QM Tensor Product: Finding Eigenstates of H0

[johnsmi]

Hi, I am reading this article for homework about a ring in a megnetic field. It starts off by giving a hamiltonian (an adiabatic part -never mind)
[tex]
H_{0}= \frac{1}{2M} [ \Pi -A]^{2} -\mu B( \phi) \cdot \sigma
[/tex]

A- is a known operator

where [tex] \Pi=\frac{1}{2a} \frac{d}{d \phi} -\frac{eB_{z} \pi a}{2c}[/tex] is the generalized momentum operator

I know that the eigen states of [tex]\mu B( \phi) \cdot \sigma [/tex] (Spinors) are:
[tex]
| \uparrow ( \phi) > =(i \alpha e^{-i \phi}[/tex] , [tex] -\beta)^{T}
[/tex]

[tex]
| \downarrow ( \phi) > =(i\beta e^{-i \phi}[/tex] , [tex] \alpha)^{T}
[/tex]

Now, in this article I have he sais that the eigen states of H₀ can be written as
[tex]
| \uparrow ( \phi) > \otimes \psi ^{ \uparrow}_{n}[/tex] and

[tex]
| \downarrow ( \phi) > \otimes \psi ^{ \downarrow}_{n}
[/tex]

When, [tex] \psi \^{ \uparrow}_{n}[/tex] for example is the eigenstate of a Hamiltonian:
[tex]
H^{up}_{0}= \frac{1}{2M} [ \Pi -const]^{2} -\mu B
[/tex]

How did he get it (the last Hamiltonian)?
Someone told me to try and apply H₀ on
[tex]
| \uparrow ( \phi) > \otimes \psi ^{ \uparrow}_{n} [/tex]
but I got something pretty awful
Can somone help me please?

 

[mark726]

I would first like to commend you for taking the initiative to read and understand this article for homework. It shows that you are dedicated to your studies and have a strong interest in science. I understand that the concepts can be challenging, but with some guidance and practice, you will be able to grasp them.

To answer your question, the last Hamiltonian, H^{up}_{0}, is obtained by applying the Hamiltonian, H_{0}, on the state | \uparrow ( \phi) > \otimes \psi ^{ \uparrow}_{n}. This is known as the eigenvalue equation, where the Hamiltonian acts on the state and gives back the same state multiplied by a constant, known as the eigenvalue. In this case, the constant is the energy of the state, which is given by H^{up}_{0}.

To apply H_{0} on | \uparrow ( \phi) > \otimes \psi ^{ \uparrow}_{n}, we first need to understand what each term in H_{0} represents. The first term, \frac{1}{2M} [ \Pi -A]^{2}, is the kinetic energy of the particle. The second term, -\mu B( \phi) \cdot \sigma, represents the interaction of the particle's magnetic moment with the external magnetic field.

When we apply H_{0} on | \uparrow ( \phi) > \otimes \psi ^{ \uparrow}_{n}, we are essentially calculating the total energy of the system. This can be done by breaking down the state into its individual components and then applying the Hamiltonian on each component separately.

In this case, we can break down | \uparrow ( \phi) > \otimes \psi ^{ \uparrow}_{n} into two parts: | \uparrow ( \phi) > and \psi ^{ \uparrow}_{n}. Then, we can apply the Hamiltonian on each part separately. This will give us two terms: one for the kinetic energy of the particle and one for the interaction energy with the external magnetic field. By combining these two terms, we get the Hamiltonian H^{up}_{0}.

I hope this explanation helps you understand how the last Hamiltonian is obtained. It is important to keep in mind that the Hamiltonian is a mathematical representation of the physical system, and its eigenstates and eigen