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Cuchulainn
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Re: Quantum Question

September 25th, 2018, 4:49 pm

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katastrofa
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Re: Quantum Question

September 25th, 2018, 5:28 pm

Cuchulainn, you really start to display signs of paranoia.
Last edited by katastrofa on September 25th, 2018, 9:00 pm
 
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Cuchulainn
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Re: Quantum Question

September 25th, 2018, 6:59 pm

katastrofa wrote:
Cuchulainn, you really starts to display signs of paranoia.

So, neither of yous know the answer, what?

A Hilbert space is just Euclidean space on steroids. LinearAlgebra++

Random thought: why no QM in Banach spaces?

infinitely dimensional Hilbert spaces.
Countably infinite I presume..Nice separable HS.
 
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Gamal
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Re: Quantum Question

September 25th, 2018, 8:44 pm

Bra-ket notation is a purely mathematical notation introduced by Grassmann some century earlier.
 
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Collector
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Re: Quantum Question

September 25th, 2018, 8:49 pm

"just because you're paranoid doesn't mean they're not after your Bra-ket"
 
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Alan
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Re: Quantum Question

September 26th, 2018, 4:09 am

Cuchulainn wrote:
Compare

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Image

Consider the outer product
[$] \phi \otimes \psi[$] vs [$] |\phi><\psi|[$].

'Slightly' more mnemonic to simply put the ket before the bra, than to switch from a parens-with-comma to a cross. Also, it's symbolically clear that this ket-bra expression is an operator (i.e., matrix) that 'wants' to act on a ket to the right (or a bra to the left).

Having said all that, the last time I needed to use the outer product (in my discussion of the spectral theorem for jump-diffusions in last book), I adopted the [$]\otimes[$] notation. After all, unless your audience is just physicists, you're going to use the more standard math notation to talk to the widest possible audience. Plus, the physics people will happily read either style, once it's clearly defined.
 
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rmax
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Re: Quantum Question

September 26th, 2018, 9:07 am

Physicists invent all sorts of notations that are not used by mathematicians. Einstein has a different form of writing tensors than the usual mathematical notation. Bra-ket is just a short hand, and is easy to see the operator that is being applied.

On the Banach space thing, I am sure N was suggesting something along those lines. He didn't believe in HUP I seem to recall and suggested it was physicists not understanding maths. I still miss N after all these years.
 
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Cuchulainn
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Re: Quantum Question

September 26th, 2018, 11:44 am

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katastrofa
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Re: Quantum Question

September 26th, 2018, 3:41 pm

Cuchulainn wrote:
katastrofa wrote:
Cuchulainn, you really starts to display signs of paranoia.

So, neither of yous know the answer, what?

A Hilbert space is just Euclidean space on steroids. LinearAlgebra++

Random thought: why no QM in Banach spaces?

infinitely dimensional Hilbert spaces.

Countably infinite I presume..Nice separable HS.


It's a convenient and compact notation. See eq. 6 in https://www.researchgate.net/publicatio ... miltonians
 
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Cuchulainn
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Re: Quantum Question

September 26th, 2018, 4:09 pm

Alan wrote:
Cuchulainn wrote:
Compare

Image
to 

Image

Consider the outer product
[$] \phi \otimes \psi[$] vs [$] |\phi><\psi|[$].

'Slightly' more mnemonic to simply put the ket before the bra, than to switch from a parens-with-comma to a cross. Also, it's symbolically clear that this ket-bra expression is an operator (i.e., matrix) that 'wants' to act on a ket to the right (or a bra to the left).

Having said all that, the last time I needed to use the outer product (in my discussion of the spectral theorem for jump-diffusions in last book), I adopted the [$]\otimes[$] notation. After all, unless your audience is just physicists, you're going to use the more standard math notation to talk to the widest possible audience. Plus, the physics people will happily read either style, once it's clearly defined.

Which page is it one, Alan?
That traffic sign symbol (tensor product I presume?) can be explained mathematically/incrementally in terms of bilinear functionals on Cartesian products of two spaces.

