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In studying systems of 3 or more identical, non-interacting particles, fermions or bosons, I have read that to create eigenfunctions of the Hamiltonian that are antisymmetric or symmetric one creates the symmetric and antisymmetric sums, respectively. Let assume the Hamiltonian is separable $H=H_1+H_+...+H_i$ and $\psi_{\alpha}(x_1,x_2,...,x_n)$ is the total Hamiltonian eigenfunction which can be written as $\psi_{\alpha}(x_1,x_2,...,x_n)=\phi_{\alpha_1}(x_1),\phi_{\alpha_2}(x_2),...,\phi_{\alpha_n}(x_n)$ where $\phi_{\alpha_i}(x_i)$ is the eigenfunction fo Hamiltonian $H_i$. The symmetric and antisymmetric sums are:

$$\psi_{\alpha}^{(s)}(x_1,x_2,...,x_n)=\sum_{i} \psi_{\alpha}(P_i(x_1,x_2,...,x_n))$$

$$\psi_{\alpha}^{(a)}(x_1,x_2,...,x_n)=\sum_{i} (-1)^{|P_i|}\psi_{\alpha}(P_i(x_1,x_2,...,x_n))$$

where $|P_i|$ assumes values $+1$ or $-1$ depending on whether the permutation $P_i$ of the coordinates is of odd or even order, respectively and $\alpha$ indicates the set of quantum numbers.

I thought this same method could be used to construct even and odd eigenfunctions for systems of only two particles as well.

For example we assume a system with total Hamiltonian $H=H_1+H_2$ where $H_1$ and $H_2$ are two non-interacting hydrogen Hamiltonians and the two particles are fermions of spin $1/2$.

We consider the physical system in its fundamental state, that is, with both particles on the ground state so that their quantum numbers are $\alpha_1=[n=1,l=0,l_z=0,s_z=1/2]$ and $\alpha_2=[n=1,l=0,l_z=0,s_z=-1/2]$, if I want to use the method of antisymmetric sum I write the quantum numbers $\alpha_1$ and $\alpha_2$ as $(n,s_z)_i$ so $(1,1/2)_1$ and $(1,-1/2)_2$ and I write $$\psi_{\alpha}(\vec{r_1},\vec{r_2})=\phi_{1,1/2}(\vec{r_1})\phi_{1,-1/2}(\vec{r_2})-\phi_{1,-1/2}(\vec{r_1})\phi_{1,1/2}(\vec{r_2})$$ which can be rewritten in the classical form $\psi_{\alpha}(\vec{r_1},\vec{r_2})=\phi_{1}(\vec{r_1})\phi_{1}(\vec{r_2})|1/2,1/2;0,0\rangle$ with $|S_1,S_2;S,S_z\rangle$. No problems till here, the problem I have is when considering this system in the first excited state when the system is in this particular arrangement: $\alpha_1=[1,-1/2]$ and $\alpha_2=[2,1/2]$ so that one particle is in the fundamental level with spin down and the other particle is in the first excited state with spin up. Again, I want to use the method of anisymmetric sum, and I obtain something wrong, in fact:

$$\psi_{\alpha}(\vec{r_1},\vec{r_2})=\phi_{1,-1/2}(\vec{r_1})\phi_{2,1/2}(\vec{r_2})-\phi_{2,1/2}(\vec{r_1})\phi_{1,-1/2}(\vec{r_2})$$

Now, from what I understand, the logic behind these mathematical formalisms is that since the particles are identical, the total state must be a linear combination of two states, one with the $1$ and $2$ indices and the other with the swapped indices, which is what I wrote via permutations, but I get the wrong result, in fact I can rewrite via spinors my wavefunction like this:

$$\psi_{\alpha}(\vec{r_1},\vec{r_2})=\phi_{1}(\vec{r_1})\begin{pmatrix} 0 \\ 1 \end{pmatrix} \phi_{2}(\vec{r_2})\begin{pmatrix} 1 \\ 0 \end{pmatrix} -\phi_{2}(\vec{r_1})\begin{pmatrix} 1 \\ 0 \end{pmatrix} \phi_{1,}(\vec{r_2})\begin{pmatrix} 0 \\ 1 \end{pmatrix} $$

