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Let $\Psi \in \mathcal{L}^{2}(\mathbb{R}^{6}; \mathbb{C}^{2}\otimes\mathbb{C}^{2})$ so that $\Psi$ represents the state of two interacting fermions.

Traditionally, we define the exchange operator $P_{12}$ via \begin{align*} P_{12}\Psi(\mathbf{r}_1, \mathbf{r}_2) = \Psi(\mathbf{r}_2, \mathbf{r}_1) \end{align*} but pretty soon we realize this is incorrect, as we know that we need to interchange the spin state as well. This is often achieved by writing \begin{align*} P_{12}\Psi(\mathbf{r}_1, \sigma_1, \mathbf{r}_2, \sigma_2) = \Psi(\mathbf{r}_2, \sigma_2, \mathbf{r}_1, \sigma_1) \end{align*} but this is still awkward, as the spin state is a dependent variable in the theory.

To avoid these problems I would like to define the exchange operator without any reference to the dummy variables $\mathbf{r}_1$ and $\mathbf{r}_2$.

This is what I have so far: Suppose $\{\psi_{n}\}$ separates $\mathcal{L}^{2}(\mathbb{R}^{3}, \mathbb{C})$. Then any function $\Psi \in \mathcal{L}^{2}(\mathbb{R}^{6}; \mathbb{C}^{2}\otimes\mathbb{C}^{2})$ can be written as \begin{align}\label{separate} \Psi &= \sum_{n} a_{nm} \begin{pmatrix} \psi_{n} \\ 0 \end{pmatrix} \otimes \begin{pmatrix} 0 \\ \psi_{m} \end{pmatrix} + b_{nm} \begin{pmatrix} \psi_{n} \\ 0 \end{pmatrix} \otimes \begin{pmatrix} \psi_{m} \\ 0 \end{pmatrix} + c_{nm} \begin{pmatrix} 0 \\ \psi_{n} \end{pmatrix} \otimes \begin{pmatrix} 0 \\ \psi_{m} \end{pmatrix} + d_{nm} \begin{pmatrix} 0 \\ \psi_{n} \end{pmatrix} \otimes \begin{pmatrix} \psi_{m} \\ 0 \end{pmatrix} \end{align} where the first part of the tensor product is assumed to act on $\mathbf{r}_1$ and the second assumed to act of $\mathbf{r}_2$. Now we define $P_{12}$ on the basis, interchanging the functions rather than the dummy variables. Exchange of the spin-coordinate is now trivial, defined by \begin{align*} P_{12}^{\mathrm{spin}} \begin{pmatrix} \psi_{n} \\ \psi_{m} \end{pmatrix} \otimes \begin{pmatrix} \psi_{j} \\ \psi_{k} \end{pmatrix} = \begin{pmatrix} \psi_{m} \\ \psi_{n} \end{pmatrix} \otimes \begin{pmatrix} \psi_{k} \\ \psi_{j} \end{pmatrix} \end{align*} However, what I would consider the natural space exchange definition \begin{align*} P_{12}^{\mathrm{space}} \begin{pmatrix} \psi_{n} \\ \psi_{m} \end{pmatrix} \otimes \begin{pmatrix} \psi_{j} \\ \psi_{k} \end{pmatrix} = \begin{pmatrix} \psi_{j} \\ \psi_{k} \end{pmatrix} \otimes \begin{pmatrix} \psi_{n} \\ \psi_{m} \end{pmatrix} \end{align*} gives results which are in disagreement with the notation found in standard quantum mechanics texts. For instance, take $\Psi(\mathbf{r}_1, \mathbf{r}_2) = \psi(\mathbf{r}_1)\psi(\mathbf{r}_2)(\uparrow \downarrow - \downarrow \uparrow)$. This is clearly antisymmetric, but is symmetric under $P_{12}^{\mathrm{space}}P_{12}^{\mathrm{spin}}$. This can be patched up by defining $P_{12}^{\mathrm{space}}$ differently on each of the four types of terms is \ref{separate}, but that seems so unnatural that I feel like I'm on the wrong track.

