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

In summary: 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,...maybe they meant the conjugate?)
  • #1
noblegas
268
0

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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  • #2
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]
 
  • #3
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?
 
  • #4
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].
 
  • #5
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.
 
  • #6
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.
 
  • #7
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.
 
  • #8
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.
 
  • #9
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?
 
  • #10
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?
 
  • #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].)
 

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

What is spin one-half of a Hamiltonian?

Spin one-half of a Hamiltonian is a mathematical operator that describes the spin properties of a quantum system with a spin value of 1/2. It is commonly used in quantum mechanics to study the behavior of particles with half-integer spin, such as electrons, protons, and neutrons.

What is the significance of spin one-half of a Hamiltonian in physics?

Spin one-half of a Hamiltonian is significant in physics because it helps us understand the intrinsic angular momentum of particles and their interactions with electromagnetic fields. It also plays a crucial role in the development of technologies such as magnetic resonance imaging (MRI) and spintronics.

How is spin one-half of a Hamiltonian represented mathematically?

Spin one-half of a Hamiltonian is represented as a 2x2 matrix in quantum mechanics. It consists of four elements representing the spin states of a particle - spin up and spin down, along with their corresponding probabilities. It is also represented using the Pauli spin matrices, which are a set of three matrices that describe the spin operators in three different directions.

What are the applications of spin one-half of a Hamiltonian in quantum computing?

Spin one-half of a Hamiltonian has several applications in quantum computing, such as in the development of qubits (quantum bits) and quantum gates. It is also used in quantum algorithms for tasks such as quantum simulation and quantum error correction. Additionally, spin one-half of a Hamiltonian is used to study the behavior of spin-based quantum systems, which can lead to advancements in quantum computing technology.

How does spin one-half of a Hamiltonian relate to the Schrödinger equation?

Spin one-half of a Hamiltonian is an essential component of the Schrödinger equation, which is a fundamental equation in quantum mechanics. The Schrödinger equation describes how the state of a quantum system changes with time, and the spin one-half of a Hamiltonian contributes to the total energy of the system, along with other terms related to the potential and kinetic energy of the particles.

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