Spin one-half of a hamiltonian

[noblegas]

Homework Statement

A single spin-one-half system has Hamiltonian

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

a) Using the representation of the spin components as Pauli spin matrices, find an expression for $$H^2$$ 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 $$H=\alpha*s_x+\beta*s_y$$ in the Pauli spin matrix representation.

d) Supposed that a t time t=0 the system is an eigenstate of $$s_z$$, with eigen value $$+\h-bar/2$$. 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 $$+\hbar/2$$

Homework Equations

$$s=(\hbar)*(\sigma)/2$$
$$(\sigma_x)$$,$$(\sigma_y)$$, and $$(\sigma_z)$$

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 $$\phi=\varphi_x+$$? is $$\varphi_x+= \hbar/2$$?

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

What do you think of my approach?

 

[jdwood983]

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:

$$\hat{H}\psi=E\psi$$

 

[noblegas]

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

$$\hat{H}\psi=E\psi$$

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

 

[jdwood983]

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 $E$.

 

[noblegas]

No, you need to find $E$.

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

 

[jdwood983]

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

$$\hat{H}$$ is not equal to $E$. $$\hat{H}$$ is an operator while $E$ is a number.

 

[noblegas]

$$\hat{H}$$ is not equal to $E$. $$\hat{H}$$ is an operator while $E$ is a number.

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

 

[jdwood983]

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

$E$ is an eigenvalue, but it is still just a number. That value of $E$ you have written looks more like the unitary operator $U$ 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.

 

[noblegas]

$E$ is an eigenvalue, but it is still just a number. That value of $E$ you have written looks more like the unitary operator $U$ 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 $$H\varphi=E\varphi$$ . Then $$H^2\varphi=H(H\varphi)=H(E\varphi)=E(H\varphi)=E^2(\varphi)$$ Am I headed in the right direction?

 

[jdwood983]

Well they say I need to use H^2 to find E. Assuming that $$H\varphi=E\varphi$$ . Then $$H^2\varphi=H(H\varphi)=H(E\varphi)=E(H\varphi)=E^2(\varphi)$$ Am I headed in the right direction?

No. $E$ is still just a number while $H$ is an operator. Do you know what an operator is?

 

[jdwood983]

For part (a), I would assume that $H^2=|H|^2=H^*H$, 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 $\alpha$ and $\beta$ to each of those matrices and then add them to form one 2x2 matrix:

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


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

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