Is there a natural definition of $P_{12}$ which doesn't make reference to the dummy variables $\mathbf{r}_1$ and $\mathbf{r}_2$?

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  • $\begingroup$ What do "space-only exchange" and "spin-only exchange" mean, and, more importantly, why do you need them at all? Why not just define the action of $P_{12}$ on a basis and be done with it? $\endgroup$ – Emilio Pisanty Feb 21 '17 at 20:36
  • $\begingroup$ @EmilioPisanty: This decomposition is from Sakurai, which I thought many readers would be familiar with. In any case, the problem remains even if you don't refer to these. $\endgroup$ – user14717 Feb 21 '17 at 20:39
  • $\begingroup$ Related (since it also deals with the proper definition of the exchange operator): physics.stackexchange.com/q/86116/50583 and its linked questions. $\endgroup$ – ACuriousMind Feb 22 '17 at 9:03
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I would normally approach this by flipping your initial formulas: take the conjugate of $$ \langle \mathbf r_1,\sigma_1,\mathbf r_2,\sigma_2| P_{12}^\mathrm{space} |\psi\rangle = \langle \mathbf r_2,\sigma_1,\mathbf r_1,\sigma_2|\psi\rangle $$ to get $$ \langle \psi|P_{12}^\mathrm{space}|\mathbf r_1,\sigma_1,\mathbf r_2,\sigma_2\rangle = \langle \psi|\mathbf r_2,\sigma_1,\mathbf r_1,\sigma_2\rangle ,$$ and then you can cancel out the $\langle\psi|$ to get $$ P_{12}^\mathrm{space}|\mathbf r_1,\sigma_1,\mathbf r_2,\sigma_2\rangle = |\mathbf r_2,\sigma_1,\mathbf r_1,\sigma_2\rangle ,$$ and similarly $$ P_{12}^\mathrm{spin}|\mathbf r_1,\sigma_1,\mathbf r_2,\sigma_2\rangle = |\mathbf r_1,\sigma_2,\mathbf r_2,\sigma_1\rangle .$$ I don't really see the $\mathbf r_i$ as dummy variables here (or indeed as any better than the $\psi_n$), since they are simply the index of a basis for the single-particle-single-spin-projection Hilbert space. However, if you want to use a basis, you can choose a set of $|\psi_n\rangle = \int \psi_n(\mathbf r)|\mathbf r\rangle\mathrm d\mathbf r$, and integrate over those; this gives you \begin{align} P_{12}^\mathrm{space}|\psi_n,\sigma_1,\psi_m,\sigma_2\rangle & = |\psi_m,\sigma_1,\psi_n,\sigma_2\rangle \\ P_{12}^\mathrm{spin}|\psi_n,\sigma_1,\psi_m,\sigma_2\rangle & = |\psi_n,\sigma_2,\psi_m,\sigma_1\rangle . \end{align} To go beyond this, you simply have to note that the $\psi_n$ and the $\sigma_j$ are arbitrary basis elements of the respective Hilbert spaces of your tensor product, so you can simply take any two $|\psi\rangle,|\varphi\rangle\in\mathcal L^2(\mathbb R^3)$ and any two $|\alpha\rangle, |\beta\rangle\in\mathbb C^2$, and get \begin{align} P_{12}^\mathrm{space}|\psi,\alpha,\varphi,\beta\rangle & = |\varphi,\alpha,\psi,\beta\rangle \\ P_{12}^\mathrm{spin}|\psi,\alpha,\varphi,\beta\rangle & = |\psi,\beta,\varphi,\alpha\rangle , \end{align} where $|\psi,\alpha,\varphi,\beta\rangle| = |\psi\rangle\otimes |\alpha\rangle \otimes |\varphi\rangle\otimes |\beta\rangle$. That's about as "without dummy variables" as it can get, honestly.

