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I am confused as to how putting a different particle with a different spin change the equation of the particle in a 3D potential well (all of same side length).

For a particle in a box, the wave equation will be $$\psi(x,y,z)_{n_x,n_y,n_z}= \left(\frac{2}{L}\right)^{3/2} \sin\left(\frac{n_x\pi x}{L}\right)\sin\left(\frac{n_y\pi y}{L}\right)\sin\left(\frac{n_z\pi z}{L}\right)$$

and the Energy eigenvalues are $$E_{n_x,n_y,n_z}=(n_x^2+n_y^2+n_z^2)\frac{{\pi}^2\hbar^2}{2mL^2}$$

Could anyone please explain how will the system be different if there is a particle of different spin?

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  • $\begingroup$ Anyone please help me $\endgroup$
    – superflash
    Feb 15 at 6:47
  • $\begingroup$ Spin of particle doesn't matter. Since there's no spin dependent term in Hamiltonian $\endgroup$ Feb 15 at 6:58
  • $\begingroup$ so there will be no difference in anything for the system if spin is changed $\endgroup$
    – superflash
    Feb 15 at 7:55
  • $\begingroup$ Yes, no change. Although if there are more than one particles, then things will be different. $\endgroup$ Feb 16 at 6:05
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A couple of things: a) In case you're new to the study of quantum mechanics, the spin of particles as far as the Schroedinger equation is considered has got no business existing. b) Now, for this discussion, the space of wavefunctions along with the tensor product of the space of the spin (If $j$ is the spin of the particle, then this is a Hilbert space of dimensionality $2j+1$) is what represents the particle in totality.

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A fermion is a particle that follows Fermi–Dirac statistics and generally has half odd integer spin $1/2, 3/2$, etc. On the other hand, a boson is a particle that follows Bose-Einstein statistics and generally has integer spin $0,1,2$, etc.

In general, the wave function is equal to the product of the partial part and the spin part $$\Psi(\boldsymbol{r},\boldsymbol{S})=\psi(\boldsymbol{r})\chi(\boldsymbol{S})$$

The consequence of the fermions is that the wave function of the particle must be anti-symmetric while for bosons the wave function of the particle must be symmetric.

Thus for fermions : $$ \Psi(\boldsymbol{r},\boldsymbol{S}) = \begin{cases} \psi_a(\boldsymbol{r})\cdot \chi_s(\boldsymbol{S}) & \\ \psi_s(\boldsymbol{r})\cdot \chi_a(\boldsymbol{S}) & \end{cases} $$

for bosons : $$ \Psi(\boldsymbol{r},\boldsymbol{S}) = \begin{cases} \psi_a(\boldsymbol{r})\cdot \chi_a(\boldsymbol{S}) & \\ \psi_s(\boldsymbol{r})\cdot \chi_s(\boldsymbol{S}) & \end{cases} $$


Edit: I'm adding a simple example for two non-interacting particles.

The symmetric and antisymmetric parts are used as follows: $$\psi_a=\psi_{n_1}(\mathbf{r}_1)\psi_{n_2}(\mathbf{r}_2)-\psi_{n_1}(\mathbf{r}_2)\psi_{n_2}(\mathbf{r}_1)$$ $$\psi_s=\psi_{n_1}(\mathbf{r}_1)\psi_{n_2}(\mathbf{r}_2)+\psi_{n_1}(\mathbf{r}_2)\psi_{n_2}(\mathbf{r}_1)$$

First If the particles are bosons then the
$$\Psi= \begin{cases} \psi_a(\boldsymbol{r})\cdot \chi_a(\boldsymbol{S}) & \\ \psi_s(\boldsymbol{r})\cdot \chi_s(\boldsymbol{S}) & \end{cases}$$ for the fermions $$\Psi= \begin{cases} \psi_a(\boldsymbol{r})\cdot \chi_s(\boldsymbol{S}) & \\ \psi_s(\boldsymbol{r})\cdot \chi_a(\boldsymbol{S}) & \end{cases}$$ If the fermion is happen to be spin-$1/2$ particle like electron you can use $\chi_a$ to be triplet and $\chi_s$ to be singlet.


Reference

  1. Identical Particles : Quantum Mechanics Concepts and Applications :-Nouredine Zettili
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  • $\begingroup$ I am new to quantum mechanics actually so it is a little difficult to understand this $\endgroup$
    – superflash
    Feb 15 at 9:12
  • $\begingroup$ It's hard to write the details but I added an example you can go through the reference books. $\endgroup$ Feb 15 at 9:28
  • $\begingroup$ So in the system we would not be able to distinguish between different spin states a particle with some spin is having and rest everything would be same? $\endgroup$
    – superflash
    Feb 18 at 5:10
  • $\begingroup$ Here we are concern with how many particles can be filled at a certain level. That's what I have done. Whether we can distinguish one state from the other is not a concern here and belong to other topics. The fact is to note the symmetric condition involved here. $\endgroup$ Feb 18 at 13:31
  • $\begingroup$ If there is only one particle then? $\endgroup$
    – superflash
    Feb 19 at 4:53

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