Re: Linear superposition: Why complex coefficients?

On Feb 23, 8:45 pm, Igor <igor...@xxxxxxxxx> wrote:
On Feb 22, 3:14 am, ste...@xxxxxxxxxxxxxxxx wrote:

On 21 Feb, 12:22, Igor <igor...@xxxxxxxxx> wrote:
Unfortunately, there is a kink in your proof, Stephen. In general, a
real anti-adjoint operator K need not have any real eigenvalues. But
you are right that if it has any real eigenvalues, they have to be
zero.
My reply assumed that Lester's question was really about why the arena
for QM is a Hilbert space rather than a real vector space. In a real
vector space the scalar product <x|y> - regarded as a functional - is
linear in <x|, whilst in Hilbert space it is anti-linear in <x|. Maybe
I should have stated this explicitly. The proof develops the
consequences of the scalar product of a real vector space being linear
in <x|. I cannot see any step in my proof which assumes that the
eigenvalues of K are purely real, so I think your criticism is off-
target in this case and the proof still looks watertight to me.

Hmm, I'm a little confused by what you mean. There are real vector
spaces and complex ones (distinguished by the kind of coefficients we
are allowed to take linear combinations with). There are also spaces
with an inner product and without. These two properties are
independent. A Hilbert space is an inner product space (with a
completeness condition thrown in for infinite dimensional cases).

Anti-linearity of the inner product <x|y> means that we must replace
<ax|y> by (a*)<x|y>, where a is any scalar and (a*) is its complex
conjugate. When the vector space is real, a can only be real, so the
conditions of linearity and anti-linearity coincide. So I don't see
the distinction you are making.

The problem I saw in your argument is that, as long as you work with a
real vector space (whether with an inner product or not), you can't
assume that the translation operators D(a) and their generator K have
any eigenvalues. As I discussed in the previous post, D(a) and K need
not be completely diagonalizable; at worst, they are only block
diagonalizable.

Igor

The only way I know of getting a translation operator on a real vector
space is to
double the dimension and use the isomorphism of
0, -1
i = 1, 0

admittedly a bit of a dodge. Essentially one is merely duplicating a
one component complex algebra with a two component real algebra.

My OP did assume a complex Hilbert space and my question concerned the
coefficients.

.

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