Difference between revisions of "Antisymmetric subspace"

From QETLAB
Jump to navigation Jump to search
m
 
(3 intermediate revisions by the same user not shown)
Line 9: Line 9:
  
 
==Construction==
 
==Construction==
The projection $P_{\mathcal{A}}$ onto the antisymmetric subspace can be constructed by averaging the signed permutation operators:<ref>R.B. Griffiths. [http://quantum.phys.cmu.edu/qm2/qmc161.pdf Systems of Identical Particles], ''33-756 Quantum Mechanics II Course Notes'', 2011.</ref>
+
The [[orthogonal projection]] $P_{\mathcal{A}}$ onto the antisymmetric subspace can be constructed by averaging the signed permutation operators:<ref>R.B. Griffiths. [http://quantum.phys.cmu.edu/qm2/qmc161.pdf Systems of Identical Particles], ''33-756 Quantum Mechanics II Course Notes'', 2011.</ref>
 
: $\displaystyle P_{\mathcal{A}} = \frac{1}{p!}\sum_{\sigma \in S_p} (-1)^{{\rm sgn}(\sigma)}P_\sigma$.
 
: $\displaystyle P_{\mathcal{A}} = \frac{1}{p!}\sum_{\sigma \in S_p} (-1)^{{\rm sgn}(\sigma)}P_\sigma$.
 +
 +
In the p = 2 case, this reduces to simply $P_{\mathcal{A}} = \frac{1}{2}(I - S)$, where $I$ is the identity operator and $S$ is the bipartite [[swap operator]]. Alternatively, we also have $P_{\mathcal{A}} = I - P_{\mathcal{S}}$ in this case (but not when $p > 2$), where $P_{\mathcal{S}}$ is the orthogonal projection onto the [[symmetric subspace]].
  
 
In order to explicitly construct an orthonormal basis of the antisymmetric subspace, first fix an orthonormal basis $\{|i\rangle\}_{i=1}^d$ of $\mathbb{C}^d$. Now let $q$ be an increasing vector with $p$ entries containing distinct elements of $\{1, 2, \ldots, d\}$ as its entries. Then define
 
In order to explicitly construct an orthonormal basis of the antisymmetric subspace, first fix an orthonormal basis $\{|i\rangle\}_{i=1}^d$ of $\mathbb{C}^d$. Now let $q$ be an increasing vector with $p$ entries containing distinct elements of $\{1, 2, \ldots, d\}$ as its entries. Then define
 
: $\displaystyle \mathbf{v}_q \triangleq \frac{1}{\sqrt{p!}}\sum_{\sigma \in S_p} (-1)^{{\rm sgn}(\sigma)} P_\sigma (| q_1 \rangle \otimes \cdots \otimes | q_p \rangle)$.
 
: $\displaystyle \mathbf{v}_q \triangleq \frac{1}{\sqrt{p!}}\sum_{\sigma \in S_p} (-1)^{{\rm sgn}(\sigma)} P_\sigma (| q_1 \rangle \otimes \cdots \otimes | q_p \rangle)$.
It is then the case that if $q$ ranges over all $\binom{d}{p}$ possible vectors $q$ then $\{\mathbf{v}_q\}_q$ is an orthonormal basis of $\mathcal{A}_p^d$.
+
It is then the case that $\{\mathbf{v}_q\}_q$ is an orthonormal basis of $\mathcal{A}_p^d$ if $q$ ranges over all $\binom{d}{p}$ possible increasing vectors with $p$ distinct entries chosen from $\{1, 2, \ldots, d\}$.
  
 
==Facts==
 
==Facts==
 
* The dimension of the antisymmetric subspace is $\binom{d}{p}$. It follows that the antisymmetric subspace only has nonzero dimension when d &ge; p. When p = 2, the dimension of the antisymmetric subspace is d(d-1)/2, from which it follows that the symmetric and antisymmetric subspaces span the whole space in this case (and this case only): $\mathcal{A}_2^d \oplus \mathcal{S}_2^d \cong \mathbb{C}^d \otimes \mathbb{C}^d$.
 
