- #1
lfqm
- 22
- 0
Hi guys, my quesion is quite simple but I think I need to give some background...
Let's suppose I have 3 qubits, so the basis of the space is:
[tex]\left\{{\left |{000}\right>,\left |{001}\right>,\left |{010}\right>,\left |{100}\right>,\left |{011}\right>,\left |{101}\right>,\left |{110}\right>,\left |{111}\right>}\right\}[/tex]
i.e. the dimesion is [tex]2^{3}=8[/tex]
I define the angular momentum-like operators:
[tex]J_{z}=\left [{\frac{1}{2}\left ({\left |{1}\right>\left <{1}\right|-\left |{0}\right>\left <{0}\right|}\right)\otimes I \otimes I}\right] + \left [{I \otimes \frac{1}{2}\left ({\left |{1}\right>\left <{1}\right|-\left |{0}\right>\left <{0}\right|}\right) \otimes I}\right] + \left [{I \otimes I \otimes \frac{1}{2}\left ({\left |{1}\right>\left <{1}\right|-\left |{0}\right>\left <{0}\right|}\right)}\right][/tex]
[tex]J_{+}=\left [{|1⟩⟨0|⊗I⊗I}\right]+\left [{I⊗|1⟩⟨0|⊗I}\right]+\left [{I⊗I⊗|1⟩⟨0|}\right][/tex]
[tex]J_{-}=\left [{|0⟩⟨1|⊗I⊗I}\right]+\left [{I⊗|0⟩⟨1|⊗I}\right]+\left [{I⊗I⊗|0⟩⟨1|}\right][/tex]
[tex]J^{2}=\frac{1}{2}\left ({J_{+}J_{-}+J_{-}J_{+}}\right)+J_{z}^{2}[/tex]
And get the right commutation relations:
[tex][J_{z},J_{\pm{}}]=\pm{}J_{\pm{}}[/tex]
[tex][J_{+},J_{-}]=2J_{z}[/tex]
[tex][J^{2},J_{z}]=0[/tex]
So I deduce [tex]J^{2}[/tex] and [tex]J_{z}[/tex] share a common set of eigenvectors [tex]\left\{{\left |{j,m}\right>}\right\}[/tex] where [tex]\left |{m}\right |\leq{j}\leq{\displaystyle\frac{3}{2}}[/tex]
But there are only 6 of this vectors:
[tex]\left\{{\left |{j=\frac{3}{2},m=\frac{3}{2}}\right>,\left |{j=\frac{3}{2},m=\frac{1}{2}}\right>,\left |{j=\frac{3}{2},m=-\frac{1}{2}}\right>,\left |{j=\frac{3}{2},m=-\frac{3}{2}}\right>,\left |{j=\frac{1}{2},m=\frac{1}{2}}\right>,\left |{j=\frac{1}{2},m=-\frac{1}{2}}\right>}\right\}[/tex]
So my question is, does the sharing set of eigenvectors between two commuting observables must be complete?
Why isn't complete in this example?
What are the other 2 missing common eigenvectors?
Thanks in advance!
Let's suppose I have 3 qubits, so the basis of the space is:
[tex]\left\{{\left |{000}\right>,\left |{001}\right>,\left |{010}\right>,\left |{100}\right>,\left |{011}\right>,\left |{101}\right>,\left |{110}\right>,\left |{111}\right>}\right\}[/tex]
i.e. the dimesion is [tex]2^{3}=8[/tex]
I define the angular momentum-like operators:
[tex]J_{z}=\left [{\frac{1}{2}\left ({\left |{1}\right>\left <{1}\right|-\left |{0}\right>\left <{0}\right|}\right)\otimes I \otimes I}\right] + \left [{I \otimes \frac{1}{2}\left ({\left |{1}\right>\left <{1}\right|-\left |{0}\right>\left <{0}\right|}\right) \otimes I}\right] + \left [{I \otimes I \otimes \frac{1}{2}\left ({\left |{1}\right>\left <{1}\right|-\left |{0}\right>\left <{0}\right|}\right)}\right][/tex]
[tex]J_{+}=\left [{|1⟩⟨0|⊗I⊗I}\right]+\left [{I⊗|1⟩⟨0|⊗I}\right]+\left [{I⊗I⊗|1⟩⟨0|}\right][/tex]
[tex]J_{-}=\left [{|0⟩⟨1|⊗I⊗I}\right]+\left [{I⊗|0⟩⟨1|⊗I}\right]+\left [{I⊗I⊗|0⟩⟨1|}\right][/tex]
[tex]J^{2}=\frac{1}{2}\left ({J_{+}J_{-}+J_{-}J_{+}}\right)+J_{z}^{2}[/tex]
And get the right commutation relations:
[tex][J_{z},J_{\pm{}}]=\pm{}J_{\pm{}}[/tex]
[tex][J_{+},J_{-}]=2J_{z}[/tex]
[tex][J^{2},J_{z}]=0[/tex]
So I deduce [tex]J^{2}[/tex] and [tex]J_{z}[/tex] share a common set of eigenvectors [tex]\left\{{\left |{j,m}\right>}\right\}[/tex] where [tex]\left |{m}\right |\leq{j}\leq{\displaystyle\frac{3}{2}}[/tex]
But there are only 6 of this vectors:
[tex]\left\{{\left |{j=\frac{3}{2},m=\frac{3}{2}}\right>,\left |{j=\frac{3}{2},m=\frac{1}{2}}\right>,\left |{j=\frac{3}{2},m=-\frac{1}{2}}\right>,\left |{j=\frac{3}{2},m=-\frac{3}{2}}\right>,\left |{j=\frac{1}{2},m=\frac{1}{2}}\right>,\left |{j=\frac{1}{2},m=-\frac{1}{2}}\right>}\right\}[/tex]
So my question is, does the sharing set of eigenvectors between two commuting observables must be complete?
Why isn't complete in this example?
What are the other 2 missing common eigenvectors?
Thanks in advance!