- #1
CharlieCW
- 53
- 5
I just began graduate school and was struggling a bit with some basic notions, so if you could give me some suggestions or point me in the right direction, I would really appreciate it.
1. Homework Statement
Given an infinite base of orthonormal states in the Hilbert space, ##|n\rangle_{n\in\N}##, and the following operators:
$$s_+=\sum_{n=0}^{\infty}|2n+1\rangle\langle2n|$$
$$s_-=\sum_{n=0}^{\infty}|2n\rangle\langle2n+1|$$
$$s_z=\sum_{n=0}^{\infty}(|2n+1\rangle\langle2n+1|+|2n\rangle\langle2n|)$$
Find the eigenvalues of the operator,
$$n_s=\vec{n}\cdot\vec{s}=cos\theta s_z +sin\theta (e^{i\phi}s_-+e^{-i\phi}s_+)$$
where ##\vec{n}=(cos\theta,sin\theta cos\phi,sin\theta sin\phi)##
$$[A,B]=AB-BC$$
$$\langle n | m \rangle=\delta_{n,m}$$
$$(|n\rangle\langle m|)^{\dagger}=|m \rangle\langle n |$$
To determine the eigenvalues of ##n_s## we just apply it to the state ##|n\rangle##:
$$n_s|n\rangle=(cos\theta s_z +sin\theta e^{i\phi}s_-+sin\theta e^{-i\phi}s_+)|n\rangle$$
We can begin by calculating the conmutators of the operators:
$$[s_+,s_-]=s_z$$
$$[s_-,s_z]=2s_+$$
$$[s_z,s_+]=2s_-$$
It is also worth noting that we directly observe that $$(s_+)^{\dagger}=s_-$$ and viceversa.
By applying the operators to the state $$|n'\rangle$$ we can also easily verify:
$$s_+|n'\rangle=\sum_{n=0}^{\infty}|2n+1\rangle\langle2n|n'\rangle=\sum_{n=0}^{\infty}|2n+1\rangle\delta_{2n,n'}=c_{n+}|n'+1\rangle$$
$$s_-|n'\rangle=\sum_{n=0}^{\infty}|2n\rangle\langle2n+1|n'\rangle=\sum_{n=0}^{\infty}|2n\rangle\delta_{2n+1,n'}=c_{n-}|n'-1\rangle$$
Where ##c_{n+}## and ##c_{n-}## are just underdetermined constants. Thus we can call ##s_+## the creator and ##s_-## the anhilitation operators for this case.
Let's begin by determining ##c_{n+}##:
$$\langle n|s_-s_+|n\rangle=c_{n+}\langle n|s_-|n+1\rangle=|c_{n+}|^{2}\langle n+1|n+1\rangle=|c_{n+}|^{2}$$
Where in the last steps I used the conjugate relationship ##(s_+)^{\dagger}=s_-##. But also we have, using the conmutator relationships:
$$\langle n|s_-s_+|n\rangle=\langle n |s_z-s_+s_-|n\rangle=\langle n |s_z|n\rangle-\langle|s_+s_-|\rangle=|c_{nz}|^2\langle n |n\rangle-|c_{n-}|^{2}=|c_{nz}|^{2}-|c_{n-}|^{2}$$
Where the last step was done by analogy:
$$|c_{n+}|^{2}=|c_{nz}|^2-|c_{n-}|^2$$
By determining ##c_{n-}## in the same way, we arrive at:
$$|c_{n-}|^{2}=|c_{nz}|^2+|c_{n+}|^2$$
However, I'm not entirely sure how to determine ##c_{nz}## as I cannot conjugate it with the other operators, i.e., only ##(s_+)^{\dagger}=s_-## holds true. Moreover, I have the impression I'm complicating this a bit, as I think there should be an easier way to determine the eigenvalues of each operator.
On the other hand, since the operators don't conmute, we can't just find the eigenvalues of each term and add them together. In this case, I don't know exactly how to work it around for an infinite basis, as all the examples I've read in Sakurai and Weinberg are for a finite basis and/or operators that conmute.