* The dimension of the antisymmetric subspace is $\binom{d}{p}$. It follows that the antisymmetric subspace only has nonzero dimension when d &ge; p. When p = 2, the dimension of the antisymmetric subspace is d(d-1)/2, from which it follows that the symmetric and antisymmetric subspaces span the whole space in this case (and this case only): $\mathcal{A}_2^d \oplus \mathcal{S}_2^d \cong \mathbb{C}^d \otimes \mathbb{C}^d$.
 +
 +
==Related [[QETLAB]] functions==
 +
* <tt>[[AntisymmetricProjection]]</tt>: Produces the [[projection]] onto the antisymmetric subspace $P_{\mathcal{A}}$
 +
* <tt>[[PermutationOperator]]</tt>: Produces a permutation operator $P_\sigma$
  
 
==See also==
 
==See also==
 
* [[Symmetric subspace]]
 
* [[Symmetric subspace]]
 
==Related [[QETLAB]] functions==
 
* <tt>[[AntisymmetricProjection]]</tt>: Produces the [[projection]] onto the antisymmetric subspace $P_{\mathcal{A}}$
 
* <tt>[[PermutationOperator]]</tt>: Produces the permutation operator $P_\sigma$
 
  
 
==References==
 
==References==
 
<references />
 
<references />

Latest revision as of 03:27, 22 November 2012

The antisymmetric subspace $\mathcal{A}_p^d$ is the subspace of $(\mathbb{C}^d)^{\otimes p}$ of all vectors that are negated by odd permutations:

$\displaystyle \mathcal{A}_p^d \triangleq \big\{ \mathbf{v} \in (\mathbb{C}^d)^{\otimes p} : \mathbf{v} = (-1)^{{\rm sgn}(\sigma)}P_\sigma \mathbf{v} \ \ \forall \sigma \in S_p \big\},$

where $S_p$ is the symmetric group, ${\rm sgn}(\sigma)$ is the parity of the permutation $\sigma$, and $P_\sigma$ is the unitary operator that permutes the $p$ subsystems according to the permutation $\sigma$:

$\displaystyle P_\sigma(\mathbf{v}_1 \otimes \cdots \otimes \mathbf{v}_p) \triangleq \mathbf{v}_{\sigma(1)} \otimes \cdots \otimes \mathbf{v}_{\sigma(p)}$.

The antisymmetric subspace plays a role quite complementary to that of the symmetric subspace (and indeed, if $\mathcal{S}_p^d$ is the symmetric subspace then $\mathcal{A}_p^d \perp \mathcal{S}_p^d$).

Construction

The orthogonal projection $P_{\mathcal{A}}$ onto the antisymmetric subspace can be constructed by averaging the signed permutation operators:[1]

$\displaystyle P_{\mathcal{A}} = \frac{1}{p!}\sum_{\sigma \in S_p} (-1)^{{\rm sgn}(\sigma)}P_\sigma$.

In the p = 2 case, this reduces to simply $P_{\mathcal{A}} = \frac{1}{2}(I - S)$, where $I$ is the identity operator and $S$ is the bipartite swap operator. Alternatively, we also have $P_{\mathcal{A}} = I - P_{\mathcal{S}}$ in this case (but not when $p > 2$), where $P_{\mathcal{S}}$ is the orthogonal projection onto the symmetric subspace.

In order to explicitly construct an orthonormal basis of the antisymmetric subspace, first fix an orthonormal basis $\{|i\rangle\}_{i=1}^d$ of $\mathbb{C}^d$. Now let $q$ be an increasing vector with $p$ entries containing distinct elements of $\{1, 2, \ldots, d\}$ as its entries. Then define

$\displaystyle \mathbf{v}_q \triangleq \frac{1}{\sqrt{p!}}\sum_{\sigma \in S_p} (-1)^{{\rm sgn}(\sigma)} P_\sigma (| q_1 \rangle \otimes \cdots \otimes | q_p \rangle)$.

It is then the case that $\{\mathbf{v}_q\}_q$ is an orthonormal basis of $\mathcal{A}_p^d$ if $q$ ranges over all $\binom{d}{p}$ possible increasing vectors with $p$ distinct entries chosen from $\{1, 2, \ldots, d\}$.

Facts

  • The dimension of the antisymmetric subspace is $\binom{d}{p}$. It follows that the antisymmetric subspace only has nonzero dimension when d ≥ p. When p = 2, the dimension of the antisymmetric subspace is d(d-1)/2, from which it follows that the symmetric and antisymmetric subspaces span the whole space in this case (and this case only): $\mathcal{A}_2^d \oplus \mathcal{S}_2^d \cong \mathbb{C}^d \otimes \mathbb{C}^d$.

Related QETLAB functions

See also

References

  1. R.B. Griffiths. Systems of Identical Particles, 33-756 Quantum Mechanics II Course Notes, 2011.