Return to Question

Although there have been a couple of questions on fermionic coherent states, I don't think any has answered the question "on what space do fermionic coherent states live?", or at least not to my understanding. Hopefully, someone with more knowledge can clarify the situation.

The usual "explanation" is that coherent states $|\psi\rangle = \exp(-\psi a^*) |0\rangle$ live on a space "larger" than the usual fermionic Fock space $\mathscr{F}$ (or the exterior algebra of the single-particle Hilbert space $\mathscr{H}$), where Grassman variables $\psi$ are the coefficients.

I understand that you can define Grassmann variables as elements in the exterior algebra $\mathscr{G}$ of an infinite-dim vector space $V$ and whether the variable is fermionic or bosonic depends on whether $\psi \in \mathscr{G}_-=\oplus \Lambda^{2k+1}(V)$ or $\in \mathscr{G}_+=\oplus \Lambda^{2k}(V)$. You can then define the Grassmann integral via an abstract algebraic generalization of the Lebesgue/Riemann integral. A rigorous explanation can be found Tao's blog.

However, my problem is how would you define/construct this "larger space"? It can't just be the tensor product $\mathscr{G} \otimes \mathscr{F}$ since we require that a fermionic Grassman variable $\psi\in \mathscr{G}_-$ anti-commute with $\mathscr{F}_- = \oplus \Lambda^{2k+1}(\mathscr{H})$ and commute with $\mathscr{F}_+ = \oplus \Lambda^{2k}(\mathscr{H})$ so that $\psi$ anti-commutes with the ladder operators $a,a^*$. It should also have a well-defined "inner product", in the sense that, it is a sesqui-linear map on this "larger space" and maps into the Grassmann variable $\mathscr{G}$.

One attempt would be to think of fermionic coherent states as anti-linear maps on the fermionic Fock space which satisfy $$ \langle m|\psi \rangle=\psi_{i_M} \cdots \psi_{i_1} $$ where $|m\rangle = (a_{i_1}^*)\cdots(a_{i_M}^*)$, and similarly think of $\langle \psi|$ as linear maps which satisfy $\langle \psi|a^* = \langle \psi|\psi$ and $\langle \psi|0\rangle = 1$. However, I'm not sure if this is the right way to think of this problem.

EDIT. After further thought, it's possible that the "larger space" is the exterior algebra $\Lambda$ of the direct sum $V\oplus \mathscr{H}$, so that $\mathscr{F},\mathscr{G}\subseteq \Lambda$. Also notice that as vector spaces, $\Lambda$ is isomorphic to $\mathscr{G}\otimes \mathscr{F}$, which is easily seen if we were to "push" all the Grassmann variables to the left and Fock space states to the right based on the (anti)-commutation relation. Therefore, the ladder operators $a,a^*$ are well-defined on $\Lambda$ as $I\otimes a, I\otimes a^*$ on $\Lambda \cong \mathscr{G}\otimes \mathscr{F}$. We can then proceed to define an "inner product" based on the anti-commutation rules. I haven't yet worked out the details though.

Although there have been a couple of questions on fermionic coherent states, I don't think any has answered the question "on what space do fermionic coherent states live?", or at least not to my understanding. Hopefully, someone with more knowledge can clarify the situation.

The usual "explanation" is that coherent states $|\psi\rangle = \exp(-\psi a^*) |0\rangle$ live on a space "larger" than the usual fermionic Fock space $\mathscr{F}$ (or the exterior algebra of the single-particle Hilbert space $\mathscr{H}$), where Grassman variables $\psi$ are the coefficients.

I understand that you can define Grassmann variables as elements in the exterior algebra $\mathscr{G}$ of an infinite-dim vector space $V$ and whether the variable is fermionic or bosonic depends on whether $\psi \in \mathscr{G}_-=\oplus \Lambda^{2k+1}(V)$ or $\in \mathscr{G}_+=\oplus \Lambda^{2k}(V)$. You can then define the Grassmann integral via an abstract algebraic generalization of the Lebesgue/Riemann integral. A rigorous explanation can be found Tao's blog.

However, my problem is how would you define/construct this "larger space"? It can't just be the tensor product $\mathscr{G} \otimes \mathscr{F}$ since we require that a fermionic Grassman variable $\psi\in \mathscr{G}_-$ anti-commute with $\mathscr{F}_- = \oplus \Lambda^{2k+1}(\mathscr{H})$ and commute with $\mathscr{F}_+ = \oplus \Lambda^{2k}(\mathscr{H})$ so that $\psi$ anti-commutes with the ladder operators $a,a^*$. It should also have a well-defined "inner product", in the sense that, it is a sesqui-linear map on this "larger space" and maps into the Grassmann variable $\mathscr{G}$.

