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In Griffth's QM text, he says in Section 5.1,

Suppose we have two noninteracting particles, both of mass $m$, in the infinite square well (Section 2.2). The one particle states are, $$ \psi_n(x)=\sqrt{\frac{2}{a}}\sin\left(\frac{n\pi}{a}x\right),\quad E_n=n^2K $$ (where $K=\pi^2\hbar^2/2ma$). If the particles are distinguishable, the composite wave functions are simple products: $$ \psi_{n_1n_2}\left(x_1,\,x_2\right)=\psi_{n_1}\left(x_1\right)\psi_{n_2}\left(x_2\right),\quad E_n=\left(n_1^2+n_2^2\right)K. $$ ...if the particles are identical fermions, there is no state with energy $2K$; the ground state is $$ \frac{\sqrt{2}}{a}\left[\sin\left(\pi x_1/a\right)\sin\left(2\pi x_2/a\right)-\sin\left(2\pi x_1/a\right)\sin\left(\pi x_2/a\right)\right], $$ and its energy is $5K$.

Griffith's says that there is no state of energy $2K$ for two fermions, but I wonder why author doesn't consider about spin. If two fermion have different spin (singlet state), they don't violate Pauli's exclusion principle and can have energy $2K$. Why am I wrong?

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  • 3
    $\begingroup$ If he had meant for the particles to have spin, don't you think he would have said so? $\endgroup$ – J. Murray Mar 25 '18 at 14:07
  • $\begingroup$ Do you mean I can ignore spin in this example? $\endgroup$ – Einstein Mar 25 '18 at 14:12
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In non-relativistic quantum mechanics, particles do not naturally have spin$^\dagger$ - you have to add that in by hand. In this example, Griffiths does not do this, so you should naturally take the particles to be spin-less.

In fact, he actually gives the single particle states to be $$\psi_n = \sqrt{\frac{2}{a}}\sin\left(\frac{n \pi x}{a}\right)$$

If the particles had spin 1/2, for example, the single particle states would be

$$\psi_{n\alpha\beta}= \sqrt{\frac{2}{a}}\sin\left(\frac{n\pi x}{a}\right) \otimes \pmatrix{\alpha \\ \beta}$$

where $|\alpha|^2 + |\beta|^2 = 1$.


As DanielC points out in his comment, this is not necessarily true. The "typical procedure" for working in nonrelativistic QM involves starting from a chosen (separable) Hilbert space and some relevant Hamiltonian. If we take this view, then we could (for instance) choose our Hilbert space to be whatever we wanted - and given two Hilbert spaces $\mathcal H_1$ and $\mathcal H_2$, we could form a new Hilbert space given by $\mathcal H_1 \otimes \mathcal H_2$.

If we want to describe a particle which exists in $3D$ space, a typical choice for the Hilbert space (in accordance with Born's statistical interpretation) is $L^2(\mathbb R^3)$. If the particle has some kind of intrinsic structure, then we could reflect that by appending some "ancillary" space to yield a total Hilbert space $L^2(\mathbb R^3) \otimes \mathcal H_{anc}$. However, there is no immediately apparent reason why we have to do this, nor is there an immediately obvious choice of $\mathcal H_{anc}$ for any given system.

However, there is alternative approach. If we declare the Galilei group to be the kinematical symmetry group of the system, a natural move would be to classify its irreducible representations. In doing so, we find that a generic irreducible representation of the Galilei group lives in the space $L^2(\mathbb R^3) \otimes \mathcal C^{2s+1}$ for some integer or half-integer $s$ - from this point of view, spin arises just as naturally as it does in QFT (where particles are seen as the irreducible representations of the Poincare group).

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  • $\begingroup$ In non-relativistic quantum mechanics, particles naturally have spin. Stop spreading this incorrect knowledge. $\endgroup$ – DanielC Mar 25 '18 at 16:27
  • $\begingroup$ @DanielC Perhaps you could educate me on this issue, then. $\endgroup$ – J. Murray Mar 25 '18 at 16:32
  • $\begingroup$ I couldn't, but Jean-Marc Levy-Leblond can: aip.scitation.org/doi/10.1063/1.1724319 $\endgroup$ – DanielC Mar 25 '18 at 16:39
  • $\begingroup$ @DanielC Particles naturally have spin if, by particle, you mean irreducible representation of the Galilei group. This is not the typical formulation of non-relativistic QM (certainly, it is not the path that Griffiths takes), in which one starts from a Hilbert space and a choice of Hamiltonian operator. $\endgroup$ – J. Murray Mar 25 '18 at 17:10
  • $\begingroup$ Ultimately it's a choice of what you consider to be fundamental. If you take the latter approach and start from a Hamiltonian which acts on $L^2(\mathbb R)$, then you are implicitly dealing with a spinless particle. OTOH if you start from a Hamiltonian which acts on $L^2(\mathbb R) \otimes \mathbb C^2$, then you are implicitly dealing with a particle with spin 1/2. There is no a priori reason to choose one over the other from this point of view, and it is in this sense that I mean that spin does not arise naturally. $\endgroup$ – J. Murray Mar 25 '18 at 17:36

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