Eigenstates in quantum mechanics

[Axiom17]

Homework Statement

1. Is state $\psi_{0,2,1}-\psi_{5,0,1}$ an eigenstate of $L_{x}$

2. Is state $\psi_{1,3,1}-\psi_{4,2,0}$ an eigenstate of $L^{2}$

Homework Equations

Stationary state of Hamiltonian defined by: [itex]\psi_{n,l,m}[/tex] where the subscripts denote quantum numbers.

The Attempt at a Solution

It's really the method I'm having problems with, just not sure how to get started

 

[MathematicalPhysicist]

For 2, I remeber by heart that L^2 acting on a $$\psi_{n,l,m}$$
gives back $$\hbar^2 l(l+1) \psi_{n,l,m}$$

So its an eigenstate iff L^2 acting on this eigenstate gives back the same eigenstate multiplied by some mathematical scalar constant.

Did that helped?

 

[Axiom17]

Did that helped?

Um, sort of. But I don't get what calculations to do to show this.

 

[MathematicalPhysicist]

If you have an opeartor A acting on its eigenstate (or eigenvector), v
then you know that Av=av where a is some numerical constant.

Now if in your cases after calculating you get that there isn't such a constant then obviously this state isn't an eigenstate of this operator.

in your case, if we have v,w eigenstates of an operator A, s.t
Av=av, Aw=bw then A(v+-w)=av+-bw
Now if it were an eigenstate then A(v+-w)=c(v+-w)
Now see if you can convince yourself why when a and b are both different than c then this state, v+-w isn't an eigenstate.

BTW, did you take a course in linear algebra?

 

[Axiom17]

OK so I need to use the equation $Av=av$ I remember that now. Let's make sure I understand, so If I have a state v which when multiplied by an operator A gives the result av where v is the input state and a is a constant, then the state v is an eigenstate of A.

So in the question:

$v=\psi_{0,2,1}-\psi_{5,0,1}$

$A=L_{x}$

.. is that correct?

Then I'd need to first calculate the result of the addition of the two wave functions, and then calculate the multiplication of the result with the operator.

1. Not sure how to add the two wave functions together.. guessing I can't just add the respective quantum numbers together. Not sure how to treat these, am used to $\psi(x)=[value]$ rather than dealing with these quantum numbers.

2. Multiplication of resultant wave function with operator.. probably can do this once sort out point 1 i.e. how to work with the wave functions.

 

[Axiom17]

.. still don't understand how to do this :grumpy:

 

[diazona]

Then I'd need to first calculate the result of the addition of the two wave functions, and then calculate the multiplication of the result with the operator.

No; just apply the operator to the sum of the two wavefunctions, and use the distributive property of addition.

 

[Axiom17]

ok.. but how to I do that?

I don't get how to sum the two wave functions together, specifically what to do with the quantum numbers, then how to multiply that result by the operator which isn't defined. If that all makes sense.

so I've got my two wave functions, p and q, and need to do Ap=ap and Aq=bq then A(p+(-q))=c(p+(-q)) to show whether p-q is an eigenstate of A.

If I just sum the quantum numbers, which I've sure is probably not the way to go but still, then I get: $\psi_{-5,2,0}$ which would then lead to $A \psi_{-5,2,0}= a\psi_{-5,2,0}$ but clearly that doesn't prove anything.

I'm stuck on how to do the calculations

 

[diazona]

You can't just add the quantum numbers of two wavefunctions. The difference of $\psi_{0,2,1}$ and $\psi_{5,0,1}$ cannot be expressed as a wavefunction with some other set of quantum numbers.

Do you know what the distributive property is?

 

[Axiom17]

You can't just add the quantum numbers of two wavefunctions. The difference of $\psi_{0,2,1}$ and $\psi_{5,0,1}$ cannot be expressed as a wavefunction with some other set of quantum numbers.

.. I thought not, but I didn't know what else to do, just thought would check that was incorrect.

Do you know what the distributive property is?

.. 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.

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}.

 

[diazona]

.. 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.

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).

 

[Axiom17]

Thanks Diazona your explanation is very good .

So I use this formula:

$$L^2\psi_{n,l,m}(x) = \hbar^2 l(l + 1)\psi_{n,l,m}(x)$$

For example:

$$L^2\psi_{3,1,1}(x) = \hbar^2 1(1 + 1)\psi_{3,1,1}(x) = 2\hbar^2\psi_{3,1,1}(x)$$

Therefore $\psi_{3,1,1}$ is an eigenstate of $L^{2}$

.. correct?

And I can substitute $L^{2}$ for $L_{x}$.. and still use the same formula?

Now if I have two wave functions, this is when I use the distributive property.

