Schrodinger's equation

[PWrong]

I'm probably not the most qualified person to attack this problem, since I'm only a first year and a lot of my understanding is self taught. But anyway, I haven't had much to do recently, so for a couple of weeks I've been trying to solve Schrodinger's equation for 4D hydrogen. I should probably get a life.

These are the main steps involved, as far as I can tell from my physics textbook.

1. Convert the 4D laplacian into glomar coordinates
2. Use separation of variables on Schrodingers equation, using the potential well of 4D hydrogen, to get four ODEs.
3. The 4 ODE's should all have an extra constant, each of which has to be an integer.
4. Use the 4 integers to find the possible electron congurations.


There's a question in my textbook on converting Laplace's equation into spherical coordinates. After I answered it, I went on to the 4D case, which is pretty similar but takes more time to solve.

I used the following glomar coordinates(except with alpha, beta and gamma instead of a,b,c). Note that I changed some conventions to make it easier to extend to higher dimensions.

x = r sin a sin b sin c
y = r sin a sin b cos c
z = r sin a cos b
w = r cos a

u_xx + u_yy +u_zz +u_ww =
u_rr+3/r u_r + u_aa/r² + u_bb/(r²sin²a)+u_cc/(r²sin²asin²b)+2 u_a cot a / r + u_b cot b / (r²sin²a)

Edit:
I just fixed the conversion, I had two extra cos^2 bits in there.

I'm working on step 2 at the moment, trying to follow this derivation.
http://scienceworld.wolfram.com/physics ... nAtom.html

[jinydu]

Solving Schrodinger's Equation for 4D hydrogen?! Whoa, that's ambitious!

Well, the easiest way would be to use a computer program :wink:

[PWrong]

Lol, good point, using mathematica would probably be easier. I don't know how well it does at solving PDE's though. It also might not mention the integers. Before you try solving it though, I made a mistake in the conversion to glomar coordinates.

I also found a much easier way to do it. The explanation is on this page, just after the scary double sum and some stuff about tensors.
http://www.mathpages.com/home/kmath214/kmath214.htm

You have to work out the "line element" ds, that is, the distance travelled by a point if you change each of the coordinates.
For instance, if you change the angle a by da, then the distance travelled is ds = r*da

The total "line element" is
ds^2 = (dr)^2 + (r da)^2 + (r sin a db)^2 + (r sin a sin b dc)^2
The scale factors for r,a,b,c are
1, r, (r sin a) and (r sin a sin b) respectively

There's a general formula for the laplacian in any coordinate system, using the scale factors of the system. So apparently the laplacian is

DEL² u = 1/r^3 *d/dr( r^3 u_r) + 1/(r^2 sin²a) * d/da (sin²a u_a) + 1/(r^2 sin a sin b) * d/db (sin b u_b) + u_cc/(r^2 sin a sin b)
which expands out to a slightly different answer to what I had before. Rather than put it here I'll just edit my first post.

[PWrong]

Ok, here's the four equations we have to solve. The first two appear in the 3D derivation, so they should be easy. The third is similar to the Legendre equation, but not quite the same, so I'll have to try to find a series solution for it. The radial one will depend on the potential function we use.

C''(c) + m²C(c) = 0
B''(b) + cot b B'(b) + ( P(p) -m²/sin²b) B(b)=0
A''(a) + 2cot a A'(a) + (L(l) - P(p)/sin²a) A(a) = 0
R''(r) + 3/r R'(r) - ( 2m/h² (U(r) -E) +L(l)/r²)R=0

Note, m is an integer, and L(l) and P(p) are functions of integers l and p.
In the 3D case, they use -m^2 and L(l) = -l(l+1)
Here, we need 3 integers, m, l and p.

I suppose U(r) = 1/r^2 is the obvious choice for the potential function. I've already assumed that it has to be a radial function, so it can't have any kind of cylindrical symmetry or anything to change it.
I just hope it doesn't ruin the whole energy levels thing.

[PWrong]

Ok, I'm still stuck on the equations for A and R.

From B, we know that P = -p(p+1), which is annoying for the A equation.
Let x=cos a and y=A, to make it look more familiar. We get:

(1-x^2) y'' - 3x y' + (L - P/(1-x^2)) y = 0

This looks similar to the associated legendre equation, except with
-3x instead of -2x

Unfortunately mathworld doesn't show how to solve the associated equation, it just gives the answer.

I can get the series solution if P=0, (it turns out L(l) = l(l+2)).
Otherwise it gets too complicated. Can mathematica can find series solutions?

I'm very worried about the radial equation. When U=1/r^2, I don't think
energy will be quantised. :? The method used on mathworld simply isn't working. Maybe you could try mathematica on this one?

[houserichichi]

I plugged your differential equation into mathematica and came up with the following solution - I'll let you try and make heads or tails of it



where P and Q are the associated Legendre polynomials of the first and second kind, respectively. C_1 and C_2 are constants, and the exponent at the bottom that's almost impossible to read is 1/4. If only they had a step-by-step solution generator...

[Batman3]

I don't believe in QM but tht's beside the point.