My idea for this case would be, once I find the eigenvalues of each operator:
$$\langle n|n_s|n\rangle=\langle n|(cos\theta s_z +sin\theta e^{i\phi}s_-+sin\theta e^{-i\phi}s_+)|n\rangle=\langle n| (cos\theta c_{nz}|n\rangle + sin\theta e^{i\phi}c_{n-}|n-1\rangle + sin\theta e^{-i\phi}c_{n+}|n+1\rangle)=\langle n |cos\theta c_{nz}|n\rangle=cos\theta c_{nz}$$
Since the other terms vanish by the orthonormality condition ##\langle n | m \rangle=\delta_{n,m}##
1. Homework Statement
Given an infinite base of orthonormal states in the Hilbert space, ##|n\rangle_{n\in\N}##, and the following operators:
$$s_+=\sum_{n=0}^{\infty}|2n+1\rangle\langle2n|$$
$$s_-=\sum_{n=0}^{\infty}|2n\rangle\langle2n+1|$$
$$s_z=\sum_{n=0}^{\infty}(|2n+1\rangle\langle2n+1|+|2n\rangle\langle2n|)$$
Find the eigenvalues of the operator,
$$n_s=\vec{n}\cdot\vec{s}=cos\theta s_z +sin\theta (e^{i\phi}s_-+e^{-i\phi}s_+)$$
where ##\vec{n}=(cos\theta,sin\theta cos\phi,sin\theta sin\phi)##
Homework Equations
$$[A,B]=AB-BC$$
$$\langle n | m \rangle=\delta_{n,m}$$
$$(|n\rangle\langle m|)^{\dagger}=|m \rangle\langle n |$$
The Attempt at a Solution
To determine the eigenvalues of ##n_s## we just apply it to the state ##|n\rangle##:
$$n_s|n\rangle=(cos\theta s_z +sin\theta e^{i\phi}s_-+sin\theta e^{-i\phi}s_+)|n\rangle$$
We can begin by calculating the conmutators of the operators:
$$[s_+,s_-]=s_z$$
$$[s_-,s_z]=2s_+$$
$$[s_z,s_+]=2s_-$$
It is also worth noting that we directly observe that $$(s_+)^{\dagger}=s_-$$ and viceversa.
By applying the operators to the state $$|n'\rangle$$ we can also easily verify:
$$s_+|n'\rangle=\sum_{n=0}^{\infty}|2n+1\rangle\langle2n|n'\rangle=\sum_{n=0}^{\infty}|2n+1\rangle\delta_{2n,n'}=c_{n+}|n'+1\rangle$$
$$s_-|n'\rangle=\sum_{n=0}^{\infty}|2n\rangle\langle2n+1|n'\rangle=\sum_{n=0}^{\infty}|2n\rangle\delta_{2n+1,n'}=c_{n-}|n'-1\rangle$$
Where ##c_{n+}## and ##c_{n-}## are just underdetermined constants. Thus we can call ##s_+## the creator and ##s_-## the anhilitation operators for this case.
Let's begin by determining ##c_{n+}##:
$$\langle n|s_-s_+|n\rangle=c_{n+}\langle n|s_-|n+1\rangle=|c_{n+}|^{2}\langle n+1|n+1\rangle=|c_{n+}|^{2}$$
Where in the last steps I used the conjugate relationship ##(s_+)^{\dagger}=s_-##. But also we have, using the conmutator relationships:
$$\langle n|s_-s_+|n\rangle=\langle n |s_z-s_+s_-|n\rangle=\langle n |s_z|n\rangle-\langle|s_+s_-|\rangle=|c_{nz}|^2\langle n |n\rangle-|c_{n-}|^{2}=|c_{nz}|^{2}-|c_{n-}|^{2}$$
Where the last step was done by analogy:
$$|c_{n+}|^{2}=|c_{nz}|^2-|c_{n-}|^2$$
By determining ##c_{n-}## in the same way, we arrive at:
$$|c_{n-}|^{2}=|c_{nz}|^2+|c_{n+}|^2$$
However, I'm not entirely sure how to determine ##c_{nz}## as I cannot conjugate it with the other operators, i.e., only ##(s_+)^{\dagger}=s_-## holds true. Moreover, I have the impression I'm complicating this a bit, as I think there should be an easier way to determine the eigenvalues of each operator.
On the other hand, since the operators don't conmute, we can't just find the eigenvalues of each term and add them together. In this case, I don't know exactly how to work it around for an infinite basis, as all the examples I've read in Sakurai and Weinberg are for a finite basis and/or operators that conmute.
My idea for this case would be, once I find the eigenvalues of each operator:
$$\langle n|n_s|n\rangle=\langle n|(cos\theta s_z +sin\theta e^{i\phi}s_-+sin\theta e^{-i\phi}s_+)|n\rangle=\langle n| (cos\theta c_{nz}|n\rangle + sin\theta e^{i\phi}c_{n-}|n-1\rangle + sin\theta e^{-i\phi}c_{n+}|n+1\rangle)=\langle n |cos\theta c_{nz}|n\rangle=cos\theta c_{nz}$$
Since the other terms vanish by the orthonormality condition ##\langle n | m \rangle=\delta_{n,m}##