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Spin one-half of a hamiltonian

  1. Nov 15, 2009 #1
    1. The problem statement, all variables and given/known data

    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]

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


    3. 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?
     
  2. jcsd
  3. Nov 15, 2009 #2
    No. The energy eigenvalues are given by E, not H:

    [tex]
    \hat{H}\psi=E\psi
    [/tex]
     
  4. Nov 15, 2009 #3
    So then , it reallly requires no effort to do this part of the problem right? Should I find the value for psi?
     
  5. Nov 15, 2009 #4
    No, you need to find [itex]E[/itex].
     
  6. Nov 15, 2009 #5
    If E is equal to H and H is given in the problem, then why wouldn't I have my H.
     
  7. Nov 15, 2009 #6
    [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.
     
  8. Nov 15, 2009 #7
    [itex]E[/itex] is an eigen value; Is E = T+V = exp(-i*E_n*t/(h-bar)) , which is what my eigenvlaues energies.
     
  9. Nov 15, 2009 #8
    [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.
     
  10. Nov 15, 2009 #9
    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?
     
  11. Nov 16, 2009 #10
    No. [itex]E[/itex] is still just a number while [itex]H[/itex] is an operator. Do you know what an operator is?
     
  12. Nov 16, 2009 #11
    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].)
     
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