I seem to remember from high school chemistry that the s2 orbital and the 3 p orbitals combine to form 4 sp orbitals. These give the second row of the periodic table(the first is H and He), with 8 possible electrons or 8 elements.. Water has 2 sp orbitals floating and 2 being shared with the 2 H. Since the 2 sp orbitals repel and the two hydrogen orbitals are not much charged, the net effect if somewhere between angle between the H of 120(for no H) and 90(for 4 H). The angle bet. the H is therefore 104 or 105 degrees. Similar things can be done in chemistry without resorting to QM.
In 4d the symmetry suggests that there would be 5 sp orbitals. (4 would stick out parallel to a 3-space and a bit down, like in our world and a 5th would stick straight up.) This would give 10 possible electrons or 10 elements. The same for the third row. Everything is pretty much the same as in 3d, until you get to Carbon(or Silicon, being similar). In 3d it is simple: 4 electrons and 4 holes. In 4d the cental Carbon has 5 elecrons and 5 holes, which instead of CH4 would give CH5. (CH4 might not exist in chemistry, I don't know, but you get the point.). For 4 holes(6 electrons), you get CH4 and for 6 holes you get CH6. There are therefore 2 or 3 Carbon elements, with new names added--perhaps C4, C5, and C6. The same for rows 3 and 4. But for row 4, the 4d10 elements in 3d are not based on symmetry so neither would they in 4d. Etc.

[PWrong]

I plugged your differential equation into mathematica and came up with the following solution - I'll let you try and make heads or tails of it

Thanks House. I've managed to work out a solution similar to what you that, although I'm not sure if it's the same. I used the substitution
y = (x^2-1)^a * u(x), then solved for "a" to make the equation simple.

I found that if L(l) = l(l+1)-3/4 and P(p) = (p-1/2)(p+1/2),
then the solution is (-1)^p (1-x^2)^(-(2p+1)/4) * P_l^p(x)

By the way, Q(x) apparently has singularities, so we can't use it in this case.

Anyway, I'm more worried about the last equation now.
R''(r) + 3/r R'(r) - ( 2m/h² (U(r) -E) +L(l)/r²)R=0

If U(r) = a/r^2, there's no 1/r term next to R. The equation is completely different to 3D. I'm beginning to suspect that energy won't even be quantised. This probably means that atoms can't form in 4D.

[jinydu]

Hopefully, I'll know enough to examine this problem myself within a few weeks.

Soon, my quantum chemistry professor will explain the hydrogen atom using Schrodinger's Equation. However, he told me that doing so requires converting the Hamiltonian to spherical coordinates, which is an extremely tedious task, then solving the equation in spherical coordinates using separation of variables. He also said that the derivation that will be presented in class will be at least in part "hand-waving" and will skip out a lot of the mathematical details.

I myself have tried a much simpler problem: A particle is trapped inside a circular (2D) box. The potential is zero inside the box and infinite outside the box.

Even in this simple case, I got bogged down with a long and bewildering equation...

[PWrong]

I myself have tried a much simpler problem: A particle is trapped inside a circular (2D) box. The potential is zero inside the box and infinite outside the box.

Even in this simple case, I got bogged down with a long and bewildering equation...

If it's a differential equation, try this site. I just found it yesterday, and it's great. http://eqworld.ipmnet.ru/en/solutions/eqindex/eqindex-ode.htm

If it's a boundary value problem, I can't help you. Most of what I know about differential equations is self taught, so I can't do all the boring stuff yet. That's what uni's for.

I haven't checked this, but in nD, I think the radius equation is:
R''(r) + (n-1)/r R'(r) - ( 2m/h2 (U(r) -E) +L(l)/r2)R=0

So they're all pretty similar. In 4D, the energy cancels straight out. There's no quantisation. So this might be bad news. I think this also happens in 2D, but I'll have to check it again.

[jinydu]

I looked up "Laplacian" on Mathworld, and I must say that the Laplacian in spherical coordinates is suspiciously simple, much simpler than my unfinished derivation for 2D polar coordinates, which I would expect to be much simpler.

As of now, I have:

Hamiltonian = -((h-bar)^2)/2m ((partial/partial r * cos theta - partial/partial theta * (sin theta)/r)(partial/partial r * cos theta - partial/partial theta * (sin theta)/r) + (partial/partial r * sin theta + partial/partial theta * 1/r * 1/(cos theta + sin theta tan theta)(partial/partial r * sin theta + partial/partial theta * 1/r * 1/(cos theta + sin theta tan theta))) + V(rcos theta, rsin theta)

which looks hopelessly complicated.

To get that far, I started off with

Hamiltonian = -((h-bar)^2)/2m (partial^2/partial x^2 + partial^2/partial y^2) + V(x, y)

I used the conversions

r = sqrt(x^2 + y^2)
theta = arctan(y/x)

And the Chain Rule

Partial/Partial x = Partial/Partial r * Partial r/Partial x + Partial/Partial theta * Partial theta/Partial x
Partial/Partial y = Partial/Partial r * Partial r/Partial y + Partial/Partial theta * Partial theta/Partial y

Can you see any obvious errors?

[PWrong]

As of now, I have:

Hamiltonian = -((h-bar)^2)/2m ((partial/partial r * cos theta - partial/partial theta * (sin theta)/r)(partial/partial r * cos theta - partial/partial theta * (sin theta)/r) + (partial/partial r * sin theta + partial/partial theta * 1/r * 1/(cos theta + sin theta tan theta)(partial/partial r * sin theta + partial/partial theta * 1/r * 1/(cos theta + sin theta tan theta))) + V(rcos theta, rsin theta)

which looks hopelessly complicated.