It's unclear to me what you mean by "spin is a dependent variable", because frankly that is at best a perspective-dependent statement. If by that you mean that you really do envision $\psi(\mathbf r_1,\mathbf r_2)$ as taking values on $\mathbb C^2\otimes\mathbb C^2$, then for one thing that's in direct conflict with notations like $\psi(\mathbf r_1,\sigma_1,\mathbf r_2,\sigma_2)$, because there you have a wavefunction whose value is a complex number, and that's in direct conflict with what you said earlier.

More damningly, it's not even clear what things like pure kets actually mean. Is $\langle \mathbf r_1,\mathbf r_2|\psi\rangle$ a vector in $\mathbb C^2\otimes\mathbb C^2$? In that case, what is $\langle\psi| \mathbf r_1,\mathbf r_2\rangle$? Something in the dual $\left(\mathbb C^2\otimes\mathbb C^2\right)^*$, some weird tensor product of row vectors? If that's the sort of Hilbert space you want to build, that's probably doable, and you will indeed need need to make $P_{12}^\mathrm{spin}$ act explicitly on that $\mathbb C^2\otimes\mathbb C^2$.

I would encourage you, however, to drop that silly notion ;-), and to simply treat spin labels much like you do position labels. In particular, this means that your Hilbert space isn't really $\mathcal L^2(\mathbb R^6;\mathbb C^2\otimes\mathbb C^2)$, and instead it's $\mathcal L^2(\mathbb R^3)\otimes\mathbb C^2\otimes \mathcal L^2(\mathbb R^3) \otimes\mathbb C^2$. But, ultimately, it's a matter of taste.

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  • $\begingroup$ Could you refer me to some literature which explicitly formulates the many-body theory in terms of $\mathcal{L}^{2}(\mathbb{R}^{3})\otimes \mathbb{C}^{2}\otimes\mathcal{L}^{2}(\mathbb{R}^{3}) \otimes \mathbb{C}^{2}$? I have only seen it formulated in the space I was using; for example www3.nd.edu/~bhall/book/quantum.html $\endgroup$ – user14717 Feb 21 '17 at 21:45
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The column vector notation can be used to express the result of exchange of spatial or spin arguments of the psi function, if we adopt this convention:

if the symbol is on the left side of the tensor product, it belongs to particle 1, if it is on the right side, it belongs to particle 2.

This means that when the symbol $\psi_n$ is moved from the left vector to the right vector as part of your operation $P_{12}^{space}$, the particle index of the spatial argument changes from 1 to 2 which is desired, but the symbol then multiplies spin vector of 2nd particle, not spin vector of the 1st one as it should. Similar problem occurs for your operator $P_{12}^{spin}$.

Consequently, one cannot represent the exchange operations for spatial or spin arguments in the way you wrote.

Using the original function notation it can be derived that the exchange operators work this way in the column vector notation:

\begin{align} P_{12}^{space} \begin{pmatrix} f_+ \\ f_- \end{pmatrix} \otimes \begin{pmatrix} g_+ \\ g_- \end{pmatrix} = \end{align}

\begin{align} = \begin{pmatrix} g_+ \\ . \end{pmatrix} \otimes \begin{pmatrix} f_+ \\ . \end{pmatrix} + \begin{pmatrix} g_- \\ . \end{pmatrix} \otimes \begin{pmatrix} . \\ f_+ \end{pmatrix} + \begin{pmatrix} . \\ g_+ \end{pmatrix} \otimes \begin{pmatrix} f_- \\ . \end{pmatrix} + \begin{pmatrix} . \\ g_- \end{pmatrix} \otimes \begin{pmatrix} . \\ f_- \end{pmatrix}. \end{align}

These 4 terms cannot be simplified into lesser number of terms, because they are linearly independent. For spin exchange the result is similar, just exchange $f$ with $g$ everywhere.

You can then try to verify that subsequent application of $P_{12}^{space}$ and $P_{12}^{spin}$ on the ket

\begin{align} \begin{pmatrix} \psi\\ . \end{pmatrix} \otimes \begin{pmatrix} . \\ \psi \end{pmatrix} - \begin{pmatrix} . \\ \psi \end{pmatrix} \otimes \begin{pmatrix} \psi\\ . \end{pmatrix} \end{align}

results in minus this ket.

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