Quantum Mechanics for Mathematicians

• ωis antisymmetric:ω(v,v′) =−ω(v′,v)


• ωis nondegenerate: ifv 6 = 0, thenω(v,·)∈V∗is non-zero.

A vector spaceV with a symplectic formωis called a symplectic vector
space. The analog of Euclidean geometry, replacing the inner product by a
symplectic form, is called symplectic geometry. In this sort of geometry, there
is no notion of length (since antisymmetry impliesω(v,v) = 0). There is an
analog of the orthogonal group, called the symplectic group, which consists of
linear transformations preservingω, a group we will study in detail in chapter
16.
Just as an inner product gives an identification ofV andV∗, a symplectic
form can be used in a similar way, giving an identification ofMandM. Using
the symplectic form Ω onM, we can define an isomorphism by identifying basis
vectors by

qj∈M↔Ω(·,qj) =−Ω(qj,·) =−

∂

∂pj

∈M

pj∈M↔Ω(·,pj) =−Ω(pj,·) =

∂

∂qj

∈M

and in general
u∈M↔Ω(·,u) =−Ω(u,·)∈M (14.6)

Note that unlike the inner product case, a choice of convention of minus sign
must be made and is done here.
Recalling the discussion of bilinear forms from section 9.5, a bilinear form on
a vector spaceVcan be identified with an element ofV∗⊗V∗. TakingV=M∗
we haveV∗= (M∗)∗=M, and the bilinear form Ω onM∗is an element of
M⊗Mgiven by

Ω =

∑d

j=1

(

∂

∂qj

⊗

∂

∂pj

−

∂

∂pj

⊗

∂

∂qj

)

Under the identification 14.6 ofMandM∗, Ω∈M⊗Mcorresponds to

ω=

∑d

j=1

(qj⊗pj−pj⊗qj)∈M∗⊗M∗ (14.7)

Another version of the identification ofMandMis then given by

v∈M→ω(v,·)∈M

In the case of Euclidean geometry, one can show by Gram-Schmidt orthog-
onalization that a basisej can always be found that puts the inner product
(which is a symmetric element ofV∗⊗V∗) in the standard form

∑n

j=1

vj⊗vj