The total wavefunction of an electron $\psi(\vec{r},s)$ can always be written as $$\psi(\vec{r},s)=\phi(\vec{r})\zeta_{s,m_s}$$ where $\phi(\vec{r})$ is the space part and $\zeta_{s,m_s}$ is the spin part of the total wavefunction $\psi(\vec{r},s)$. In my notation, $s=1/2, m_s=\pm 1/2$.

Question 1 Is the above statement true? I am asking about any wavefunction here. Not only about energy eigenfunctions.

Now imagine a system of two electrons. Even without any knowledge about the Hamiltonian of the system, the overall wavefunction $\psi(\vec{r}_1,\vec{r}_2;s_1,s_2)$ is antisymmetric. I think (I have this impression) under this general conditions, it is not possible to decompose $\psi(\vec{r}_1,\vec{r}_2;s_1,s_2)$ into a product of a space part and spin part. However, if the Hamiltonian is spin-independent, only then can we do such a decomposition into space part and spin part.

Question 2 Can someone properly argue that how this is so? Please mention about any wavefunction of the system and about energy eigenfunctions.


Your claim

[any arbitrary] wavefunction of an electron $\psi(\vec{r},s)$ can always be written as $$\psi(\vec{r},s)=\phi(\vec{r})\zeta_{s,m_s} \tag 1$$ where $\phi(\vec{r})$ is the space part and $\zeta_{s,m_s}$ is the spin part of the total wavefunction $\psi(\vec{r},s)$

is false. It is perfectly possible to produce wavefunctions which cannot be written in that separable form - for a simple example, just take two orthogonal spatial wavefunctions, $\phi_1$ and $\phi_2$, and two orthogonal spin states, $\zeta_1$ and $\zeta_2$, and define $$ \psi = \frac{1}{\sqrt{2}}\bigg[\phi_1\zeta_1+\phi_2\zeta_2 \bigg]. $$

Moreover, to be clear: the hamiltonian of a system has absolutely no effect on the allowed wavefunctions for that system. The only thing that depends on the hamiltonian is the energy eigenstates.

The result you want is the following:

If the hamiltonian is separable into spatial and spin components as $$ H = H_\mathrm{space}\otimes \mathbb I+ \mathbb I \otimes H_\mathrm{spin},$$ with $H_\mathrm{space}\otimes \mathbb I$ commuting with all spin operators and $\mathbb I \otimes H_\mathrm{spin}$ commuting with all space operators, then there exists an eigenbasis for $H$ of the separable form $(1)$.

To build that eigenbasis, simply diagonalize $H_\mathrm{space}$ and $H_\mathrm{spin}$ independently, and form tensor products of their eigenstates. (Note also that the quantifiers here are crucial, particularly the "If" in the hypotheses and the "there exists" in the results.)

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    $\begingroup$ @SRS The claim is specifically that there exists a separable eigenbasis. There is no claim that all eigenbases for such a hamiltonian are separable, because that claim is false. Please read more carefully. $\endgroup$ – Emilio Pisanty Mar 25 '19 at 13:23
  • $\begingroup$ @SRS "H atom problem" is undefined. (I.e.: are you including fine structure and spin-orbit coupling?) If you're only talking about the Keplerian hamiltonian (i.e. kinetic energy plus electrostatic potential energy, with a frozen proton) then yes, a separable eigenbasis exists; there the $\zeta_j$ are arbitrary (as there is no spin hamiltonian). If you're including spin-orbit coupling, then the hamiltonian does not satisfy the hypotheses I laid out, and the result is false. $\endgroup$ – Emilio Pisanty Mar 25 '19 at 13:37
  • $\begingroup$ Comments are not for back-and-forth - particularly about another user's question. If you have further queries, take them to chat or ask separately. $\endgroup$ – Emilio Pisanty Mar 25 '19 at 13:38
  • $\begingroup$ Is the statement that all the $H_{space}\otimes I$ must commute with all the $I\otimes H_{spin}$ not redundant? From the way you have written them, it seems like they must commute, no? $\endgroup$ – user1936752 Mar 25 '19 at 14:55
  • $\begingroup$ @user1936752 Yes, this is redundant, but I don't think it hurts. $\endgroup$ – Emilio Pisanty Mar 25 '19 at 14:56

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