Derivative of a particle’s energy

[Guineafowl]

TL;DR

Given an expression for the energy of a particle’s energy in a Coulomb well, how to take the derivative (then set zero, then solve)

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?

 

[PeterDonis]

Given the expression

Posting equations in images is not allowed. Please post your equations using the PF LaTeX feature. You will find a "LaTeX Guide" link at the bottom left of the post window.

 

[Nugatory]

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?

You don't have a $a^2$ term, you have an $a^{-2}$ term that differentiates to $-2a^{-3}$, and similarly you have an $a^{-1}$ not an $a$. Get this right and the algebra will come home for you.

 

[Guineafowl]

$E \approx \frac {\hbar^2} {2ma^2} - \frac {e^2} {4\pi\varepsilon_0 a}$ Original equation

$\frac {dE} {da} = \frac {\hbar^2 (-2a^-3)} {2m} - \frac {e^2 (-a^-2)} {4\pi\varepsilon_0} = 0$ Take derivative and set to zero

$\frac {\hbar^2 (-2a^-3)} {2m} = \frac {e^2 (-a^-2)} {4\pi\varepsilon_0}$ Now (try to) solve for a

$\frac {\hbar^2 4\pi\varepsilon_0} {2me^2} = \frac {-a^-2} {-2a^-3}$ Is as far as my feeble math(s) (last used 24 years ago) will get me.

 

[PeroK]

Okay so far. What's next?

 

[Guineafowl]

Okay so far. What's next?

I’d like to multiply through by 2 to change the $2me^2$ into $me^2$.

$\frac {\hbar^2 4\pi\varepsilon_0} {me^2} = \frac {-2a^-2} {-2a^-3}$ Is that ok?

$\frac {\hbar^2 4\pi\varepsilon_0} {me^2} = \frac {-2a^3} {-2a^2}$ Can I invert the right-hand side?

$\frac {\hbar^2 4\pi\varepsilon_0} {me^2} = a$ Cancel the -2, and subtract the exponents?

 

[PeroK]

$\frac {\hbar^2 4\pi\varepsilon_0} {me^2} = a$ Cancel the -2, and subtract the exponents?

Is that the Bohr radius, then?

 

[Guineafowl]

Is that the Bohr radius, then?

Yes, about 0.5nm.

The next stage, apparently, is the minimum energy of 13.6 eV, given by:

$E_{min} \approx - \frac {1} {2} \frac {me^4} {(4\pi\varepsilon_0)^2 \hbar^2}$

Now, taking the derivative of the initial equation, and setting it to zero is what I would call finding the minimum, and we’ve done that. So how is the $E_{min}$ derived?

 

[PeroK]

Yes, about 0.5nm.

The next stage, apparently, is the minimum energy of 13.6 eV, given by:

$E_{min} \approx - \frac {1} {2} \frac {me^4} {(4\pi\varepsilon_0)^2 \hbar^2}$

Now, taking the derivative of the initial equation, and setting it to zero is what I would call finding the minimum, and we’ve done that. So how is the $E_{min}$ derived?

Strictly speaking you haven't shown that the energy is a minimum and not a maximum at the Bohr radius. Although that's not too hard to do.

To get $E_{min}$, you simply plug in the value of $a$ you found.

 

[Guineafowl]

Strictly speaking you haven't shown that the energy is a minimum and not a maximum at the Bohr radius. Although that's not too hard to do.

Take the second derivative:
$\frac {d^2E} {da^2} = \frac {\hbar^2(6a^{-4})} {2m} - \frac {e^2(2a^{-3})} {4\pi\varepsilon_0}$

Plug in the known values, and see if it’s positive (increasing bend)? I get $+3.805x10^{68}$ so I guess it is a minimum.

To get Emin, you simply plug in the value of a you found

How is the expression derived? There’s a $1/2 {term}^2$ involved, so is there some integration?

 

[PeroK]

Take the second derivative:
$\frac {d^2E} {da^2} = \frac {\hbar^2(6a^{-4})} {2m} - \frac {e^2(2a^{-3})} {4\pi\varepsilon_0}$

Plug in the known values, and see if it’s positive (increasing bend)? I get $+3.805x10^{68}$ so I guess it is a minimum.