One attempt would be to think of fermionic coherent states as anti-linear maps on the fermionic Fock space which satisfy $$ \langle m|\psi \rangle=\psi_{i_M} \cdots \psi_{i_1} $$ where $|m\rangle = (a_{i_1}^*)\cdots(a_{i_M}^*)$, and similarly think of $\langle \psi|$ as linear maps which satisfy $\langle \psi|a^* = \langle \psi|\psi$ and $\langle \psi|0\rangle = 1$. However, I'm not sure if this is the right way to think of this problem.

Best attempt so far. After further thought, it's possible that the "larger space" is the exterior algebra $\Lambda$ of the direct sum $V\oplus \mathscr{H}$, so that $\mathscr{F},\mathscr{G}\subseteq \Lambda$. Also notice that as vector spaces, $\Lambda$ is isomorphic to $\mathscr{G}\otimes \mathscr{F}$, which is easily seen if we were to "push" all the Grassmann variables to the left and Fock space states to the right based on the (anti)-commutation relation. Therefore, the ladder operators $a,a^*$ are well-defined on $\Lambda$ as $I\otimes a, I\otimes a^*$ on $\Lambda \cong \mathscr{G}\otimes \mathscr{F}$. We can then proceed to define an "inner product" based on the anti-commutation rules. I haven't yet worked out the details though.

Although there have been a couple of questions on fermionic coherent states, I don't think any has answered the question "on what space do fermionic coherent states live?", or at least not to my understanding. Hopefully, someone with more knowledge can clarify the situation.

The usual "explanation" is that coherent states $|\psi\rangle = \exp(-\psi a^*) |0\rangle$ live on a space "larger" than the usual fermionic Fock space $\mathscr{F}$ (or the exterior algebra of the single-particle Hilbert space $\mathscr{H}$), where Grassman variables $\psi$ are the coefficients.

I understand that you can define Grassman variables as elements in the exterior algebra $\mathscr{G}$ of an infinite-dim vector space $V$ and whether the variable is fermionic or bosonic depends on whether $\psi \in \mathscr{G}_-=\oplus \Lambda^{2k+1}(V)$ or $\in \mathscr{G}_+=\oplus \Lambda^{2k}(V)$. You can then define the Grassman integral via an abstract algebraic generalization of the Lebesgue/Riemann integral. A rigorous explanation can be found Tao's bog.

However, my problem is how would you define/construct this "larger space"? It can't just be the tensor product $\mathscr{G} \otimes \mathscr{F}$ since we require that a fermionic Grassman variable $\psi\in \mathscr{G}_-$ anti-commute with $\mathscr{F}_- = \oplus \Lambda^{2k+1}(\mathscr{H})$ and commute with $\mathscr{F}_+ = \oplus \Lambda^{2k}(\mathscr{H})$ so that $\psi$ anti-commutes with the ladder operators $a,a^*$. It should also have a well-defined "inner product", in the sense that, it is a sesqui-linear map on this "larger space" and maps into the Grassman variable $\mathscr{G}$.

One attempt would be to think of fermionic coherent states as anti-linear maps on the fermionic Fock space which satisfy $$ \langle m|\psi \rangle=\psi_{i_M} \cdots \psi_{i_1} $$ where $|m\rangle = (a_{i_1}^*)\cdots(a_{i_M}^*)$, and similarly think of $\langle \psi|$ as linear maps which satisfy $\langle \psi|a^* = \langle \psi|\psi$ and $\langle \psi|0\rangle = 1$. However, I'm not sure if this is the right way to think of this problem.

EDIT. After further thought, it's possible that the "larger space" is the exterior algebra $\Lambda$ of the direct sum $V\oplus \mathscr{H}$, so that $\mathscr{F},\mathscr{G}\subseteq \Lambda$. Also notice that as vector spaces, $\Lambda$ is isomorphic to $\mathscr{G}\otimes \mathscr{F}$, which is easily seen if we were to "push" all the Grassman variables to the left and Fock space states to the right based on the (anti)-commutation relation. Therefore, the ladder operators $a,a^*$ are well-defined on $\Lambda$ as $I\otimes a, I\otimes a^*$ on $\Lambda \cong \mathscr{G}\otimes \mathscr{F}$. We can then proceed to define an "inner product" based on the anti-commutation rules. I haven't yet worked out the details though.