For example:

$$\psi_{3,5,1}-i\psi_{4,1,0}$$

So first..

$$L_{z][\psi_{3,5,1} -i\psi_{4,1,0}] = L_{z}(\psi_{3,5,1}) + L_{z}(-i\psi_{4,1,0})$$

Then..

$$L_{z}(\psi_{3,5,1}) = \hbar^2 l(l + 1)\psi_{n,l,m}(x) = \hbar^2 5(5 + 1)(\psi_{3,5,1})=30\hbar^2(\psi_{3,5,1})$$

$$L_{z}(-i\psi_{4,1,0})= \hbar^2 l(l + 1)-i\psi_{n,l,m}(x)=\hbar^2 1(1+ 1)(-i\psi_{4,1,0})=2\hbar^2(-i\psi_{4,1,0})$$

.. correct?

(Not sure if I've treated the $i$ correctly?)

So both wavefunctions are eigenfunctions of the operator $L_{z}$. But I need to show the state comprising of both of them is an eigenstate of $L_{z}$.. so do I need to do any more calculations, or is the steps above enough to prove this?Hopefully I've understood the method now!

 

[diazona]

For example:

$$L^2\psi_{3,1,1}(x) = \hbar^2 1(1 + 1)\psi_{3,1,1}(x) = 2\hbar^2\psi_{3,1,1}(x)$$

Therefore $\psi_{3,1,1}$ is an eigenstate of $L^{2}$

.. correct?

Yes, that's correct. The fact that you can write
$$L^2\psi_{3,1,1}(x) = (\text{some constant})\psi_{3,1,1}(x)$$
means that $\psi_{3,1,1}(x)$ is an eigenfunction of $L^2$.

And I can substitute $L^{2}$ for $L_{x}$.. and still use the same formula?

Well, no. There are a couple of reasons why: first of all, the formula for the eigenvalue is different for different operators. For example, if you used $L_z$ instead of $L^2$, it would be
$$L_z \psi_{n,l,m}(x) = \hbar m \psi_{n,l,m}(x)$$
The other reason is that the states $\psi_{n,l,m}(x)$ are not eigenstates of $L_x$. When you apply $L_x$ to a single one of these states, what you get will not be a multiple of the original state. It takes a linear combination of different $\psi$'s to make an eigenstate of $L_x$. Anyway, since that's slightly more involved, let's just focus on part (b) for now.

Now if I have two wave functions, this is when I use the distributive property.

For example:

$$\psi_{3,5,1}-i\psi_{4,1,0}$$

So first..

$$L_{z][\psi_{3,5,1} -i\psi_{4,1,0}] = L_{z}(\psi_{3,5,1}) + L_{z}(-i\psi_{4,1,0})$$

Then..

$$L_{z}(\psi_{3,5,1}) = \hbar^2 l(l + 1)\psi_{n,l,m}(x) = \hbar^2 5(5 + 1)(\psi_{3,5,1})=30\hbar^2(\psi_{3,5,1})$$

$$L_{z}(-i\psi_{4,1,0})= \hbar^2 l(l + 1)\psi_{n,l,m}(x)=\hbar^2 1(1+ 1)(\psi_{4,1,0})=2\hbar^2(\psi_{4,1,0})$$

.. correct?

You got one thing wrong in there: when you calculate $L_{z}(-i\psi_{4,1,0})$, the -i is a constant factor which you can pull out in front.
$$L_{z}(-i\psi_{4,1,0}) = -i L_z\psi_{4,1,0}$$
This is a feature common to all linear operators. In fact, it's right in the definition of what it means to be linear:
$$A[c_1 f(x) + c_2 g(x)] = c_1 A f(x) + c_2 A g(x)$$
If (and only if) operator A satisfies this equation for all possible choices of constants $c_1$ and $c_2$ and for all functions $f(x)$ and $g(x)$, then A is linear.

So both wavefunctions are eigenfunctions of the operator $L_{z}$. But I need to show the state comprising of both of them is an eigenstate of $L_{z}$.. so do I need to do any more calculations, or is the steps above enough to prove this?

You'll need to do a little more. For any wavefunction $\psi$ (not just one of the numbered hydrogen eigenfunctions, but any wavefunction at all), the eigenvalue equation reads
$$A\psi(x) = (\text{some constant})\psi(x)$$
Now, if $\psi$ is a linear combination of two other wavefunctions, $\psi(x) = \psi_a(x) + \psi_b(x)$ for example, that equation becomes
$$A[\psi_a(x) + \psi_b(x)] = (\text{some constant})[\psi_a(x) + \psi_b(x)]$$
That's what you need to show (or show that it can't be done). What you have so far is
$$A[\psi_a(x) + \psi_b(x)] = (\text{some constant})\psi_a(x) + (\text{some other constant})\psi_b(x)$$
You should be able to figure out under what conditions this last equation reduces to the preceding one.

