Eigenvalue problem what am i doing wrong?

[qqchico]

This is for a spin 1 particle. I can't get the determinant to come out right. Can someone show me what i am doing wrong

 

[qqchico]

S*n=[tex]\hbar/\sqrt{2}[/tex][tex]\left(\begin{array}{ccc}\sqrt{2}·COS(\theta)&SIN(\theta)·COS(\phi) - iSIN(\theta)·SIN(\phi)&0\\SIN(\theta)·COS(\phi) + i·\SIN(\theta)·SIN(\phi)&0&SIN(\theta)·\COS(\phi) - i·SIN(\theta)·SIN(\phi)\\0&\SIN(\theta)·COS(\phi) + i·SIN(\theta)·SIN(\phi)&-\sqrt{2}·COS(\theta)\end{array}\right)[/tex]

 

[qqchico]

this is supposed to be a 3x3 matrices but it wouldn't fit

 

[Nylex]

Where you have {cc} in the code, it should be {ccc} for 3 columns. Also, you use \cos \sin, etc for those functions.

 

[qqchico]

[tex]\hbar/\sqrt{2}[/tex][tex]\left(\begin{array}{ccc}\sqrt{2}·COS(\theta)&e^(^-^\imath^\phi^)SIN(\theta)&0\\e^(^\imath^\phi^)SIN(\theta)&0&e^-^(^\imath^\phi^)SIN(\theta)\\0&SIN(\theta)e^(^\imath^\phi^)&-\sqrt{2}·COS(\theta)\end{array}\right )[/tex][tex]\left(\begin{array}{ccc}\(A&\\B\\C\end{array}\right)[/tex]=[tex]\hbar/\sqrt{2}[/tex] [tex]*\lambda[/tex][tex]\left(\begin{array}{ccc}\(A&\\B\\C\end{array}\right)[/tex]


I then set the determinant equal to 0 subtracted lamda across the diagonals to get

det [tex]\left(\begin{array}{ccc}\sqrt{2}·COS(\theta)-\lambda&e^(^-^\imath^\phi^)SIN(\theta)&0\\e^(^\imath^\phi^)SIN(\theta)&-\lambda&e^-^(^\imath^\phi^)SIN(\theta)\\0&SIN(\theta)e^(^\imath^\phi^)&-\sqrt{2}·COS(\theta)-\lambda\end{array}\right )[/tex]=0

but i get the wrong answer. where am i messing up?













1

 

[Nylex]

Post the actual wording of the question and the rest of your working, so we can see.

 

[qqchico]

Solve the eigenvalue problem Sn[tex]|\lambda>[/tex]=[tex]\lambda|\lambda>[/tex] for a spin 1 particle. Find the eigenvectors. I actually have the eigenvectors. I just need to show how to get them

 

[qqchico]

When I solve the determinant i get [tex]2\lambda-\lambda^3=0[/tex]

 

[Galileo]

What happened to the [itex]\frac{\hbar}{\sqrt{2}}[/itex]? It's not in your expression for the determinant. Also remember to multiply each entry by that factor before subtracting [itex]\lamba I[/itex].

EDIT: I got [itex]\lambda \hbar^2-\lambda^3=0[/itex] as the characteristic equation.

 

[ayalam]

but wouldn't the [tex]\hbar/\sqrt{2}[/tex] cancel out because of the other across the equal sign

 

[qqchico]

Anyone? I really need help this is due tomorrow.

 

[dextercioby]

What does that matrix represent ?

Daniel.

 

[Galileo]

Why is there another one across the equal sign? It shouldn't be there if you were just calculating the eigenvalues of the matrix Sn.

Please post the wording of the question.

 

[qqchico]

S= sin operator S*n for a spin 1 particle where Shat is equal to Sxihat+Syjhat+Szkhat

and nhatis equal to [tex]\sin\theta\cos\phi(i)+sin\theta\sin\phi(j)+\cos\theta\\(k)[/tex]

multiply the together to get

[tex]\sin\theta\cos\phi(Sx)+sin\theta\sin\phi(Sy)+\cos\theta\\(Sz)[/tex]

[tex]Sx=[/tex] [tex]\left(\begin{array}{ccc}0&1&0\\1&0&1\\0&1&0\end{array}\right)[/tex]

[tex]Sy=[/tex] [tex]\left(\begin{array}{ccc}0&i&0\\i&0&-i\\0&i&0\end{array}\right)[/tex]

[tex]Sz=[/tex] [tex]\left(\begin{array}{ccc}1&0&0\\0&0&0\\0&0&-1\end{array}\right)[/tex]

plugged into [tex]\sin\theta\cos\phi(Sx)+sin\theta\sin\phi(Sy)+\cos\theta\\(Sz)[/tex]


and solved to get my initial equation

[tex]\left(\begin{array}{ccc}\sqrt{2}·COS(\theta)&SIN(\theta)·COS(\phi) - iSIN(\theta)·SIN(\phi)&0\\SIN(\theta)·COS(\phi) + i·\SIN(\theta)·SIN(\phi)&0&SIN(\theta)·\COS(\phi) - i·SIN(\theta)·SIN(\phi)\\0&\SIN(\theta)·COS(\phi) + i·SIN(\theta)·SIN(\phi)&-\sqrt{2}·COS(\theta)\end{array}\right)[/tex]


