Explanation of spin degrees of freedom with respect to the x axis

In summary, the question is asking for the simultaneous eigenstates of energy and spin along the x-axis, which is not possible. You need to find the energies and corresponding eigenstates for the Hamiltonian that they give you.
  • #1
Somali_Physicist
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Hey Guys/Gals i understand the general premise of this question and can calculate the solution but i am a bit confused.

I am supposed to represent a generic state as a linear combination of the |-,x> , |+,x> basis vectors. However i don't know why, is the question actually asking for the simultaneous eigenstates of energy and the spin observable along the x axis. If that is what it is asking that makes sense as we would be dealing with a state which must be a linear combination of |+,x> , |+,x> then make sure its an eigen state.
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-cheers sarinle
 
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  • #2
Is not asking for the simultaneous eigenstates of spin in the x-direction and the energy (the two operators doesn't commute so it's impossible to find them). You simply have to find the energies and the corresponding eigenstates for the Hamiltonian that they give you.
 
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  • #3
Gaussian97 said:
Is not asking for the simultaneous eigenstates of spin in the x-direction and the energy (the two operators doesn't commute so it's impossible to find them). You simply have to find the energies and the corresponding eigenstates for the Hamiltonian that they give you.

But in the solution we represent the generic energy eigen vector as a linear combination of |+,x> and |-,x> spin basis states.

|ψ> = a|+,x> + b|-,x>
|ψ>(HMatrix) = c|ψ> : a,b, c are constants

The first statement implies its made up of |+,x> and |-,x> basis states which implies that its an eigen function ... right?
 
  • #4
Somali_Physicist said:
But in the solution we represent the generic energy eigen vector as a linear combination of |+,x> and |-,x> spin basis states.

|ψ> = a|+,x> + b|-,x>
|ψ>(HMatrix) = c|ψ> : a,b, c are constants

The first statement implies its made up of |+,x> and |-,x> basis states which implies that its an eigen function ... right?
No, a linear combination of two eigenstates is not necessarily an eigenstate. Since ##\left|+,x\right>## and ##\left|-,x\right>## form a base of the Hilbert space, ANY state can be written as a linear combination of those two (by definition of base).
 
  • #5
Gaussian97 said:
No, a linear combination of two eigenstates is not necessarily an eigenstate. Since ##\left|+,x\right>## and ##\left|-,x\right>## form a base of the Hilbert space, ANY state can be written as a linear combination of those two (by definition of base).
I don't wish to be pedantic about this but isn't the hilbert space an infinite dimensional vector space? since the number of basis represent the dimension how can 2 = infinity?
 
  • #6
No, in the case of spin-1/2 particles the Hilbert space containing the spin states is only 2-dimensional. That's why ##H## is a 2x2 matrix and why you only need two orthonormal states to form a base.
 
  • #7
Gaussian97 said:
No, in the case of spin-1/2 particles the Hilbert space containing the spin states is only 2-dimensional. That's why ##H## is a 2x2 matrix and why you only need two orthonormal states to form a base.
So the hilbert space in the case of spin-1/2 particles are 2 dimensional. And i am assuming a new hilbert space can be made for every observable. For instance if we have some X property which has 4 base states we can represent particles with X property construct an hilbert space of 4 dimensions ?

Thanks a lot for the help, i have an exam tomorrow.
 
  • #8
Well, it's not that you can make a Hilbert space for every new observable, one of the postulates of QP tells you that for any observable there exists an associated hermitian operator that acts on a Hilbert space. But of course, there are several (in fact, infinite) observables that act on the same Hilbert space.
And yes, if you can find a base of only 4 states then your observable is acting on a dimension 4 Hilbert space.
 
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  • #9
Gaussian97 said:
Well, it's not that you can make a Hilbert space for every new observable, one of the postulates of QP tells you that for any observable there exists an associated hermitian operator that acts on a Hilbert space. But of course, there are several (in fact, infinite) observables that act on the same Hilbert space.
And yes, if you can find a base of only 4 states then your observable is acting on a dimension 4 Hilbert space.
Wow!

so this let's us do some neat trick thanks mate.
 
  • #10
Just to be sure. The exercise disregard any spatial degree of freedom and focus only on the Spin. For a general case, the total Hilbert space will be the tensor product of spatial degrees of freedom and the Spin.

What the x mean in the exercise is the following:
There is a spin operator for each axis $$
s_{x},s_{y},s_{z}
$$
But, they don't commute with each other. So, you can't find a common basis of eigenvectors. The operator $$
S^{2}=s_{x}^{2}+s_{y}^{2}+s_{z}^{2}$$
do commute with them. Therefore, you can find a basis of vector that are eigenvectors for , let say
$$S^{2},s_{z}$$
let's write this base as $$
\left\{ \left|+,z\right\rangle ,\left|-,z\right\rangle \right\}

$$
where the z means they are eigenvectors of the sz operator. This is normally what is done, and usually the z is missing because it is implcitly understood. But, there is nothing sacred about the z axis, the same can be done with the x axis. There exists a basis
$$ \left\{ \left|+,x\right\rangle ,\left|-,x\right\rangle \right\} $$
of common eigenvectors of
$$ S^{2},s_{x} $$

The two basis are different, but they are related by a linear transformation.
 

1. What is meant by "spin degrees of freedom"?

Spin degrees of freedom refer to the intrinsic angular momentum of a particle. In quantum mechanics, particles are described not only by their position and momentum, but also by their spin, which can take on discrete values. This spin is a fundamental property of particles and plays a crucial role in many physical phenomena.

2. How does spin relate to the x axis?

The x axis is one of the three spatial axes in a Cartesian coordinate system. In quantum mechanics, particles can have spin along any of these axes, including the x axis. This means that the spin of a particle can be aligned or anti-aligned with the x axis, and this orientation can affect its behavior in certain situations.

3. Why is the x axis specifically used to explain spin degrees of freedom?

The choice of the x axis to explain spin degrees of freedom is arbitrary and could be substituted with any other axis. However, the x axis is often used as a reference because it is a commonly used coordinate system and allows for a clear and consistent explanation of spin in relation to other physical quantities.

4. How is spin measured along the x axis?

Spin is quantized, meaning it can only take on certain discrete values. Along the x axis, spin is typically measured in units of ħ/2, where ħ is the reduced Planck's constant. This value represents the minimum amount of angular momentum that a particle can have along the x axis.

5. What are some examples of how spin degrees of freedom along the x axis are relevant in physics?

Spin along the x axis is relevant in a variety of physical phenomena, including the behavior of atoms in a magnetic field, the polarization of light, and the properties of certain materials such as semiconductors. It also plays a crucial role in quantum computing and quantum information processing.

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