\partial\psi}{\partial r}|^{2} = \frac{-\hbar^{2}}{2m}[ \int_{0}^{R}A^{2}\sin^{2}{k_{1}}dr + \int_{R}^{\infty}C^{2}e^{-2k_{2}r}dr]##

My question why are we not supposed to apply the Laplacian? The answers works out but I am not sure why I do not need to use the Laplacian.