In classical physics if you have a particle with energy E, mass m and momentum p then you can break the momentum down into its x, y and z components, so the energy can be expressed as

Extending this to the de Broglie relations [4] we get

We can also break apart the wave equation into its x, y and z components so instead of just δ^{2}Ψ/δx^{2} we get

Writing out the bit in the brackets every time would get quite tedious, luckily there is a simple operator that means the same thing (An operator is a mathematical device that acts on whatevers to the right of it, there will be loads more operators in this section)

The operator we use is called Del-Squared and is defined as

So whenever you put something to the right of del-squared you take its second differentials with respect to x, y and z

So our Schrödinger equation is now

As before we are only interested in the time independent situation so we can forget the left hand side of the equation for now and we are left with

[14]

Which is just the **Time Independent Schrödinger Equation in 3 Dimensions** (equation [8] with a del-squared term).

The first thing to consider is what happens if the Potential that the particle is in can be split into x, y and z components like so

Once again using the separation of variables method (which by now you must be an expert at given the amount of times it’s appeared) we can split up the wave function into the combination of 3 one dimensional functions

Substituting the 2 above relationships into [14] and using separation of variables you get

As the three terms added together must equal E, and they are all independent of each other then once again we say that they are each equal to a constant, which in this case we call E_{1}, E_{2} and E_{3}

Each of the separate E's can be thought of as the Energy in that plane and the total energy is just the sum of these

To calculate this total energy we just add up 3 lots of the equation we derived in the case of the infinite square well (equation [11]), however each one can have a different width and n value, so we get

We can now obtain an equation for the wave function in the 3 dimensional case (from equation [12])

Where cos is used if n is odd and sin is used if n is even, notice this is extremely similar to the wave function of the one dimensional infinite well.

It is worth noting that something interesting happens if two sides of the infinite well are equal. If for example we make a and b the same length the equation becomes

In this equation there are a number of different combinations of n_{1} and n_{2} that give the same energy. For example, if the state of the wave function was [n_{1}=2, n_{2}=1, n_{3}=1] this would give an identical energy to [n_{1}=1, n_{2}=2, n_{3}=1]
However if you look at the wavefunctions of the two situations you find that they are different

When you get two or more quantum states with the same energy you call it **degeneracy**, and it is closely related to the symmetry of the system. If you took measurement of a system you knew was degenerate and you got a value for the energy you would not be able to tell which actual system it was, you would have to assume that each one was equally probable. So the appropriate expression for the wave function would be the average of |u_{211}|^{2} and |u_{121}|^{2}.

For the case of the Harmonic Oscillator the equation in 3 dimensions for the energy and the wave function are just

Where ω_{1}, ω_{2} and ω_{3} are the angular frequencies of the x, y and z components respectively, and x'=(mω_{1}/ℏ)^{1/2} etc. This system will experience degeneracy if all three ω's are equal causing states with equal (n_{1} + n_{2} + n_{3}) to have the same energy.

Cartesian coordinates are all well and good but in the case of atoms it makes things easier (or so I've been told) to work in a different coordinate system. As atoms are generally spherically symmetrical we will use a spherical polar coordinate system where a position isn't in terms of (x,y,z) but in terms of r,φ and θ.

The relationship between cartesian and polar coordinates is given by the following expression

We want to be able to write out our equation in the same form as we did in cartesian coordinates i.e.

