Stupid noob question about state vectors

[Denver Dang]

Homework Statement

I got a Problem that says:
In this Problem we consider the angular momentum, $$\hat{L}$$, which normed common eigenstates for $$\hat{L}^{2}$$ and $$\hat{L}_{z}$$ is described by $\left|l,m\right\rangle$, where $l$ and $m$ is the corresponding quantum numbers:

$$\hat{L}^{2}\left|l,m\right\rangle = l\left(l+1\right)\hbar^{2}\left|l,m\right\rangle$$

$$\hat{L}_{z}\left|l,m\right\rangle = m\hbar\left|l,m\right\rangle$$

As usual we introduce the operators:

$$\hat{L}_{+} = \hat{L}_{x} + i\hat{L}_{y}$$ and $$\hat{L}_{-} = \hat{L}_{x} - i\hat{L}_{y}$$

Now consider the three-dimensional subspace, $\mathcal{H}$, corresponding to the quantum number $l = 1$, ie. the states $\left|1,-1\right\rangle$, $\left|1,0\right\rangle$ and $\left|1,1\right\rangle$.

Now, express the operators $$\hat{L}^{2}$$ and $$\hat{L}_{z}$$ as matrices in $\mathcal{H}$.

Homework Equations

The Attempt at a Solution

Well, my problem is that I'm not sure how to interpret the states $\left|1,-1\right\rangle$, $\left|1,0\right\rangle$ and $\left|1,1\right\rangle$. Usually when there are fx. 1 in one of them, it's the first basisvector etc. But what is it now ? Is it still a vector, in which case what kind ?
I've been flipping through my entire book to see if it showed anything similar, but I haven't been able to find anything that could tell me how I have to interpret the states "vectorwise".
So I was hoping someone could help me :)

I know it's kinda noobish, but I have been stuck here for a while, and can't seem to figure out what the matrix is before I know how to interpret the states.Regards.

 

[kuruman]

Well, my problem is that I'm not sure how to interpret the states $\left|1,-1\right\rangle$, $\left|1,0\right\rangle$ and $\left|1,1\right\rangle$.

Interpret them as three basis vectors spanning a three-dimensional Hilbert space, much like i, j and k span three-dimensional Euclidean space. You can order them any way you please, but conventionally the order is |1,1>, |1,0>, |1,-1>.

I don't know if I have answered your question, but I don't understand the source of your confusion and exactly what you are asking. If you can explain exactly what troubles you, I might be able to give you a better answer.

 

[Denver Dang]

Hmmm, I mean, if it was just |1> I know it would be a basis vector in three dimensions (This case at least), which I would write: (1,0,0).
But how does this look like in vector-form, that's my question and problem :)

 

[kuruman]

Hmmm, I mean, if it was just |1> I know it would be a basis vector in three dimensions (This case at least), which I would write: (1,0,0).
But how does this look like in vector-form, that's my question and problem :)

OK. Note that the basis states are written in the form |L,M_L> where L = 1 for all three states. You can toss that out since it is the same for everybody and write the states as
|1>, |0>, |-1>. So

|1> = (1,0,0)
|0> = (0,1,0)
|-1> = (0,0,1)

 

[Denver Dang]

OK. Note that the basis states are written in the form |L,M_L> where L = 1 for all three states. You can toss that out since it is the same for everybody and write the states as
|1>, |0>, |-1>. So

|1> = (1,0,0)
|0> = (0,1,0)
|-1> = (0,0,1)

Ahhh, I see :)

But what if L = 2, how would that look fx. ?

 

[kuruman]

Ahhh, I see :)

But what if L = 2, how would that look fx. ?

The dimensionality of the Hilbert space is n = 2*L+1. For L =2, therefore, you are in 5-dimensional Hilbert space. If you denote the basis vectors just by their M_L value, you get column vectors
|2> = (1,0,0,0,0)
|1> = (0,1,0,0,0)
|0> = (0,0,1,0,0)
...
and so on.

 

[Denver Dang]

The dimensionality of the Hilbert space is n = 2*L+1. For L =2, therefore, you are in 5-dimensional Hilbert space. If you denote the basis vectors just by their M_L value, you get column vectors
|2> = (1,0,0,0,0)
|1> = (0,1,0,0,0)
|0> = (0,0,1,0,0)
...
and so on.

Great :)

Thank you very much kuruman.