Diagonalizing of Hamiltonian of electron and positron system

In summary, the operators ##\hat{\vec{S}}_1\cdot\hat{\vec{S}}_2##, ##\hat{S}_{1_z}##, and ##\hat{S}_{2_z}## do not commute, so the basis states ##\ket{s_1,s_2;m_1,m_2}## and ##\ket{s,m}## are not eigenfunctions of the Hamiltonian. To find the eigenvectors, one can use a convenient set of basis states, such as the four direct-product states ##\ket{\uparrow \uparrow}##, ##\ket{\downarrow \uparrow}##, ##\ket{\uparrow \downarrow
  • #1
Davidllerenav
424
14
Homework Statement
The spin-dependant Hamiltonian of an electron-positron system in the presence of a uniform magnetic field in the z-direction ##(\vec{B}=B\vec{k})## can be written as

##\hat{H}=\lambda \hat{\vec{S}}_1\cdot\hat{\vec{S}}_2+\left(\frac{eB}{mc}\right)(\hat{S}_{1_z}-\hat{S}_{2_z})##

##\lambda## is a constant and ##\hat{\vec{S}}_1## and ##\hat{\vec{S}}_2## are the Spin operators of the electro and positron respectively. Find the energy eigenvalues and eigenvectors by diagonalizing the Hamiltonian
Relevant Equations
##\hat{\vec{S}}=\hat{\vec{S}}_1+\hat{\vec{S}}_2##
##s_1=1/2##
##s_2=1/2##
##m_1=\pm 1/2##
##m_2=\pm 1/2##
##m=m_1+m_2##
##s=0,1##
What I did was first noting that ##\hat{\vec{S}}_1\cdot\hat{\vec{S}}_2=\frac{1}{2}(\hat{\vec{S}}^2-\hat{\vec{S}}_1^2-\hat{\vec{S}}_2^2)##, but these operators don't commute with ##\hat{S}_{1_z}## and ##\hat{S}_{2_z}##, this non the decoupled basis ##\ket{s_1,s_2;m_1,m_2}## nor the coupled one ##\ket{s,m}## are eigenfunctions of this Hamiltonian. So I need to find the eigenvectors.

I tried to find the matrix of the Hamiltonian, but I'm confused about how should I write the components of the matrix, since I've only done it when I have a given ##s##, but here I have ##s=0,1##. All I know is that the matrix must be a ##4\times 4## one.
 
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  • #2
Davidllerenav said:
I tried to find the matrix of the Hamiltonian, but I'm confused about how should I write the components of the matrix, since I've only done it when I have a given ##s##, but here I have ##s=0,1##. All I know is that the matrix must be a ##4\times 4## one.
You can choose any convenient set of basis states for the system. For example, you could choose the four direct-product states

##\ket{1} = \ket{\uparrow}_e \ket{\uparrow}_p \equiv \ket{\uparrow \uparrow}##
##\ket{2} = \ket{\downarrow}_e \ket{\uparrow}_p \equiv \ket{\downarrow \uparrow}##
##\ket{3} = \ket{\uparrow}_e \ket{\downarrow}_p \equiv \ket{\uparrow \downarrow}##
##\ket{4} = \ket{\downarrow}_e \ket{\downarrow}_p \equiv \ket{\downarrow \downarrow}##

The arrows denote spin up or down along the z-axis.

Can you find the matrix elements of ##H## in this basis, such as ##H_{32} = \bra{3} H \ket{2}##?

Note, for example, that $$\hat S_{1_y} \hat S_{2_y} \ket{2} = \hat S_{e_y} \hat S_{p_y}\ket{2} = \hat S_{e_y}\hat S_{p_y}\ket{\downarrow \uparrow} \equiv \left( \hat S_{e_y} \ket{\downarrow}_e \right) \left( \hat S_{p_y} \ket{\uparrow}_p \right) = \left(- i \frac{\hbar}{2}\ket{\uparrow}_e \right) \left(i \frac{\hbar}{2} \ket{\downarrow}_p \right) = \frac{\hbar^2}{4}\ket{\uparrow \downarrow} = \frac{\hbar^2}{4}\ket{3} $$
 
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FAQ: Diagonalizing of Hamiltonian of electron and positron system

What is the Hamiltonian of an electron and positron system?

The Hamiltonian of an electron and positron system is a mathematical operator that describes the total energy of the system. It takes into account the kinetic energy and potential energy of both particles.

Why is it important to diagonalize the Hamiltonian of an electron and positron system?

Diagonalizing the Hamiltonian allows us to find the eigenvalues and eigenvectors of the system, which represent the allowed energy states and corresponding wavefunctions. This is important because it allows us to make predictions about the behavior of the system and understand its quantum mechanical properties.

How is the Hamiltonian diagonalized for an electron and positron system?

The Hamiltonian can be diagonalized using techniques from linear algebra, such as finding the eigenvalues and eigenvectors of the matrix representation of the operator. In the case of an electron and positron system, the Hamiltonian can be written in terms of the creation and annihilation operators for the particles.

What is the significance of the eigenvalues and eigenvectors in diagonalizing the Hamiltonian of an electron and positron system?

The eigenvalues represent the allowed energy states of the system, while the corresponding eigenvectors represent the wavefunctions associated with those energy states. By diagonalizing the Hamiltonian, we can determine the energy levels and wavefunctions of the system.

How does diagonalizing the Hamiltonian of an electron and positron system relate to the Schrödinger equation?

The Schrödinger equation describes the time evolution of a quantum system, including the behavior of the wavefunction. By diagonalizing the Hamiltonian, we can determine the energy levels and wavefunctions of the system, which are necessary for solving the Schrödinger equation and predicting the behavior of the system over time.

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