As before we will only be considering the case of time independence so we will get rid of the time dependent bit of the equation, set it equal to the total energy of the system and replace the del-squared term

However in polar coordinates ∇^{2} is not as simple as before, in fact it looks horrible

So when we substitute this in to our wave equation we get the following

[15]

I think you will agree that this looks horrible, so once again it's time to use our old friend separation of variables. First of all we assume we can separate the radial part of the equation from the angular part of the equation, like so

So we go through and replace all of the u's with RY's

We then divide throughout by RY, which separates the R and Y terms inside the square brackets, and times throughout by r^{2}, which gets rid of all of the r^{2}'s from the square brackets. So we now have

So we have one term in the square bracket that only depends on R and two that only depend on Y, and then outside the square brackets we have two r^{2} terms. So if we rearrange and group all the R's and r's together and all the Y's together we get

Once again, if the two bits of the equation are independent and equal they must be equal to a constant, in this case the constant is λ. So for the radial part we have

Now if we times by R and divide by r^{2} we get

Now we do a bit or rearranging we get

Now this bears the same form to equation [8], we have a differential of R term, a potential of R term all equal to the energy of R. However we can't solve it yet as we don't know what λ is.

So if we now look at the angular dependent term we have

Just to make things easier for ourselves we times throughout by Y to get

Now we once again use separation of variables. We assume that Y can be split into a function of theta and a function of phi, like so

So we go through and replace all of the Y's with ΦΘ's and we get

We now divide throughout by ΦΘ to get

Now what we have in the above equation is nearly completely separated into phi and theta, except for the sin^{2}(θ) in the phi term, so we just times by sin^{2}(θ) and rearrange the λ so it's with the theta terms (as it will now have a sin^{2}(θ) term from the multiplication) we get

So now, as with all the other variable separable equations, we set both bits equal to a constant, in this case we call it ν (Note: this is not meant to be a v, it is ment to be a greek lower case nu, however HTML doesn't seem to want to let me do that). So for the theta dependent equation you get

Again we rearrange to make things easier and nicer, first by multiplying by Θ

Then by moving the νΘ to the left hand side

For the phi dependent term we just do the same thing. Set equal to ν

Multiply by Φ

And rearrange to get

So from the one horrible equation [15] we now have 3 not so bad equations, one dependent on the radius, one on the theta angle and one on the phi angle

[16]

[17]

[18]

Notice that the Φ and Θ equations do not contain V. This means the solutions to those two equations will apply to **ANY** spherical potential, so then we would 'only' have to solve for the radial part to get the complete wave-function.

At the moment we can't solve equation [16] as we don't know what λ is, and we can't solve equation [17] as we don't know what λ and v are. So we start off solving [18] for v.

The solution to equation [18] is straight-forward, we just need something that when differentiated twice with respect to φ and multiplied by ℏ^{2}/2m_{e} is equal to vΦ. It turns out the solution is

Where A is a constant. One of the boundary conditions for Quantum mechanics is that the wave-function must be single valued, this means Φ must be single valued. As the limits of φ are 0 to 2π then the following condition must be true

This puts constraints on the exponential part of the equation which give rise to the following relationship

[19]

This m **Is Not Mass**. It is a quantum number, the value of which is always a positive or negative integer or zero. Our equation for phi is now

Where the factor at the front of the equation before the exponential is just in order to normalise the wave-function so that

So now we have solved one of the 3 differential equations. Yay for us! And we have also obtained an important quantum condition (equation [19]). From equation [19] we can also get a value of v

Which we can then substitute into equation [17] and get

Once again we rearrange to get it in a nicer form (at least it's not separation of variables again) by multiplying through by 2m_{e} and dividing by ℏ^{2} to get

If we now make the following substitution

[20]

Then the equation becomes

[21]

Solving this become easier (although still insanely difficult) if we make yet another substitution

This leads to a replacement for d/dθ

Then substituting this back into [20] gives

If we now consider the simple case of m=0 then our equation becomes

[22]

Now, in order to solve this equation we have to write it out and solve it in terms of a power series. To me this bit seems tough, and by tough I mean horrendously impossible. I can't fully get my head around it however some people reading this might be able to, so I have decided to include it with the best explanation I can manage. If you want to see it then just click the 'Horrible Maths' button, if not just carry on reading.