Although there have been a couple of questions on fermionic coherent states, I don't think any has answered the question "on what space do fermionic coherent states live?", or at least not to my understanding. Hopefully, someone with more knowledge can clarify the situation.

The usual "explanation" is that coherent states $|\psi\rangle = \exp(-\psi a^*) |0\rangle$ live on a space "larger" than the usual fermionic Fock space $\mathscr{F}$ (or the exterior algebra of the single-particle Hilbert space $\mathscr{H}$), where Grassman variables $\psi$ are the coefficients.

I understand that you can define Grassmann variables as elements in the exterior algebra $\mathscr{G}$ of an infinite-dim vector space $V$ and whether the variable is fermionic or bosonic depends on whether $\psi \in \mathscr{G}_-=\oplus \Lambda^{2k+1}(V)$ or $\in \mathscr{G}_+=\oplus \Lambda^{2k}(V)$. You can then define the Grassmann integral via an abstract algebraic generalization of the Lebesgue/Riemann integral. A rigorous explanation can be found Tao's blog.

However, my problem is how would you define/construct this "larger space"? It can't just be the tensor product $\mathscr{G} \otimes \mathscr{F}$ since we require that a fermionic Grassman variable $\psi\in \mathscr{G}_-$ anti-commute with $\mathscr{F}_- = \oplus \Lambda^{2k+1}(\mathscr{H})$ and commute with $\mathscr{F}_+ = \oplus \Lambda^{2k}(\mathscr{H})$ so that $\psi$ anti-commutes with the ladder operators $a,a^*$. It should also have a well-defined "inner product", in the sense that, it is a sesqui-linear map on this "larger space" and maps into the Grassmann variable $\mathscr{G}$.

One attempt would be to think of fermionic coherent states as anti-linear maps on the fermionic Fock space which satisfy $$ \langle m|\psi \rangle=\psi_{i_M} \cdots \psi_{i_1} $$ where $|m\rangle = (a_{i_1}^*)\cdots(a_{i_M}^*)$, and similarly think of $\langle \psi|$ as linear maps which satisfy $\langle \psi|a^* = \langle \psi|\psi$ and $\langle \psi|0\rangle = 1$. However, I'm not sure if this is the right way to think of this problem.

EDIT. After further thought, it's possible that the "larger space" is the exterior algebra $\Lambda$ of the direct sum $V\oplus \mathscr{H}$, so that $\mathscr{F},\mathscr{G}\subseteq \Lambda$. Also notice that as vector spaces, $\Lambda$ is isomorphic to $\mathscr{G}\otimes \mathscr{F}$, which is easily seen if we were to "push" all the Grassmann variables to the left and Fock space states to the right based on the (anti)-commutation relation. Therefore, the ladder operators $a,a^*$ are well-defined on $\Lambda$ as $I\otimes a, I\otimes a^*$ on $\Lambda \cong \mathscr{G}\otimes \mathscr{F}$. We can then proceed to define an "inner product" based on the anti-commutation rules. I haven't yet worked out the details though.

Although there have been a couple of questions on fermionic coherent states, I don't think any has answered the question "on what space do fermionic coherent states live?", or at least not to my understanding. Hopefully, someone with more knowledge can clarify the situation.

The usual "explanation" is that coherent states $|\psi\rangle = \exp(-\psi a^*) |0\rangle$ live on a space "larger" than the usual fermionic Fock space $\mathscr{F}$ (or the exterior algebra of the single-particle Hilbert space $\mathscr{H}$), where Grassman variables $\psi$ are the coefficients.

I understand that you can define Grassman variables as elements in the exterior algebra $\mathscr{G}$ of an infinite-dim vector space $V$ and whether the variable is fermionic or bosonic depends on whether $\psi \in \mathscr{G}_-=\oplus \Lambda^{2k+1}(V)$ or $\in \mathscr{G}_+=\oplus \Lambda^{2k}(V)$. You can then define the Grassman integral via an abstract algebraic generalization of the Lebesgue/Riemann integral. A rigorous explanation can be found Tao's bog.