instead, in this case, I should get 4 of these "blocks", namely: $$\psi_{\alpha}(\vec{r_1},\vec{r_2})=\phi_{1}(\vec{r_1})\begin{pmatrix} 0 \\ 1 \end{pmatrix} \phi_{2}(\vec{r_2})\begin{pmatrix} 1 \\ 0 \end{pmatrix} -\phi_{2}(\vec{r_1})\begin{pmatrix} 0 \\ 1 \end{pmatrix} \phi_{1,}(\vec{r_2})\begin{pmatrix} 1 \\ 0 \end{pmatrix} +\phi_{1}(\vec{r_1})\begin{pmatrix} 1 \\ 0 \end{pmatrix} \phi_{2}(\vec{r_2})\begin{pmatrix} 0 \\ 1 \end{pmatrix} -\phi_{2}(\vec{r_1})\begin{pmatrix} 1 \\ 0 \end{pmatrix} \phi_{1,}(\vec{r_2})\begin{pmatrix} 0 \\ 1 \end{pmatrix}$$

Where do I go wrong?

Important, I know how to build the states correctly, my problem is that I can't build them correctly with this other method that I've seen is only used with 3 particles or more.

I would like to show what I am doing wrong in the construction of the states with this second method, if it can be used in the case of only two particles, and if it is right the reasoning that I have set out in the sentence before the last equation (Now, from what I understand, the logic behind these mathematical formalisms...)

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1 Answer 1

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This is too long for a comment...

Your first statements are not always correct. For three or more particles, one must proceed with extreme care because there are presentations of $S_3$ or $S_n$ that are not 1-dimensional, and neither symmetric nor antisymmetric: a state can be partially symmetric and not return to a multiple of itself after every permutation.

To construct antisymmetric states, one must combine the multidimensional spatial and spin states of representations that transform by conjugate representations of $S_n$. For $S_2$ is collapses to symmetric and antisymmetric representations, respectively but not for $n>3$ or greater. See for instance

Harvey M. The symmetric group and its relevance to fermion physics. CM-P00067619; 1981.

In particular, it is clearly not possible to create fully antisymmetric states of three spin-1/2 particles, so in such a case multidimensional irreps of $S_3$ (or more generally of $S_n$) will have to be combined. In particular the spin and space parts of the states are no longer separable.

This is a very inefficient way, and one usually prefers Slater determinants of space and spin states, which are automatically antisymmetric, as a construction method for fermionic states.


Edit:

In the case of two particles you have in mind, it is possible to separate the spatial and spin degrees of freedom. Thus: \begin{align} \Psi_\pm (x_1,x_2)=\left(\phi_1(x_1)\phi_2(x_2)\pm \phi_2(x_1)\phi_1(x_2)\right)\left(\vert +\rangle_1\vert -\rangle_2 \mp \vert -\rangle_1\vert +\rangle_2\right) \end{align}
are examples of a fully antisymmetric states with total energy $E_1+E_2$