It was simpler to note that as $a \rightarrow 0$ (with $a > 0$), then $E(a) \to +\infty$. And, as $a \to \infty$, then $E(a) \to 0$. As you have only one turning point, it must be a minimum.

I'll also show you a trick for handling cases like this. We have a function of the form:
$$E(a) = \frac A {a^2} - \frac B a$$Differentiating this gives:
$$E'(a) = -\frac{2A}{a^3} + \frac{B}{a^2}$$Now, $E'(a) = 0$ implies that:
$$a = \frac{2A}{B}$$Note that you can now substitute $A = \frac{\hbar^2}{2m}$ and $B = \frac{e^2}{4\pi\epsilon_0}$ to get the expression for $a$ at the turning point.

To see that this is a minimim, we can find the second derivative:
$$E''(a) = \frac{6A}{a^4} - \frac{2B}{a^3} = \frac{2}{a^4}\big (3A - Ba \big )$$And
$$E''(a = \frac{2A}{B}) = \frac{2}{a^4}\big (3A - 2A \big ) = \frac{2A}{a^4} > 0$$And we see that replacing those more complicated coefficients with $A$ and $B$ has made things a lot simpler to manage.

How is the expression derived? There’s a $1/2 {term}^2$ involved, so is there some integration?

I don't know why there would be integration?! From the above, we have:
$$E(a = \frac{2A}{B}) = \frac{AB^2}{4A^2} - \frac{B^2}{2A} = -\frac{B^2}{4A}$$
Another tip now is to evaluate $A$ and $B$ using the known constants and then evaluate that expression. I would use a spreadsheet, although some students still prefer pen and paper as you can't take a spreadsheet into an exam.

 

[Guineafowl]

It was simpler to note that as $a \rightarrow 0$ (with $a > 0$), then $E(a) \to +\infty$. And, as $a \to \infty$, then $E(a) \to 0$. As you have only one turning point, it must be a minimum.

I'll also show you a trick for handling cases like this. We have a function of the form:
$$E(a) = \frac A {a^2} - \frac B a$$Differentiating this gives:
$$E'(a) = -\frac{2A}{a^3} + \frac{B}{a^2}$$Now, $E'(a) = 0$ implies that:
$$a = \frac{2A}{B}$$Note that you can now substitute $A = \frac{\hbar^2}{2m}$ and $B = \frac{e^2}{4\pi\epsilon_0}$ to get the expression for $a$ at the turning point.

To see that this is a minimim, we can find the second derivative:
$$E''(a) = \frac{6A}{a^4} - \frac{2B}{a^3} = \frac{2}{a^4}\big (3A - Ba \big )$$And
$$E''(a = \frac{2A}{B}) = \frac{2}{a^4}\big (3A - 2A \big ) = \frac{2A}{a^4} > 0$$And we see that replacing those more complicated coefficients with $A$ and $B$ has made things a lot simpler to manage.

I don't know why there would be integration?! From the above, we have:
$$E(a = \frac{2A}{B}) = \frac{AB^2}{4A^2} - \frac{B^2}{2A} = -\frac{B^2}{4A}$$
Another tip now is to evaluate $A$ and $B$ using the known constants and then evaluate that expression. I would use a spreadsheet, although some students still prefer pen and paper as you can't take a spreadsheet into an exam.

Thanks once again. I‘ll go through that when I next have a chance.

 

[Guineafowl]

I'll also show you a trick for handling cases like this. We have a function of the form:
$$E(a) = \frac A {a^2} - \frac B a$$Differentiating this gives:
$$E'(a) = -\frac{2A}{a^3} + \frac{B}{a^2}$$Now, $E'(a) = 0$ implies that:
$$a = \frac{2A}{B}$$Note that you can now substitute $A = \frac{\hbar^2}{2m}$ and $B = \frac{e^2}{4\pi\epsilon_0}$ to get the expression for $a$ at the turning point.

To see that this is a minimim, we can find the second derivative:
$$E''(a) = \frac{6A}{a^4} - \frac{2B}{a^3} = \frac{2}{a^4}\big (3A - Ba \big )$$And
$$E''(a = \frac{2A}{B}) = \frac{2}{a^4}\big (3A - 2A \big ) = \frac{2A}{a^4} > 0$$And we see that replacing those more complicated coefficients with $A$ and $B$ has made things a lot simpler to manage.