However, my problem is how would you define/construct this "larger space"? It can't just be the tensor product $\mathscr{G} \otimes \mathscr{F}$ since we require that a fermionic Grassman variable $\psi\in \mathscr{G}_-$ anti-commute with $\mathscr{F}_- = \oplus \Lambda^{2k+1}(\mathscr{H})$ and commute with $\mathscr{F}_+ = \oplus \Lambda^{2k}(\mathscr{H})$ so that $\psi$ anti-commutes with the ladder operators $a,a^*$. It should also have a well-defined "inner product", in the sense that, it is a sesqui-linear map on this "larger space" and maps into the Grassman variable $\mathscr{G}$.

One attempt would be to think of fermionic coherent states as anti-linear maps on the fermionic Fock space which satisfy $$ \langle m|\psi \rangle=\psi_{i_M} \cdots \psi_{i_1} $$ where $|m\rangle = (a_{i_1}^*)\cdots(a_{i_M}^*)$, and similarly think of $\langle \psi|$ as linear maps which satisfy $\langle \psi|a^* = \langle \psi|\psi$ and $\langle \psi|0\rangle = 1$. However, I'm not sure if this is the right way to think of this problem.

EDIT. After further thought, it's possible that the "larger space" is the exterior algebra $\Lambda$ of the direct sum $V\oplus \mathscr{H}$, so that $\mathscr{F},\mathscr{G}\subseteq \Lambda$. Also notice that as vector spaces, $\Lambda$ is isomorphic to $\mathscr{G}\otimes \mathscr{F}$, which is easily seen if we were to "push" all the Grassman variables to the left and Fock space states to the right based on the (anti)-commutation relation. We can then proceed to define an "inner product" based on the anti-commutation rules. I haven't yet worked out the details though.

Although there have been a couple of questions on fermionic coherent states, I don't think any has answered the question "on what space do fermionic coherent states live?", or at least not to my understanding. Hopefully, someone with more knowledge can clarify the situation.

The usual "explanation" is that coherent states $|\psi\rangle = \exp(-\psi a^*) |0\rangle$ live on a space "larger" than the usual fermionic Fock space $\mathscr{F}$ (or the exterior algebra of the single-particle Hilbert space $\mathscr{H}$), where Grassman variables $\psi$ are the coefficients.

I understand that you can define Grassman variables as elements in the exterior algebra $\mathscr{G}$ of an infinite-dim vector space $V$ and whether the variable is fermionic or bosonic depends on whether $\psi \in \mathscr{G}_-=\oplus \Lambda^{2k+1}(V)$ or $\in \mathscr{G}_+=\oplus \Lambda^{2k}(V)$. You can then define the Grassman integral via an abstract algebraic generalization of the Lebesgue/Riemann integral. A rigorous explanation can be found Tao's bog.

However, my problem is how would you define/construct this "larger space"? It can't just be the tensor product $\mathscr{G} \otimes \mathscr{F}$ since we require that a fermionic Grassman variable $\psi\in \mathscr{G}_-$ anti-commute with $\mathscr{F}_- = \oplus \Lambda^{2k+1}(\mathscr{H})$ and commute with $\mathscr{F}_+ = \oplus \Lambda^{2k}(\mathscr{H})$ so that $\psi$ anti-commutes with the ladder operators $a,a^*$. It should also have a well-defined "inner product", in the sense that, it is a sesqui-linear map on this "larger space" and maps into the Grassman variable $\mathscr{G}$.

One attempt would be to think of fermionic coherent states as anti-linear maps on the fermionic Fock space which satisfy $$ \langle m|\psi \rangle=\psi_{i_M} \cdots \psi_{i_1} $$ where $|m\rangle = (a_{i_1}^*)\cdots(a_{i_M}^*)$, and similarly think of $\langle \psi|$ as linear maps which satisfy $\langle \psi|a^* = \langle \psi|\psi$ and $\langle \psi|0\rangle = 1$. However, I'm not sure if this is the right way to think of this problem.

EDIT. After further thought, it's possible that the "larger space" is the exterior algebra $\Lambda$ of the direct sum $V\oplus \mathscr{H}$, so that $\mathscr{F},\mathscr{G}\subseteq \Lambda$. Also notice that as vector spaces, $\Lambda$ is isomorphic to $\mathscr{G}\otimes \mathscr{F}$, which is easily seen if we were to "push" all the Grassman variables to the left and Fock space states to the right based on the (anti)-commutation relation. Therefore, the ladder operators $a,a^*$ are well-defined on $\Lambda$ as $I\otimes a, I\otimes a^*$ on $\Lambda \cong \mathscr{G}\otimes \mathscr{F}$. We can then proceed to define an "inner product" based on the anti-commutation rules. I haven't yet worked out the details though.