Actually, I think this is correct, it just looks ugly.
The main problem is that dtheta/dy isn't simplified. I'll just use "th" for theta from here on.
1/r *1/(cos th + sin th tan th) = 1/r*cos th/(cos^2 th+ sin^2 th) = cos th/r

You can do these conversions easily using "scale factors". I'm not sure exactly why they work, but they're very useful for simple coordinate systems, like spheres.

Basically, the scale factor for each coordinate is the distance travelled if you change that coordinate, divided by the change. For instance, if you change r by dr, the distance travelled is just dr. So h<r> = 1.
But if you move around the circle, that is, change theta by d th, holding r constant, the distance is r*d th. So h<th> = r

There's a simple formula for the laplacian in any coordinate system, given the scale factors. In this case, it's
1 / (h<r> h<th>) *(d/dr ( h<th>/h<r> d/dr) + d/dth ( h<r>/h<th> d/dth))

Substituting the scale factors, that's
1/r *(d/dr (r d/dr) + d/dth ( 1/r d/dth))
=1/r*d/dr (r d/dr) + 1/r^2 d²/d th²

[jinydu]

Note: At this moment, I don't have my calculations with me, so I'm going on memory. The equations I type may not be exactly right.

I stubbornly persisted with what I was doing and ultimately managed to find the Hamiltonian in polar coordinates.

Then, I set my sights on the simplest quantum system that is suitable for polar coordinates: a particle in a circular box. The Hamiltonian is given by:

H = 0 when r1 < r < r2
H = infinity otherwise

I managed to apply separation of variables and the differential equation for theta wasn't so hard. But I've been stumped by the differential equation for r:

r^2 (partial^2 psi_r)/(partial r^2) + r (partial psi_r)/(partial r) + ((2mE)/(h-bar)^2)*r^2 - (n_theta)^2)psi_r = 0
where n_theta is an integer [it comes from solving the differential equation for theta]

The angular part of the wavefunction is:

psi_theta (theta) = A_(theta) * cos((n_theta)*theta + phi)
where A_(theta) and phi are arbitrary constants

I asked my professor about the differential equation for psi_r and he recognized it immediately. He said that the solution takes the form of a Bessel function, which is a non-elementary function.

Apparently then, what I should do next is write down the general solution, apply the boundary conditions, solve for the undetermined constants, and hope that this leads to quantization of energy. Unfortunately, I don't really know how to work with Bessel functions. My professor suggested that I first look at the special case where r1 = 0, but I was unable to deal with even that simplified case, other than to say that in that case, either the angular quantum number is zero or the radial part of the wavefunction is zero at the origin.

Of course, if I could solve this, it may give some hints about how to attack 2D hydrogen. If I used the Born-Oppenheimer approximation (fix the proton at the origin and neglect its kinetic energy), all I would have to do is add an extra term to the Hamiltonian. Since electrostatic force varies inversely with distance in 2D, the potential function would be logarithmic.

[PWrong]

Apparently then, what I should do next is write down the general solution, apply the boundary conditions, solve for the undetermined constants, and hope that this leads to quantization of energy. Unfortunately, I don't really know how to work with Bessel functions.

I think you have to show that the function will only continuous if E is a function of an integer. I know this is true for the spherical wavefunction (a laguerre polynomial), but I'm not sure about Bessel functions. I'm not much good with special functions either, although apparently I have a gift for solving differential equations (lucky me ). I'm pretty sure the 4D radial function is a bessel function too. Maybe it's an odd-even thing.

Of course, if I could solve this, it may give some hints about how to attack 2D hydrogen. If I used the Born-Oppenheimer approximation (fix the proton at the origin and neglect its kinetic energy), all I would have to do is add an extra term to the Hamiltonian. Since electrostatic force varies inversely with distance in 2D, the potential function would be logarithmic.

Hmm, I've wondered about that. A logarithmic potential seems really strange :?. It might be possible to solve the equation eventually if we use the taylor series for ln x.

[jinydu]

Mathematica seems unable to solve for everything through exactly, but it is able to give "approximate" (>40 decimal places, and I could easily push it much farther) values for the energy levels whenever I give it specific values for the inner radius, outer radius and angular quantum number (n_theta). In all the cases I've seen, the energy levels are indeed quantized and for any fixed value of n_theta, it appears that in the limit of very large n_r (the radial quantum number), the energy levels appear to be of the form E = constant * (n_r)^2, which (reassuringly) reminds me of the classic particle in a (rectangular) box.

Apparently (I say apparently because there is no way to get Mathematica to understand that (n_theta)^2 is a perfect square; so I just had to plug in (n_theta)^2 = 121 and hope that this holds true for all perfect squares) Mathematica claims that the general solution is:

(psi_r)(r) = A_r * BesselJ[n_theta, sqrt(2mE)/(h-bar) * r] + B_r * BesselY[n_theta, sqrt(2mE)/(h-bar) * r]
where A_r and B_r are arbitrary constants, BesselJ is the Bessel function of the first kind ( http://mathworld.wolfram.com/BesselFunc ... tKind.html ) and BesselY is the Bessel function of the second kind ( http://mathworld.wolfram.com/BesselFunc ... dKind.html )

In the special case where the inner radius is 0 and the particle really is confined inside a circle, B_r = 0. This is because, according to the second link, BesselY goes to negative infinity as r goes to zero, and we clearly can't allow the wavefunction to go to any kind of infinity.