I don't know why there would be integration?! From the above, we have:
$$E(a = \frac{2A}{B}) = \frac{AB^2}{4A^2} - \frac{B^2}{2A} = -\frac{B^2}{4A}$$
Another tip now is to evaluate $A$ and $B$ using the known constants and then evaluate that expression. I would use a spreadsheet, although some students still prefer pen and paper as you can't take a spreadsheet into an exam.

I think the message is, don’t give up my day job! There are yawning gaps in my knowledge. For example, I would not have known to factor out as you have done, or been particularly confident with it here:
$E''(a) = \frac{6A}{a^4} - \frac{2B}{a^3} = \frac{2}{a^4}\big (3A - Ba \big )$

And would not have known to leave the $a^4$ term alone, but would have got bogged down trying to evaluate ${\frac{2A}{B}}$ to the fourth power here:

$E''(a = \frac{2A}{B}) = \frac{2}{a^4}\big (3A - 2A \big ) = \frac{2A}{a^4} > 0$

 

[PeroK]

I think the message is, don’t give up my day job! There are yawning gaps in my knowledge. For example, I would not have known to factor out as you have done, or been particularly confident with it here:
$E''(a) = \frac{6A}{a^4} - \frac{2B}{a^3} = \frac{2}{a^4}\big (3A - Ba \big )$

It's a common idea. The trick is to check that what you end up with reduces to what you started with:
$$\frac{2}{a^4}\big (3A - Ba \big ) = \frac{2(3A)}{a^4} - \frac{2Ba}{a^4} = \frac{6A}{a^4} - \frac{2B}{a^3}$$It's always worth a quick backward check!

And would not have known to leave the $a^4$ term alone, but would have got bogged down trying to evaluate ${\frac{2A}{B}}$ to the fourth power here:

Which is precisely what I wanted to avoid.

And I can’t even follow this:
$\frac{AB^2}{4A^2} - \frac{B^2}{2A} = -\frac{B^2}{4A}$

It's just cancelling the $A$ in the first fraction then using the lowest common denominator:
$$\frac{AB^2}{4A^2} - \frac{B^2}{2A} = \frac{B^2}{4A} - \frac{2B^2}{4A} = -\frac{B^2}{4A}$$

 

[Guineafowl]

Yes, I did get that last bit just after I posted!

Anyway, this has been very helpful and interesting, and thanks for so patiently flogging this dead horse.

 

[PeroK]

Yes, I did get that last bit just after I posted!

Anyway, this has been very helpful and interesting, and thanks for so patiently flogging this dead horse.

On serious note, QM is full of hard algebra and messy equations. And, even when you escape those it's only into the world of formal linear algebra.

There is no way to avoid putting in the hard yards of all those practice problems at every level of physics and mathematics to hone your maths skills.

 

[Guineafowl]

It certainly looks that way. I was hoping to go a stage or two beyond the pop sci accounts, as without the mathematics, they just read like science fiction.

I did physics to ‘A’ level at school (one stage below an undergrad degree in UK), but we steered clear of QM other than de Broglie wavelengths and such. Instead of taking maths to the same level, I did Latin, which might have been regrettable, but I did enjoy it.

 

[PeroK]

It certainly looks that way. I was hoping to go a stage or two beyond the pop sci accounts, as without the mathematics, they just read like science fiction.

I did physics to ‘A’ level at school (one stage below an undergrad degree in UK), but we steered clear of QM other than de Broglie wavelengths and such. Instead of taking maths to the same level, I did Latin, which might have been regrettable, but I did enjoy it.

If you are not committed to your current course of study, there is an extremely insightful introduction to QM here that is lighter on the mathematics generally and is probably the most accessible treatment at undergraduate level:

http://physics.mq.edu.au/~jcresser/Phys304/Handouts/QuantumPhysicsNotes.pdf

 

[Guineafowl]

We’ve come full circle, because that’s what I am reading, and the equations discussed are from there! As you’ve seen, it presupposes a maths level beyond mine, but it is very good, so I’ll continue through it.

Many thanks again.