# Statement in Cohen-Tanoudji I don't understand

Homework Helper
Gold Member
He writes, talking about the ket space X of a physical system,

In general, the dual space $X^*$ and the state space $X$ are not isomorphic, except of course, if $X$ is finite dimensional.

Can someone explain why they are isom. in the event of finite dimensionality?

For instance, say X is finite. To every |x> in X is associated the linear functional $<x|$ (the "inner product functional"). But also, given any (non-identity) linear operator A on X, <x|A is a new linear functionnal on X, is it not*? So that makes the cardinality of $X^*$ greater than the cardinality of X, so there can be no bijection between them.

What's wrong?

*I should say potentially new, because if A maps every |x1> towards |x2>'s such that, by miracle, <x|x1>=<x|x2>, then <x| and <x|A really are the same functional...

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dextercioby
Homework Helper
I don't get it. That dual is the algebraic dual or the topological dual ? (In the case of finite dim. LTS they coincide, so it doesn't matter.)

How can you infer that " the cardinality of X^{*} is greater than the cardinality of X" based on what you've written ?

Daniel.

Homework Helper
Gold Member
Ok, my counter-exemple's not good. But my main question remains. Why are X and it's dual isomorphic in the even that X is finite dimensional?

(The dual X^* is the set of all linear functionals on X.. I guess that's the algebraic dual?)

StatusX
Homework Helper
In a Hilbert space you can show every linear functional arises as the inner product with some element in the space. There's your bijection. Of course, having the same cardinality does not entail having the same dimension, but you can show the image of a basis in the space under this bijection is a basis in the dual space.

dextercioby