 

[Axiom17]

For example, if you used $L_z$ instead of $L^2$, it would be
$$L_z \psi_{n,l,m}(x) = \hbar m \psi_{n,l,m}(x)$$

Ok sure, good. I take it this would be the same method if I had just a '2' or a '3i' or some term like that as part of the wave function.

So I have this:

$$L_{z}(\psi_{3,5,1}) = \hbar^2 l(l + 1)\psi_{n,l,m}(x) = \hbar^2 5(5 + 1)(\psi_{3,5,1})=30\hbar^2(\psi_{3,5,1})$$
$$L_{z}(-i\psi_{4,1,0})= -i\hbar^2 l(l + 1)\psi_{n,l,m}(x)=-i\hbar^2 1(1+ 1)(\psi_{4,1,0})=-2i\hbar^2(\psi_{4,1,0})$$

Hence:
$$L_{z}(\psi_{3,5,1}-i\psi_{4,1,0}) = 30\hbar^2(\psi_{3,5,1})-2i\hbar^2(\psi_{4,1,0})$$

Hence:

$$L_{z}(\psi_{3,5,1}-i\psi_{4,1,0}) = ((30-2i)\hbar)((\psi_{3,5,1})-i(\psi_{4,1,0}))$$

So the state is an eigenstate of $L_{z}$. The constant is just $(30-2i)\hbar$.

I think that's all correct now and have done what's required to show it.

Now how would I go about showing a state (combination of wavefunctions similar to already working with) was an eigenstate of the Hamiltonian? I assume that it's a similar method, different formula? Perhaps it's this: $-i\hbar \frac{d\psi}{dx}=p\psi$ or something similar.

 

[diazona]

So I have this:

$$L_{z}(\psi_{3,5,1}) = \hbar^2 l(l + 1)\psi_{n,l,m}(x) = \hbar^2 5(5 + 1)(\psi_{3,5,1})=30\hbar^2(\psi_{3,5,1})$$
$$L_{z}(-i\psi_{4,1,0})= -i\hbar^2 l(l + 1)\psi_{n,l,m}(x)=-i\hbar^2 1(1+ 1)(\psi_{4,1,0})=-2i\hbar^2(\psi_{4,1,0})$$

No, that's not correct - you're still using the eigenvalue for $L^2$. Look at the formula that you quoted from me.

Hence:
$$L_{z}(\psi_{3,5,1}-i\psi_{4,1,0}) = 30\hbar^2(\psi_{3,5,1})-2i\hbar^2(\psi_{4,1,0})$$

If the operator were $L^2$ instead of $L_z$, that would be correct.

Hence:

$$L_{z}(\psi_{3,5,1}-i\psi_{4,1,0}) = ((30-2i)\hbar)((\psi_{3,5,1})-i(\psi_{4,1,0}))$$

No! That doesn't follow from the previous formula at all.

Now how would I go about showing a state (combination of wavefunctions similar to already working with) was an eigenstate of the Hamiltonian? I assume that it's a similar method, different formula? Perhaps it's this: $-i\hbar \frac{d\psi}{dx}=p\psi$ or something similar.

If you have a wavefunction $\psi$, in order to show that it is an eigenfunction of the Hamiltonian, you would have to show that when you apply the Hamiltonian to the function, you get back a multiple of the same function. Mathematically, that would be expressed as
$$H\psi = E\psi$$
where E is some constant, the eigenvalue. It's the same as with the angular momentum operators, except that, again, the formula for the eigenvalue E will be different. What the formula is depends on what the Hamiltonian is.

 

[Axiom17]

I think it's not helping me messing up latex coding on here a bit! :shy:

Let's try again..

Formula:

$$L_z \psi_{n,l,m}(x) = \hbar m \psi_{n,l,m}(x)$$

State - Wavefunction 1:

$$L_{z}(\psi_{3,5,1}) = \hbar (1)\psi_{3,5,1}= \hbar \psi_{3,5,1}$$

State - Wavefunction 2:

$$L_{z}(-i\psi_{4,1,0})= -i \hbar (0) \psi_{4,1,0} = 0$$

Hence:

$$L_z (\psi_{3,5,1}-i\psi_{4,1,0})=\hbar \psi_{3,5,1}$$

So the state $\psi_{3,5,1}-i\psi_{4,1,0}$ is a eigenstate of $L_{z}$

(though not sure on that now, since calculation doesn't result in [constant][original state] just part of the state)

.. how's that?

 

[diazona]

(though not sure on that now, since calculation doesn't result in [constant][original state] just part of the state)

"Part of the state" doesn't count. It has to be a multiple of the original state exactly.

Other than that, looks good

 

[Axiom17]

.. yes so as I thought, in that example the state wouldn't be an eigenstate of the operator.

Thanks for you help with this!