Horrible Maths

From the horrible maths solution we get an infinite value at v=1, which is not allowed. This can be avoided if the series terminates at a specific value p, lets call it *l*. This leads to the following condition

[24]

Where *l* is an integer greater than or equal to zero. We can now solve P, now written as P_{l}, as a polynomial which contains only odd or even powers of v. These polynomials are called the Legendre Polynomials and can be calculated using equations [23] and [24] like so

The general solution to P for the case of nonzero values of m is obtained by differentiating equation [22] |m| times using (Leibniz's generalized product rule), the solution of which is

**Note: the |m| is not a power it just signifies that P has a dependence on it**. As P_{l} is a polynomial of degree *l* then the |m|^{th} derivative will be zero if |m| is bigger than *l*, therefore we have the following condition

So for example if *l* is 4 then m is -4, -3, -2, -1, 0, 1, 2, 3, 4

We now have solutions for the θ and φ parts so we can just combine them both for the full angular solution, which is written as Y_{lm}(θ,φ)

The factor in the square brackets is just a normalizing factor so that

The functions Y_{lm} are known as **Spherical Harmonics**. Equations for spherical harmonics with *l* from 0 to 2

Now lets go back to the radial part of the equation. We had

However now we can write λ in terms of *l* from equations [20] and [24] to get

We can then simplify by making the substitution χ(r)=rR(r), which gives

So now we have a differential of χ term, a potential of χ term all equal to the energy of χ

Quantum numbers are used in quantum mechanics to describe values of conserved numbers in a system. They often describe specifically the energies of electrons in atoms, angular momentum and spin etc.

*n*The main quantum number is

*n*and is called the**Principle Quantum Number**.*n*is always an integer and is always greater than 0.*n*determines the energy level or shell that an electron is in.*l*The next quantum number is

*l*, which is called the**Azimuthal Quantum Number**or**Orbital Angular Momentum Quantum Number**and it determines the orbital angular momentum of the system. like*l*,*l*is an integer however it can take the value 0. The value of*l*is constrained by the value of*n**l*can take any of the values between 0 and*n*-1, so for example, if*n*was 3 then*l*could be -2, -1, 0, 1 or 2.*m*_{l}The next quantum number is

*m*and is called the_{l}**Magnetic Quantum Number**. A basic way of describing it is to say that it refers to the direction of the angular momentum vector. Given a value of*l*,*m*can be integer from*-l*to*l*. This means that for a given orbital momentum value,*l*, there are 2*l*+ 1, integral magnetic quantum numbers m which determine the fraction of the total angular momentum.So if

*l*was 3 than*m*could be -3, -2, -1, 0, 1, 2 or 3The quantum number

*l*relates to the angular momentum of the system (**L**) by the following formulaYou can also find the projection of

**L**onto the z-axis using the*m*quantum number_{l}So the value of the angular momentum has only one value and its value depends on

*l*, however its projection onto the z-axis is dependent on*m*which takes multiple values depending on_{l}*l*. So for example if*l*= 1 then*m*= -1, 0, +1_{l}*s*The next quantum number is

*s*and is the**Spin Quantum Number**and refers to the spin of the particle in the system. For example all electrons have s = 1/2 as do positrons, neutrinos and quarks. Photons have s=1 and the hypothetical graviton is predicted to have s=2. The Higgs boson is unique among elementary particles as it is thought to have a spin of s=0.Another interesting thing to point out is that particles with half-integer spin are fermions(matter particles) and particles with integer spin are bosons (force carriers). For more info have a look here.

*m*_{s}Just like

*l*had an associated "projection" number,*m*, so too does_{l}*s*. This is called**The Spin Projection Quantum Number**and is represented by*m*. The value of_{s}*m*is limited by the value of_{s}*s*The value of

*m*ranges from −s to +s in steps of one._{s}

One problem that arises when dealing with all these numbers is spin-orbit interaction. This is where a particles motion and its spin start to affect each other. This effect causes *l*, *m _{l}*,

*j**j*is the symbol that represents the**Total Angular Momentum Quantum Number**

This page was written by The Rev, please direct any queries you have regarding it to him.

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