Also, according to http://en.wikipedia.org/wiki/Bessel_function (you'll have to scroll down a lot), in the limit of very large E, we would expect the wavefunction to behave like a sinusoidal wave, with an extra amplitude-decaying factor proportional to sqrt(2/(pi *x)). And of course, we know that a sinusoidal wavefunction will give rise to quantized energy levels that are proportional to the square of the energy level; this probably explains my observation at the beginning of this post.

I'm not finished with this problem yet, so stay tuned!

[jinydu]

First, I'll look at the simplified case where the inner radius is 0. Let's have the outer radius be r2. As mentioned in my previous post, the radial parts of the eigenfunctions are of the form:

(psi_r)(r) = A_r * BesselJ[n_theta, sqrt(2mE)/(h-bar) * r)]

Applying the second boundary condition, (psi_r)(r2) = 0 leads to:

BesselJ[n_theta, sqrt(2mE)/(h-bar) * r2)] = 0

Thus, the energy levels are:

E = j(n_r, n_theta)^2 * (h-bar)^2 / 2m(r2)^2

and the radial part of the wavefunction is:

(psi_r)(r) = A_r * BesselJ[n_theta, j(n_r, n_theta) * r/r2]

where j(n_r, n_theta) is the (n_r)th zero of the (n_theta)th Bessel function of the first kind.

An important question is: Can j be zero? If n_theta is nonzero, BesselJ[n_theta, 0 * r/r2] = 0 for all r, so the wavefunction would always be zero, which is impossible. But what happens when n_theta = 0? In that case, BesselJ[0, j * r/r2] does not have a root at r = 0. Thus, j cannot be zero.

It is interesting to know whether n_r can take on negative values (clearly, it can be positive). As can be seen, if j(n_(-r), n_theta) = j(n_r, n_theta) or -j(n_r, n_theta), and BesselJ[n_theta, -x] = BesselJ[n_theta, x] or -BesselJ[n_theta, x], changing n_r to n_(-r) will only change the wavefunction by a factor of -1 or not at all. Since the wavefunction is itself only ever defined up to a (complex) constant with absolute value 1, that would be essentially no change at all, and there would be no need to distinguish between n_r and n_(-r).

I'm looking at equation 70 of the first Mathworld link. Since cos(-z*cos(u)) = cos(z*cos(u)), the only difference between BesselJ[n, -z] and BesselJ[n, z] is in the z^n factor. But (-z)^n = z^n (when n is even) and (-z)^n = -(z^n) (when n is odd). Thus, when n is a positive integer, BesselJ[n, -z] differs from BesselJ[n, z] by a factor of -1 or not at all, whenever n is a positive integer.

Now, I'm looking at equation 49 on the same page. It implies that BesselJ[-n, z] differs from BesselJ[n, z] by a factor of -1 or not at all. Thus, the preceding conclusion can be generalized to all nonzero integers n.

Since BesselJ[n, -z] = +-BesselJ[n, z],
it is clear that BesselJ[n, -z] is a root if and only if BesselJ[n, z] is a root. Thus, we really do have j(n_(-r), n_theta) = +- j(n_r, n_theta).

So there really is no need to distinguish between j(n_r, n_theta) and j(n_(-r), n_theta) and we might as well restrict n_r to be positive.

Thus, the total wavefunction is:

psi(r, theta) = A * BesselJ[n_theta, j * r/r2] * cos((n_theta) * theta + phi)
where I've combined A_theta and A_r into a single constant A

I could try to normalize that by assuming that:

Integral from 0 to r2 (Integral from 0 to 2pi(|psi|^2)) r dr d(theta) = 1

I have no idea what the integral of the square of the Bessel function is. Instead, I ask Mathematica to do the job, and I find that one possible solution is:

A = sqrt(2/pi) * 1/r2 * ((BesselJ[n_theta, j])^2 - BesselJ[(n_theta) - 1, j]*BesselJ[(n_theta) + 1, j])^(-1/2)

So the eigenfunctions and eigenvalues are:

psi(r, theta) = sqrt(2/pi) * 1/r2 * 1/sqrt(BesselJ[(n_theta) - 1, j]*BesselJ[(n_theta) + 1, j]) * BesselJ[n_theta, j * r/r2] * cos((n_theta) * theta + phi) when n_theta is nonzero

psi(r, theta) = sqrt(1/pi) * 1/r2 * 1/(BesselJ[1, j]) * BesselJ[0, j * r/r2] when n_theta = 0

E = j^2 * (h-bar)^2 / 2m(r2)^2

n_r = 1, 2, 3 ...
n_theta = 0, +-1, +-2, +-3 ...

where j(n_r, n_theta) is the (n_r)th zero of the (n_theta)th Bessel function of the first kind(just a reminder)

(I needed a separate expression for the case where n_theta = 0 because in that case, the angular part reduces to a constant, which affects normalization).

