Straightforward integration…

[Guineafowl]

TL;DR

…keeps coming out wrong, despite each step being seemingly right, as checked on an online integral calculator. It may simply be a silly mistake with algebra. Self study.

Solving the integral for a wave function:
$$1 = |C|^2 \int_0^\infty \frac{x^2}{a^2} e^{-2x/a} dx$$

Shunting out the $\frac{1}{a^2}$ term, I went for integration by parts, a couple of times, first one being:$u=x^2, \frac {du}{dx}=2x, \frac {dv}{dx}=e^{-2x/a}, v=\frac {-ae^{-2x/a}}{2}$

Now, I come up with:
$$\frac {-ax^2e^{-2x/a}}{2} - \frac {2a^2 xe^{-2x/a}+a^3e^{-2x/a}}{4}$$

Multiplying through by the $\frac{1}{a^2}$ term, and simplifying (ignoring the $|C|^2$ term and definite integral for now):
$$\frac {-(2x^2-2ax+a^2)e^{-2x/a}}{4a}$$

The right answer has a $+2ax$ instead of the $-2ax$ and I can’t find where it’s coming from, despite checking each integration step individually on the integral calculator. If someone could point out my idiocy, I’d be grateful.

Also, some guidance on evaluating the integral between 0 and $\infty$, which I’m not familiar with, would be appreciated.

 

[PeroK]

TL;DR: …keeps coming out wrong, despite each step being seemingly right, as checked on an online integral calculator. It may simply be a silly mistake with algebra. Self study.

Solving the integral for a wave function:
$$1 = |C|^2 \int_0^\infty \frac{x^2}{a^2} e^{-2x/a} dx$$

Shunting out the $\frac{1}{a^2}$ term, I went for integration by parts, a couple of times, first one being:$u=x^2, \frac {du}{dx}=2x, \frac {dv}{dx}=e^{-2x/a}, v=\frac {-ae^{-2x/a}}{2}$

Now, I come up with:
$$\frac {-ax^2e^{-2x/a}}{2} - \frac {2a^2 xe^{-2x/a}+a^3e^{-2x/a}}{4}$$

Multiplying through by the $\frac{1}{a^2}$ term, and simplifying (ignoring the $|C|^2$ term and definite integral for now):
$$\frac {-(2x^2-2ax+a^2)e^{-2x/a}}{4a}$$

The right answer has a $+2ax$ instead of the $-2ax$ and I can’t find where it’s coming from, despite checking each integration step individually on the integral calculator. If someone could point out my idiocy, I’d be grateful.

Also, some guidance on evaluating the integral between 0 and $\infty$, which I’m not familiar with, would be appreciated.

I'm struggling to see what you are doing with integration by parts. I applied parts twice, with the initial tern vanishing at both end points in both cases, and the remaining integrands being:
$$\frac x a e^{-\frac{2x}{a}}$$And$$\frac 1 2 e^{-\frac{2x}{a}}$$

 

[Guineafowl]

Ok, how about simplifying to:
$$\int x^2e^{-2x/a}dx$$

Using the integration by parts as above, ie $u=x^2, \frac {du}{dx}=2x, \frac {dv}{dx}=e^{-2x/a}, v=\frac {-ae^{-2x/a}}{2}$
The IBP formula $uv-\int v\frac{du}{dx}dx$ gives me:
$$\frac {-ax^2e^{-2x/a}}{2} - \int -ae^{-2x/a} x dx$$

Call this {equation 1}.

I can’t use the preview, so let’s see what you think of that, maybe?