One can also write (for instance) \begin{align} &\hbox{Det}\left\vert \begin{array}{cc} \phi_1(x_1)\vert +\rangle_1 &\phi_2(x_1)\vert -\rangle_1 \\ \phi_1(x_2)\vert +\rangle_2 &\phi_2(x_2)\vert -\rangle_2 \\ \end{array}\right\vert = \phi_1(x_1)\phi_2(x_2)\vert +\rangle_1\vert -\rangle_2 - \phi_1(x_2)\phi_2(x_1)\vert -\rangle_1\vert +\rangle_2\, ,\\ &\hbox{Det}\left\vert \begin{array}{cc} \phi_1(x_1)\vert -\rangle_1 &\phi_2(x_1)\vert +\rangle_1 \\ \phi_1(x_2)\vert -\rangle_2 &\phi_2(x_2)\vert +\rangle_2 \\ \end{array}\right\vert = \phi_1(x_1)\phi_2(x_2)\vert -\rangle_1\vert +\rangle_2 - \phi_1(x_2)\phi_2(x_1)\vert +\rangle_1\vert -\rangle_2\, ,\\ &\hbox{Det}\left\vert \begin{array}{cc} \phi_1(x_1)\vert +\rangle_1 &\phi_2(x_1)\vert +\rangle_1 \\ \phi_1(x_2)\vert +\rangle_2 &\phi_2(x_2)\vert +\rangle_2 \\ \end{array}\right\vert = \left(\phi_1(x_1)\phi_2(x_2)-\phi_2(x_1)\phi_1(x_2)\right)\vert +\rangle_1\vert +\rangle_2\\ &\hbox{Det}\left\vert \begin{array}{cc} \phi_1(x_1)\vert -\rangle_1 &\phi_2(x_1)\vert -\rangle_1 \\ \phi_1(x_2)\vert -\rangle_2 &\phi_2(x_2)\vert -\rangle_2 \\ \end{array}\right\vert = \left(\phi_1(x_1)\phi_2(x_2)-\phi_2(x_1)\phi_1(x_2)\right)\vert -\rangle_1\vert -\rangle_2\end{align} all of which are properly antisymmetrized with total energy $E_1+E_2$.

Note that any linear combination of antisymmetric states is also an antisymmetric states, so additional combinations are possible.

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  • $\begingroup$ I've got your point, let me ask you one thing. With reference to the Hamiltonian in my question, if we consider the two fermions in the first excited state, a state like: $\psi_1(\vec{r_1},\vec{r_2})=\phi_{11/2}(\vec{r_1})\phi_{2-1/2}(\vec{r_2})-\phi_{2-1/2}(\vec{r_1})\phi_{11/2}(\vec{r_2})$ with $\phi_{n,S_z}$ and $n$ is the eigenvalue of $\hat{H}$ is a valid eigenfunction? It seems antisymmetric to me. Is this the equivalent of $\psi_1(\vec{r_1},\vec{r_2})=\phi_{2}(\vec{r_1})\phi_{1}(\vec{r_2})-\phi_{1}(\vec{r_1})\phi_{2}(\vec{r_2})|1/2,1/2;1,0\rangle$? $\endgroup$
    – Salmone
    Mar 2 at 22:08
  • $\begingroup$ I'll try to simplify as much as possible what I don't understand. Landau: "Because of the identity of the particles, the states of the system obtained from each other by merely interchanging the two particles must be completely equivalent physically." Now, what does it mean physically interchanging the two particles since in quantum mechanics particles aren't "concrete" objects and in which way particles are identical? In the case of two fermions in two energy levels with different $S_z$, if I interchange the two particles I obtain a new state different from the original, right? $\endgroup$
    – Salmone
    Mar 3 at 15:40
  • $\begingroup$ Let's see if I've understood, in the case of two fermions in two energy levels with different $S_z$, let's say $1/2$ and $-1/2$, if I don't write the total energy eigenfunction as antisymmetric, when I apply the exchange operator I get a different state, that is: $\phi_{1 1/2}(x_1)\phi_{2 -1/2}(x_2)$ if I apply exchange operator I get $\phi_{2 -1/2}(x_1)\phi_{1 1/2}(x_2)$. The two states I've got are different BUT the particles are identical so I don't like this result. SECOND PART ON NEXT COMMENT $\endgroup$
    – Salmone
    Mar 3 at 19:36
  • $\begingroup$ What do I do? I write the total energy eigenfunction as a linear combination of the two different states so that when I apply the exchange operator I get the same exact state. The fact that the eigenfunction is antisymmetric gives me not the same state but −initialstate, however this is not a problem cause $|ψ_{initial}|^2=|ψ_{afterexchange}|^2$, right? $\endgroup$
    – Salmone
    Mar 3 at 21:53
  • $\begingroup$ Thanks but I've given up, I can't find an answer. I understand your point but, why do we have to do linear combinations of two states, yes the linear comb. of anisymmtetric states is antisymmetric but, for the same reason, we should do linear comb. of all the 4 states, for me it doesn't make any sense. Instead the things I wrote in the last two comments do you think are right? $\endgroup$
    – Salmone
    Mar 4 at 13:48

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