The lowest energy level occurs when n_r = 1 and n_theta = 0. In that case, j ~ 2.404826

Interestingly, if we were to change j --> n*pi (where n is any positive integer) and r2 --> L, we would have exactly the energy levels of the standard "particle in a (rectangular, 1D) box" with length L.

Now that I have the wavefunction, I can proceed to "play around" by calculating things such as expectation values and uncertainties.

For instance, the expectation value of r in the ground state (n_r = 1, n_theta = 0) is about 0.424057625938838 r2.

The uncertainty in r in the ground state is about 0.195529411 r2.

The expectation value of r at n_r = 1, n_theta = 10 is about 0.800789 r2.

The expectation value of r at n_r = 3, n_theta = 10 is about 0.682233 r2.

But these values should be taken with a (big) grain of salt. For different values of (n_r, n_theta), my computer sometimes gives ridiculous answers (such as 2.5 * 10^ 19 r2 when n_r = 10, n_theta = 10, which is obviously impossible).

EDIT: Argh! I forgot that there's no reason why the wavefunction has to be zero at the center of the circle, since there is no infinite potential region "inside the center". I'll have to revise this post...

I think it's fixed now

[PWrong]

Thus, the energy levels are:

E = j(n_r, n_theta)^2 * (h-bar)^2 / 2m(r2)^2

That's a surprise. I didn't think you'd get any energy levels at all. The 4D wavefunctions are bessel functions too, so maybe there's still hope.

But these values should be taken with a (big) grain of salt. For different values of (n_r, n_theta), my computer sometimes gives ridiculous answers (such as 2.5 * 10^ 19 r2 when n_r = 10, n_theta = 10, which is obviously impossible).

It might not be impossible. Unless it breaks some important rule, maybe it's common for electrons to be ridiculously far away from the nucleus in 2D.

If you feel like trying real 2D hydrogen when you've finished, with the logarithmic potential, I've softened up the equation a bit.

x^2 y'' + x y' + ( 2m/h^2 (x^2 ln x - E x^2) + n)*y = 0
First, we'll choose units so that 2m/h^2 = 1.

Now if we let t = 2 ln x, we get
4 y'' + ((t/2 +E) e^t + n)*y = 0

Then, let k^2 = -n/4, and let y(t) = u(t)*e^(kt)
This gives us
u'' + 2 k u' + 1/4 (t/2 - E) e^t u = 0

That looks a bit nicer, but from here I think we need a series solution.

[jinydu]

It might not be impossible. Unless it breaks some important rule, maybe it's common for electrons to be ridiculously far away from the nucleus in 2D.

Actually, it is impossible. Remember that the potential is infinite outside r2, so there is zero probability of finding the particle outside of r2.

You should no less capable of attacking these problems than me though. Don't you have access to some kind of computer program that can deal with differential equations and the like? (EDIT: Mathematica 4 is unable to solve the problem directly).

Also, it seems that you've decided to drop into a certain other forum, where I have a healthy number of posts :wink: . Since you're there already, do you know the answer to the question I posed in the "Separation of Variables in Quantum Theory" thread in the Physics section of that forum? Thanks.

[PWrong]

You should no less capable of attacking these problems than me though. Don't you have access to some kind of computer program that can deal with differential equations and the like? (EDIT: Mathematica 4 is unable to solve the problem directly).

I got Mathematica for Christmas . Now I can try to figure out 4D hydrogen properly.

If the potential function is U(r) = k/r^2, then we have
r^2 R'' + 3 r R' + ( k - L(l) - E r^2) R = 0

The solutions are based on bessel functions, just like in 2D.

R = 1/x * BesselJ[ sqrt(1-k + L(l)) , x*sqrt(-E)]

I guess this tells us that sqrt(1-k + L(l)) has to be an integer, i.e.
L(l) = l^2 + k - 1

But I don't see where the energy levels come in. :?

[bo198214]

And thats what I mean when saying that the titles sometimes dont match the contents: 4D planet orbits are discussed in Fourth Dimension Calculation and Stability of 4D atoms and 4D atoms are discussed in Schrodinger's Equation

Nonetheless its an interesting topic and I read yesterday a german quantum mechanics text, which states (translated):

The solution of Schrodinger's equation shows, that for central forces, where the force does not increase more than with (the reciprocal of) the 3rd power of the distance to the center, a wave form with minimal energy exists (the so called base state).

So we are on the lucky side because in 4D the force increases with the reciprocal of the 3rd power of the radius. And this should assure the existence of 4D atoms (though it does not provide for 4D solar systems).

I am not really familiar with quantum mechanics, so Id like someone to explain the details to me

[PWrong]

And thats what I mean when saying that the titles sometimes dont match the contents: 4D planet orbits are discussed in Fourth Dimension Calculation and Stability of 4D atoms and 4D atoms are discussed in Schrodinger's Equation

I know what you mean, but in this case the title matches the contents perfectly. This thread is entirely about solving Schrodinger's equation in other dimensions. Schrodinger's equation is basically all about atoms.

The solution of Schrodinger's equation shows, that for central forces, where the force does not increase more than with (the reciprocal of) the 3rd power of the distance to the center, a wave form with minimal energy exists (the so called base state).