The question is solve the eigenvalue problem to show that the eigen vectors are given by and then they give me what the eigenvectors are supposed to be for the spin one particle

 

[qqchico]

we did some for spin 1/2 particles but it was rushed I've had to teach myself and there are no tutors for this since there are only 3 people at my school who have taken it me and 2 girls and they are as lost as i am. The way i did it was the way we did it for the spin 1/2

 

[qqchico]

[tex]|1,1>\mapsto[/tex][tex]\left(\begin{array}{ccc}(e^(^-^i^\phi^)(1+cos\theta))/2\\(sin\theta)/\sqrt{2}\\e^i^\phi(1-cos\theta)/2\end{array}\right)[/tex]



[tex]|1,0>\mapsto[/tex][tex]\left(\begin{array}{ccc}-e^(^i^\phi^)(sin\theta)/\sqrt{2}\\(cos\theta)\\(e^i^\phi\sin\theta/\sqrt{2}\end{array}\right)[/tex]


[tex]|1,-1>\mapsto[/tex][tex]\left(\begin{array}{ccc}e^(^-^i^\phi^)(1-cos\theta)/2\\-(sin\theta)/\sqrt{2}\\e^i^\phi(1+cos\theta)/2\end{array}\right)[/tex]

these are what my answer should turn out to be

 

[dextercioby]

Hmm,okay,no wonder it doesn't appear in the formal theory of angular momentum.
The spin matrices are missing hbar & the sqrt of 2.It's less relevant.

[tex] \left(\hat{\vec{S}}\cdot\vec{n}\right)|\psi\rangle =\lambda|\psi\rangle [/tex]

Which means

[tex] \left(\hat{\vec{S}}\cdot\vec{n}-\lambda \hat{1}\right)|\psi\rangle=0 [/tex]

,i.e.

[tex]\frac{\hbar}{\sqrt{2}}\left(\begin{array}{ccc}\left\sqrt{2}\cos\theta -\frac{\sqrt{2}}{\hbar}\lambda & e^{-i\phi}\sin\theta & 0\\e^{i\phi}\sin\theta & -\frac{\sqrt{2}}{\hbar}\lambda & e^{-i\phi}\sin\theta \\0 & e^{i\phi}\sin\theta} & -\sqrt{2}\cos\theta -\frac{\sqrt{2}}{\hbar}\lambda \end{array}\right)\left(\begin{array}{c}\psi_{1}\\ \psi_{2}\\ \psi_{3}\end{array}\right)=\left(\begin{array}{c} 0 \\ 0 \\ 0\end{array}\right) [/tex]

with the characteristic polynomial

[tex] P(\lambda)=-\lambda^{3}\frac{2\sqrt{2}}{\hbar^{3}}+\lambda\frac{2\sqrt{2}}{\hbar}=0 [/tex]

whose eigenvalues are

[tex] \left\{\begin{array}{c}\lambda_{1}=+\hbar\\ \lambda_{2}=0 \\ \lambda_{3}=-\hbar\end{array} \right [/tex]

Daniel.

 

[CarlB]

Another way of putting what Daniel just said is to note that your [tex]S_x[/tex] and [tex]S_y[/tex] matrices did not have the proper eigenvalues for spin-1 matrices. Their eigenvalues were instead [tex]-\sqrt{2}, 0, +\sqrt{2}[/tex].

Your approach to the problem was correct. Redo it with [tex]S_x[/tex] and [tex]S_y[/tex] divided by [tex]\sqrt{2}[/tex], and you will get the right answers.

Carl

 

[dextercioby]

The initial vectors if the stansard basis were

[tex] |1,1\rangle =\left(\begin{array}{c} 1\\0\\0\end{array}\right) [/tex]

[tex] |1,0\rangle =\left(\begin{array}{c} 0\\1\\0\end{array}\right) [/tex]

[tex] |1,-1\rangle =\left(\begin{array}{c} 0\\0\\1\end{array}\right) [/tex]

Now,the operator [itex] \hat{\vec{S}}\cdot\vec{n} [/itex] accomplishes a rotation of each of those vectors.Follwing [1],the rotation matrix is

[tex] \mathcal{D}^{(1)}_{m,m'}\left(\alpha,\beta,\gamma\right)=\left(\begin{array}{ccc} \frac{1+\cos\beta}{2}e^{-i(\alpha+\gamma)} & -\frac{\sin\beta}{\sqrt{2}}e^{-i\alpha} & \frac{1-\cos\beta}{2}e^{-i(\alpha-\gamma)} \\ \frac{\sin\beta}{\sqrt{2}} e^{-i\gamma} & \cos\beta & -\frac{\sin\beta}{\sqrt{2}} e^{i\gamma} \\ \frac{1-\cos\beta}{2}e^{i(\alpha-\gamma)} & \frac{\sin\beta}{\sqrt{2}}e^{i\alpha} & \frac{1+\cos\beta}{\sqrt{2}} e^{i(\alpha+\gamma)} \end{array}\right) [/tex]

with the adjustments

[tex] \left\{\begin{array}{c} \alpha\longrightarrow \phi \\ \beta\longrightarrow \theta \\ \gamma\longrightarrow 0 \end{array} \right [/tex]

Apply the rotation matrix on each of the standard basis vectors and you'll get the new vectors.

Check they are eigenvectors for the initial matrix.

Daniel.

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[1]Galindo & Pascual,"Quantum Mechanics I",Springer Verlag,1990.