Axiom17 said:
.. is that where you can do multiplication by multiplying each part seperately then adding the products. So the calculation is say 3x17 so can do 3x10=30 + 3x7=21 hence 3x17=30+21=51.
Yep, that's the one.
Axiom17 said:
Ok so what I can do (psi1)x(L_{x})+(psi2)x(L_{x})=(psi1+psi2)x(L_{x})?
Still leaves me with the issue of how to actually multiply the wavefunctions (taking into account the quantum numbers) with the value of the operator L_{x}.
ah... I think you may be lacking a proper understanding of the way operators work.
Think of an operator as representing some action that you apply to some object, which is usually a function. A very simple example of an operator would be just multiplication by
x. You may know this as the position operator in QM. In general, multiplication is the simplest kind of operator. But there are more complicated actions that you can perform on a function, such as taking the derivative. This is also an operator - in fact, except for a constant factor, it's just the momentum operator. But it's not actually multiplication.
Operators always act on the things that appear to their right. So if I write
\hat{p}\psi(x)
where I'm using a hat over the
p just to remind us that it's an operator, that represents the momentum operator acting on the wavefunction \psi. I can't write it the other way around, as
\psi(x)\hat{p}
because that means something different. (If you're curious: it's a new operator that represents the action "apply the momentum operator, then multiply by \psi(x)." When you create one operator by composing other operators like this, the component operators are applied from right to left.)
When you apply an operator to a function, you get a new function. There are various ways to figure out what that function is. In principle, you can always just apply the operator directly, but sometimes that takes a lot of effort. For example, if you wanted to apply L^2 to the wavefunction \psi_{3,1,1}, you would have to calculate
-\hbar^2\biggl[\frac{1}{\sin\theta}\frac{\partial}{\partial\theta}\biggl(\sin\theta\frac{\partial}{\partial\theta}\biggr) + \frac{1}{\sin^2\theta}\frac{\partial^2}{\partial\phi^2}\biggr]\frac{1}{81\sqrt{\pi}}\biggl(\frac{1}{a_0}\biggr)^{\frac{3}{2}}\biggl(6 - \frac{r}{a_0}\biggr)\frac{r}{a_0}e^{-\frac{r}{3a_0}}\sin\theta e^{i\phi}
if my sources are correct. That would be quite a pain to work out manually.
But there is an easier way. The particular wavefunctions that are given the symbols \psi_{n,l,m} are chosen for a reason: when you apply one of the angular momentum operators L^2 or L_z to them, the new function you get back is a constant factor times the original wavefunction. Functions with this property for a particular operator are called
eigenfunctions of that operator, and the corresponding constant value is called the
eigenvalue. (Maybe you already knew that) In my example above, you could use this fact to say that applying L^2 to \psi_{3,1,1} will just give you C\psi_{3,1,1}, and all you have to do is figure out what the constant
C is. It should come as no surprise that the value of the constant is related to the quantum numbers of the wavefunction. MathematicalPhysicist gave you the formula above:
L^2\psi_{n,l,m}(x) = \hbar^2 l(l + 1)\psi_{n,l,m}(x)
or in this example specifically,
L^2\psi_{3,1,1}(x) = \hbar^2 1(1 + 1)\psi_{3,1,1}(x) = 2\hbar^2\psi_{3,1,1}(x)
Now, you might be wondering, what if you have a sum or difference of two of these eigenfunctions, as in your problem? As I said, you can't just convert that sum or difference into a single eigenfunction with a single set of quantum numbers (because the eigenfunctions are linearly independent). But you can use the distributive property. You're familiar with this property for multiplication over addition, but it also works for "operation" over addition. So if you have the operator
L^2, and you apply it to a sum of two states \psi_{n,l,m}(x) + \psi_{n',l',m'}(x), it works roughly the same way as with multiplication:
L^2[\psi_{n,l,m}(x) + \psi_{n',l',m'}(x)] = L^2\psi_{n,l,m}(x) + L^2\psi_{n',l',m'}(x)
even though the operator L^2 does not actually represent multiplication. The same thing works for any other operator instead of L^2, and for any other sum or difference or linear combination of any wavefunctions. (Well, technically: it's only true for
linear operators. All common QM operators are linear, so don't worry about that exception.)
Once you've applied the distributive property, you have two terms of the form L^2\psi, where \psi is an eigenfunction of L^2. You can evaluate those with the formula for the eigenvalues, as I did above. Then look at what you get and see if it can be simplified into the form (eigenvalue)x(eigenfunction).
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