That's odd, I wonder how they derived that. In any case, it's not just the force that changes. The radial part of Schrodinger's equation is different in 4D, because we're using glomar coordinates, not spherical coordinates.

I am not really familiar with quantum mechanics, so Id like someone to explain the details to me

I doubt I could explain quantum mechanics in sufficient detail, partly because I don't yet understand it myself. Jinydu could probably do a much better job, and it's easy to find introductions on the net, but you'd have to get a degree in physics for a complete understanding. But I'll explain roughly what this is about, and what I'm trying to do here. Unfortunately, some basic calculus is a prerequsite. Some physicists might be able to explain quantum mechanics without maths, but I can't.

Basically, in quantum mechanics, a particle isn't simply a point with a specified position and velocity, it's spread out over all space as a "wavefunction". The wavefunction is related to the probability of detecting the particle at some position. Schrodinger's equation is a differential equation that tells you how to find the wavefunction, given the potential energy function.

The actual equation looks a bit like this:
Laplacian [ w(x,y,z)] + U(x,y,z) w(x,y,z) = E w(x,y,z)
where U = potential energy at (x,y,z),
E is the total energy of the system, and the laplacian is a kind of nD version of the second derivative.

Here's the general idea. You have an electron sitting in some kind of potential well. It sort of spreads out in waves, sometimes spilling over the edge, even if it doesn't have enough energy to do so. You solve the equation to get an expression for the wavefunction of the electron, which usually depends on E in some special way.

Now, you also have to apply some conditions. One of these conditions is that all the probabilities sum to one, a basic feature of probability.
It turns out that for most potential functions, this condition is only satisfied for special values of E.

For instance, in a hydrogen atom in spherical coordinates, the wavefunction is based on Laguerre functions. The laguerre function only converges for certain discrete values of E. This means that an electron can only exist at these energy levels. It turns out they are exactly the same energy levels as found in experiments.

The problem I have is that the wavefunction for a 4D electron seems to converge regardless of E. So there are no energy levels, and we lose the interesting, discrete structure of atoms. I suspect/hope I've made a mistake somewhere.

There's a few examples of simple potential wells on scienceworld, like the
Finite Square Potential Well. There's also a page on 3D hydrogren there, but it's a bit complex. There's a simpler explanation at http://www.kw.igs.net/~jackord/bp/i6.html.

[jinydu]

Now, you also have to apply some conditions. One of these conditions is that all the probabilities sum to one, a basic feature of probability.
It turns out that for most potential functions, this condition is only satisfied for special values of E.

Your post is mostly correct, but you seem to have slipped up here. What you wrote there (about the probabilities summing to one) is known as the Normalization Condition. In this case, it (more precisely, the condition that the wavefunction approaches zero as the radial coordinate goes to infinity) does to lead to quantization, but this isn't always necessarily the case. For example, in the (1D) "particle-in-a-box" system, quantization arises not from normalization, but from continuity of the wavefunction at the endpoints of the box.

Also, when you wrote down Schrodinger's equation, you forgot the - [(h-bar)^2]/2m factor in front of the kinetic energy term.

But otherwise, not a bad job. Unfortunately, I'm too caught up with other problems and preparing to start a new quarter at university to attempt an attack on 4D hydrogen. Still, I may be able to give some advice:

A quantum mechanics book that my professor gave me as a gift provides a thorough derivation of the hydrogen atom (of course, in 3D). Jumping right into the radial equation part (it is assumed that you have already solved all the non-radial differential equations):

1) Make the substitution u = R(r)

You should now have a differential equation identical in form to that of the 1D Schrodinger equation.

2) "Peel off" asymptotic behavior. First take the limit of the effective potential energy term as the radial coordinate goes to infinity (that is, drop all effective potential energy terms except the highest order one), solve this (simpler) differential equation, and apply the boundary condition that this function cannot go to infinity as the radial coordinate goes to infinity. Second, take the limit of the effective potential energy term as the radial coordinate goes to zero (this time, keep only the lowest order term), and solve this differential equation as well. Apply the boundary condition that this second function cannot go to infinity as the radial coordinate approaches zero.

3) Finally, assume that

u(r) = v(r) * first asymptotic function * second asymptotic function

and finish up using a series solution (solve for the coefficients recursively). You should find that in order to prevent the u(r) from diverging at infinity, the series must terminate.

[PWrong]

Thanks, but unfortunately I don't think that's going to help. I think I can prove that energy isn't quantised even without using Mathematica's solution.

The equation to solve is:
R'' + 3/r R' - (k/r^2 - L(l)/r^2 - E) R = 0
r^2 R'' + 3 r R' - (k - L(l) - E r^2) R = 0

where k is a constant related to the electric field,
L(l) = l (l + 2) where l is an integer, and E is energy.

Let u = sqrt(E)*r
Then
dR/dr = dR/du * sqrt(E)
d^2R/dr^2 = d^2R/du^2 * E

So the equation becomes:
u^2 R'' + 3 u R' - (k - L(l) - u^2) R = 0

The solution to this is of the form R = f(u) = f(sqrt(E)*r)
The energy doesn't change the function except to make it wider. It has no effect on whether the integral converges.

