How Do You Solve a Spin-One-Half Hamiltonian Problem?

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Homework Statement



A single spin-one-half system has Hamiltonian

[tex]H=\alpha*s_x+\beta*s_y[/tex], where [tex]\alpha[/tex] and [tex]\beta[/tex] are real numbers, and [tex]s_x[/tex] and [tex]s_y[/tex] are the x and y components of spin .

a) Using the representation of the spin components as Pauli spin matrices, find an expression for [tex]H^2[/tex] in termms of the above parameters.

b) used the result from part(a) to find the energy eigenvalues.

c) Find the eigenvectors of H in equation [tex]H=\alpha*s_x+\beta*s_y[/tex] in the Pauli spin matrix representation.

d) Supposed that a t time t=0 the system is an eigenstate of [tex]s_z[/tex], with eigen value [tex]+\h-bar/2[/tex]. Find the state vector as a function of time in the Pauli spin matrix representation.

e) Suppose the z-component of the spin in the state found in part d) is measured at time t>0 . Find probability that the result is [tex]+\hbar/2[/tex]

Homework Equations


[tex]s=(\hbar)*(\sigma)/2[/tex]
[tex](\sigma_x)[/tex],[tex](\sigma_y)[/tex], and [tex](\sigma_z)[/tex]


The Attempt at a Solution



a) Just multiply H twice right? but just need to insert matrix of x-component and y component for spin x and spin y

b) No idea what the energy eigenvalue is; Wouldn't it be H ? could they mean : U=exp(-i*H*t/(h-bar))?

c)Do they want me to just write the equation H out explicitly, i.e. with the matrix components of x and y ?

d) No idea what the state vector is; is it [tex]\phi=\varphi_x+[/tex]? is [tex]\varphi_x+= \hbar/2[/tex]?

e) I probably need to square the state vector which would be [tex](\hbar^2)/4[/tex] if my state vector in d is correct.

What do you think of my approach?
 
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noblegas said:
b) No idea what the energy eigenvalue is; Wouldn't it be H ? could they mean : U=exp(-i*H*t/(h-bar))?

No. The energy eigenvalues are given by E, not H:

[tex] \hat{H}\psi=E\psi[/tex]
 
jdwood983 said:
No. The energy eigenvalues are given by E, not H:

[tex] \hat{H}\psi=E\psi[/tex]

So then , it reallly requires no effort to do this part of the problem right? Should I find the value for psi?
 
noblegas said:
So then , it reallly requires no effort to do this part of the problem right? Should I find the value for psi?

No, you need to find [itex]E[/itex].
 
jdwood983 said:
No, you need to find [itex]E[/itex].

If E is equal to H and H is given in the problem, then why wouldn't I have my H.
 
noblegas said:
If E is equal to H and H is given in the problem, then why wouldn't I have my H.

[tex]\hat{H}[/tex] is not equal to [itex]E[/itex]. [tex]\hat{H}[/tex] is an operator while [itex]E[/itex] is a number.
 
jdwood983 said:
[tex]\hat{H}[/tex] is not equal to [itex]E[/itex]. [tex]\hat{H}[/tex] is an operator while [itex]E[/itex] is a number.

[itex]E[/itex] is an eigen value; Is E = T+V = exp(-i*E_n*t/(h-bar)) , which is what my eigenvlaues energies.
 
noblegas said:
[itex]E[/itex] is an eigen value; Is E = T+V = exp(-i*E_n*t/(h-bar)) , which is what my eigenvlaues energies.

[itex]E[/itex] is an eigenvalue, but it is still just a number. That value of [itex]E[/itex] you have written looks more like the unitary operator [itex]U[/itex] for the time-evolution of the Hamiltonian, but it would not be a solution of the eigenvalues for the Hamiltonian in general. You may want to review your textbook for definitions of operators, eigenvalues, and wave functions for further help.
 
jdwood983 said:
[itex]E[/itex] is an eigenvalue, but it is still just a number. That value of [itex]E[/itex] you have written looks more like the unitary operator [itex]U[/itex] for the time-evolution of the Hamiltonian, but it would not be a solution of the eigenvalues for the Hamiltonian in general. You may want to review your textbook for definitions of operators, eigenvalues, and wave functions for further help.

Well they say I need to use H^2 to find E. Assuming that [tex]H\varphi=E\varphi[/tex] . Then [tex]H^2\varphi=H(H\varphi)=H(E\varphi)=E(H\varphi)=E^2(\varphi)[/tex] Am I headed in the right direction?
 
noblegas said:
Well they say I need to use H^2 to find E. Assuming that [tex]H\varphi=E\varphi[/tex] . Then [tex]H^2\varphi=H(H\varphi)=H(E\varphi)=E(H\varphi)=E^2(\varphi)[/tex] Am I headed in the right direction?

No. [itex]E[/itex] is still just a number while [itex]H[/itex] is an operator. Do you know what an operator is?
 
For part (a), I would assume that [itex]H^2=|H|^2=H^*H[/itex], that is you need to multiply the complex conjugate to the original matrix. You know what the spin matrices are, I'm assuming, so you multiply [itex]\alpha[/itex] and [itex]\beta[/itex] to each of those matrices and then add them to form one 2x2 matrix:

[tex] H=\frac{\hbar}{2}\left(\begin{array}{cc}0&\alpha\\ \alpha&0\end{array}\right)+\frac{\hbar}{2}\left(\begin{array}{cc}0&-i\beta\\ i\beta&0\end{array}\right)[/tex]


[tex] H^*=\frac{\hbar}{2}\left(\begin{array}{cc}0&\alpha\\ \alpha&0\end{array}\right)+\frac{\hbar}{2}\left(\begin{array}{cc}0&i\beta\\ -i\beta&0\end{array}\right)[/tex]

Add the matrices of each and then multiply the two matrices and you will have [itex]H^2[/itex]. You can then use the characteristic equation, [itex]\det(H-\lambda\mathbb{I})=0[/itex] to find the eigenvalues, [itex]\lambda[/itex], in terms of [itex]\hbar,\,\alpha[/itex] and [itex]\beta[/itex]. (Though I'm not sure why your professor suggested using the solution to [itex]H^2[/itex] to find the eigenvalues and not the original matrix, [itex]H[/itex].)