 

[Guineafowl]

Tackling:
$$\int -ae^{-2x/a} x dx$$
$$-a \int xe^{-2x/a}dx$$

IBP again: $u=x, \frac {du}{dx}=1, \frac {dv}{dx}=e^{-2x/a}, v=\frac {-ae^{-2x/a}}{2}$

$$-a[\frac{-axe^{-2x/a}}{2}-(\frac {-a}{2}\frac{-ae^{-2x/a}}{2})]$$

$$ \frac{2a^2xe^{-2x/a} + a^3e^{-2x/a}{4} $$

 

[PeroK]

Ok, how about simplifying to:
$$\int x^2e^{-2x/a}dx$$

Using the integration by parts as above, ie $u=x^2, \frac {du}{dx}=2x, \frac {dv}{dx}=e^{-2x/a}, v=\frac {-ae^{-2x/a}}{2}$
The IBP formula $uv-\int v\frac{du}{dx}dx$ gives me:
$$\frac {-ax^2e^{-2x/a}}{2} - \int -ae^{-2x/a} x dx$$

I can’t use the preview, so let’s see what you think of that, maybe?

I'm not sure why you are calculating an indefinite integral. The first rule of mathematical physics is that the first term in a parts expansion always vanishes!

 

[Guineafowl]

I'm not sure why you are calculating an indefinite integral. The first rule of mathematical physics is that the first term in a parts expansion always vanishes!

Beyond my pay grade, I think ;)

 

[Guineafowl]

Combining the result in post #4, which for some reason isn’t displaying, with Equation 1 of post #3:

$$\frac{-ax^2 e^{-2x/a}}{2} - \frac {2a^2xe^{-2x/a} + a^3e^{-2x/a}}{4}$$

Which gives the extra minus sign that’s causing all the trouble.

 

[PeroK]

Combining the result in post #4, which for some reason isn’t displaying, with Equation 1 of post #3:

$$\frac{-ax^2 e^{-2x/a}}{2} - \frac {2a^2xe^{-2x/a} + a^3e^{-2x/a}}{4}$$

Which gives the extra minus sign that’s causing all the trouble.

Why do you think that's wrong?

 

[Guineafowl]

Why do you think that's wrong?

The integral-calculator.com has it like this:
$$-\frac{\left(2x^{2} + 2ax + a^{2}\right) \mathrm{e}^{-\frac{2x}{a}}}{4a}$$

Versus my answer:
$$-\frac {(2x^2-2ax+a^2)e^{-2x/a}}{4a}$$

And when asked to do the definite integral, it’s correct: $|C|^2\frac {a}{4}=1$ according to the notes, (J Cresser, Quantum Physics Notes).

 

[PeroK]

The integral-calculator.com has it like this:
$$-\frac{\left(2x^{2} + 2ax + a^{2}\right) \mathrm{e}^{-\frac{2x}{a}}}{4a}$$

Versus my answer:
$$-\frac {(2x^2-2ax+a^2)e^{-2x/a}}{4a}$$

What you posted in post #7 is consistent with the integral calculator answer. It looks like you got a minus sign confused in the final step.

 

[Guineafowl]

What you posted in post #7 is consistent with the integral calculator answer. It looks like you got a minus sign confused in the final step.

Strewth, are we back to basics again? You may remember me from such threads as:

G

High School Thread 'Derivative of a particle’s energy'

May 10, 2022

I’d appreciate some help with a mathematical block that I’m sure is trivial to most of you.

Given the expression (1):

Take the derivative of E with respect to a, set to zero and solve for a. Answer is shown at the bottom.

This is not homework; I’m following an account of the development of quantum physics. I’m out of practice with my calculus.

Now, I can see an a^2 term that should differentiate to 2a, and another a that should go to 1, but several attempts have not led to the answer. Would someone mind taking me through it?

• Guineafowl
• Derivative Energy
• Replies: 18
• Forum: Quantum Physics

Or, could have blanked from your mind the experience of dragging a biologist through simple algebra and calculus. I’ll have another look later; cheers for now.

 

[PeroK]

Combining the result in post #4, which for some reason isn’t displaying, with Equation 1 of post #3:

$$\frac{-ax^2 e^{-2x/a}}{2} - \frac {2a^2xe^{-2x/a} + a^3e^{-2x/a}}{4}$$

Which gives the extra minus sign that’s causing all the trouble.

All three terms have minus signs there, when you expand it out.