I guess this isn't a big surprise. The electron is simply unstable in 4D, just like planets. In both cases, the problem is the inverse cube law.

[bo198214]

For me (as absolute quantum novice) it sounds that the electron can orbiting anywhere around the core :?
What exactly would mean that it is unstable?

Can you also please make the computations for 2D?!
For planet orbits 2D is stable (meaning that it doesnt fly away or fall into the sun, when slightly changing the starting condition) at least
it looks so in my numerical computations with maple 1, 2, and 3 (more explanations).

[jinydu]

I still feel a little optimistic about 4D hydrogen. My course did claim that quantization is a feature of almost all quantum systems (although admittedly, they didn't have 4D hydrogen in mind).

Regarding 2D hydrogen, you have to solve a 2D version of Schrodinger's equation. The potential energy will be logarithmic.

By the way, PWrong, have you taken a look at my question on differential equations in the Question and Answers section? I'm still looking for a proof...

http://tetraspace.alkaline.org/forum/vi ... .php?t=442

Basically, the question boils down to this:

Let psi(r, theta, phi) be a function that satisfies:

arg(Laplacian(psi)) = arg(psi)
where arg is the complex argument

for every point in space.

Show that the integral over all space of |psi|^2 cannot be finite and nonzero.

[PWrong]

I just realised that 4D might be the only dimension without quantisation. Everywhere else, even in 5D, there should be some kind of quantisation.

I think it's possible to prove that only the potential function affects whether quantisation can happen, not the dimension or the coordinate system. I guess this makes sense, since looking from a different angle shouldn't get rid of energy levels.

The general Schrodinger's equation is:
Laplacian[ w(x) ] + U( x ) w(x) = E w(x)
where x is the position in any coordinate system. (By the way, I'm choosing my system of units so that h/2m = 1, so I don't have to write it).

Now suppose we make a substitution, where each x_i is multiplied by sqrt(E). So u = sqrt(E) x

Now from the definition of the Laplacian, which easily extends to n dimensions, we should find that the whole laplacian gets multiplied by E.

E Laplacian[ w(u) ] + U( u/sqrt(E) ) w( u ) = E w( u )
Divide through by E:
Laplacian[ w(u) ] +1/E * U[ u/sqrt(E) ] w( u ) = w( u )
Now E only appears in the potential energy, so lets look at the U function.

In n dimensions, U(r) = k/r^(n-2)
After the substitution and some algebra, we end up with:

k/r^(n-2) * E^((n-4)/2)

Clearly, when n=4, energy cancels out and becomes a useless scaling factor. Note that k/r^2 is the only potential function that has this effect.

I haven't proved that all other systems are quantised, but I'm pretty sure they will be. I might even try 5D to see what happens.

[thigle]

hey you tetronians sorry to bother you with such a basic question, but, can you please explain what is "quantisation" ? what does it mean that '4d might be the only dimension without quantisation", or more generally what does it mean when "somewhere there is/isn't some kind of quantisation" ?

[jinydu]

In this case, quantization means that only certain discrete values of some quantity are allowed. This is in contrast to continuous variables which can take on any real value (although possibly only in some interval).

A basic feature of quantum mechanics is that most systems with boundary conditions have quantized energy.

[thigle]

so then the fact that there is no energy quantization in 4D can be interpreted as meaning that "NOT only certain discrete values of some quantity are allowed", or that 4d systems with boundary conditions have not quantized energy and are thus exceptional with regard to majority of quantum systems ?

1. how does energy get quantized ?
2. what is agent of quantization, what is way of quantization ?
3. and what is object of quantization ?

(answer to last one - object of quantisation - seems to me to be the energy of the system. ? is the agent that 'quantizes' the energy the system ? is the way it does so the way the system is structured internally ?)

also, what do you mean by "systems with boundary conditions" ? dissipative systems ? or general system definition (whole over sum of parts + emergence) ?

i am just a hobbyist asking (this time :wink:)

[jinydu]

No, PWrong's derivation (if it is correct, and I haven't checked it yet) states that 4D hydrogen does not have quantized energy. Of course, some 4D systems do have energy quantization; for instance, the "particle-in-a-box" has quantized energy in any number of dimensions.

First though, you have understand what a system is. Simply put, in quantum mechanics, a system is a collection of physical objects.

The short answer to your questions is that quantization of energy comes from boundary conditions on the wavefunction.

Of course, in order to understand that statement, you also have to understand the definitions of "energy", "boundary conditions" and "wavefunction", which requires an understanding of quantum mechanics. I can try to briefly explain it here, but if you really want to understand it correctly, it is best to consult an expert or an educational website. Furthermore, it will require a solid grasp of mathematics, especially calculus.

One of the fundamental postulates of quantum mechanics is that everything that can be known about a system can be derived from the system's corresponding wavefunction. A wavefunction is a function that associates a complex number to every point in space and time. To simplify the discussion, let's concentrate on systems that have only one particle (it is not difficult to generalize to arbitarily large numbers of particles, once you understand how things work for one particle).

The wavefunction is usually denoted as Psi(x, t) or Psi(x, y, z, t) (or something else), depending on the number of dimensions the system has. If you take the absolute square of the wavefunction, you get the probability density of finding the particle at a particular point in space at a particular time. To find the probability of finding the particle in some finite region at some particular time, integrate the absolute square of the wavefunction over that region.

The (time-dependent) Schrodinger Equation, the "master equation" of quantum mechanics, allows you to predict the wavefunction of a system at any time in the past or future, provided that you know the system's wavefunction at one specific instant and that no observations are made on the system in the meantime. The (time-dependent) Schrodinger Equation is:

i h-bar partial(Psi)/partial(t) = [-((h-bar)^2)/2m Laplacian(Psi)] + V Psi

(It would be much easier to read if this forum supported math symbols).

In that equation, i is the imaginary unit, h-bar is a universal physical constant, m is the mass of the particle and V is the classical potential energy the particle would have if it were in some location (in general, V is a function of position).

In general, in any quantum mechanical problem, the main goal is to solve for Psi because once you have done so, you can extract all the information about the system that you could ever know. From one system to the next, the only things that can change in that equation (besides Psi, of course) are the mass (which you can determine experimentally), the Laplacian (which depends only the number of dimensions the system has) and V (which is essentially what makes every different system unique, and depends only on position). Even if you know the all those things, however, there will in general be infinitely many wavefunctions Psi that satisfy the equation. To determine which wavefunction actually describes the system, you have to perform an experiment; a very large number of measurements on many identically prepared systems, at one particular instant in time, suffices. This is analogous to the situation in Newtonian mechanics, where the mass, force (at all times) and position and velocity of a particle (at some particular time) determine its position at all past and future times.

In studying quantum systems, we are interested in knowing what the possible wavefunctions are. Of special interest are the wavefunctions of the form:

Psi(x, t) = psi(x) T(t)

that is, wavefunctions that are the product of two functions, one that depends only on position (psi) and another that depends only on time (T). If you plug the above functional form into the time-dependent Schrodinger equation, you eventually get two separate equations:

[-((h-bar)^2)/2m Laplacian(psi)] + V psi = E psi

T = e^(-i ((h-bar)/E) t)

where E is a constant called energy (in quantum mechanics, this is the definition of energy).

The first of those equations is called the time-independent Schrodinger equation. If we let

H = [-((h-bar)^2)/2m Laplacian(Psi)] + V Psi

the time-independent Schrodinger equation takes the simple-looking form:

H(psi) = E psi

H is called the Hamiltonian operator, also known as the total energy operator. It contains all the information you can ever have about a system if you don't know any initial conditions. H is different for any two systems that differ by more than just their initial conditions.

There are further restrictions on what kinds of wavefunctions a system can have; these are known as the Born conditions, and they correspond to the probabilistic interpretation of the wavefunction. Basically, the Born conditions state that the wavefunction must be "well-behaved"; that is, they must be continuous, finite (even when the spatial coordinate approaches infinity) and quadratically normalizable (the integral of the absolute square of the wavefunction over all space must be equal to 1). If you impose these conditions on a system's possible wavefunctions (derived from the Schrodinger equation), you almost always find that almost all of the originally derived wavefunctions are rejected, leaving behind only wavefunctions whose corresponding energies are quantized (can take on only certain discrete values, which of course depend on the details of the particular system). This then, is the origin of energy quantization.

In this thread, PWrong was discussing a particular quantum system called the 4D hydrogen atom. A 4D hydrogen atom is made up of two particles: a proton (with mass m_p and charge +e) and an electron (with mass m_e and charge -e). PWrong was attempting to the find the solutions to the time-independent Schrodinger equation for this system. A simple extension of the ideas I presented above show that the time-independent Schrodinger equation for this system (in Cartesian coordinaes) is:

- ((h-bar)^2)/2m_p) (Laplacian of proton position) psi - ((h-bar)^2)/2m_e) (Laplacian of electron position) psi + V psi = E psi

where V = - (e^2)/(4pi^2 epsilon-nought ((x_p-x_e)^2 + (y_p-y_e)^2 + (z_p-z_e)^2 + (w_p-w_e)^2)) [letting (x_p, y_p, z_p, w_p) be the proton coordinates and (x_e, y_e, z_e, w_e) be the electron coordinates] (this can easily be derived from Gauss' Law)

and psi is a function of six coordinates (3 for the proton and 3 for the electron).

Using a technique called "transformation to reduced mass coordinates", it is possible to show that psi can be written in the form

psi = psi_cm (x, y, z, w) psi_rm (x, y, z, w)

where x = x_p - x_e, y = y_p - y_e, z = z_p - z_e, w = w_p - w_e and

- ((h-bar)^2)/2(m_p + m_e) Laplacian in (x, y, z, w) (psi_cm) = E psi_cm

- ((h-bar)^2)/2m Laplacian in (x, y, z, w) psi_rm - (e^2)/(4pi^2 epsilon-nought (x^2 + y^2 + z^2 + w^2)) = E psi_rm

where we have defined m = (m_p)(m_e)/(m_p + m_e)

The first equation is in fact that of a free particle; it behaves badly and is usually dealt with using classical mechanics instead. The second equation is meat of the problem; it contains all the interesting information about the 4D hydrogen atom. After converting it to hyperspherical coordinates, you get precisely the problem that PWrong has been trying to solve.

If there is anything that I hope you have gained from this, it is an appreciation for the rigor and precision of a real physics theory.