Let [$]U[$] and [$]V[$] be two vector spaces over the same field [$]K[$] (usually real or complex). Then define [$]L(U,V;K)[$] to be the vector space of all bilinear mappings [$]\alpha: U\times V\rightarrow K[$].

The tensor product of [$]U[$] and [$]V[$] is the dual(*) of  [$]L(U,V;K)[$]. 

(*) The dual of a vector space [$]V[$] is the mapping of each element of [$]V[$] to an element of [$]K[$]. It is denoted by [$]L(V;K)[$] or [$]V{'}[$].

// This way of thinking is similar to C++ template programming.
Last edited by Cuchulainn on September 26th, 2018, 5:30 pm
 
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Cuchulainn
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Re: Quantum Question

September 26th, 2018, 4:57 pm

So, we get a composite definition

[$]U \otimes V = L{'}(U,V;K) = L(L(U,V;K),K)[$]
Last edited by Cuchulainn on September 26th, 2018, 4:59 pm
 
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Alan
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Re: Quantum Question

September 26th, 2018, 5:49 pm

Cuchulainn wrote:
Alan wrote:
Cuchulainn wrote:
Compare

Image
to 

Image

Consider the outer product
[$] \phi \otimes \psi[$] vs [$] |\phi><\psi|[$].

'Slightly' more mnemonic to simply put the ket before the bra, than to switch from a parens-with-comma to a cross. Also, it's symbolically clear that this ket-bra expression is an operator (i.e., matrix) that 'wants' to act on a ket to the right (or a bra to the left).

Having said all that, the last time I needed to use the outer product (in my discussion of the spectral theorem for jump-diffusions in last book), I adopted the [$]\otimes[$] notation. After all, unless your audience is just physicists, you're going to use the more standard math notation to talk to the widest possible audience. Plus, the physics people will happily read either style, once it's clearly defined.

Which page is it one, Alan?
That traffic sign symbol (tensor product I presume?) can be explained mathematically/incrementally in terms of bilinear functionals on Cartesian products of two spaces.

Let [$]U[$] and [$]V[$] be two vector spaces over the same field [$]K[$] (usually real or complex). Then define [$]L(U,V;K)[$] to be the vector space of all bilinear mappings [$]\alpha: U\times V\rightarrow K[$].

The tensor product of [$]U[$] and [$]V[$] is the dual(*) of  [$]L(U,V;K)[$]. 

(*) The dual of a vector space [$]V[$] is the mapping of each element of [$]V[$] to an element of [$]K[$]. It is denoted by [$]L(V;K)[$] or [$]V{'}[$].

// This way of thinking is similar to C++ template programming.

Yes, tensor product aka outer product. Actually in my book (Vol II, pg 44), I call it a Hilbert space outer product and define it in terms of a Hilbert space inner product (see below). 

To elaborate, in my discussion, I consider the Hilbert space  [$]\mathcal{H} = L^2(\Omega) [$]. Then, for the Hilbert space inner product, I write [$] \langle f,g \rangle \equiv \int_{\Omega} \bar{f}(x) g(x) \, dx[$], where the overbar is complex conjugation. With that notation, I then define [$]\otimes[$] by the property that

[$] (a \otimes b) c = \langle b,c \rangle a[$] for any three vectors in [$]\mathcal{H}[$].

For finite-dimensional real-valued vectors, it's just the usual outer product of a column and row vector that produces a matrix, as per the the wikipedia link.
 
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Cuchulainn
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Re: Quantum Question

September 27th, 2018, 7:39 pm

rmax wrote:
Physicists invent all sorts of notations that are not used by mathematicians. Einstein has a different form of writing tensors than the usual mathematical notation. Bra-ket is just a short hand, and is easy to see the operator that is being applied.

On the Banach space thing, I am sure N was suggesting something along those lines. He didn't believe in HUP I seem to recall and suggested it was physicists not understanding maths. I still miss N after all these years.

I did a course on QM as undergrad. I was a bit underwhelmed to be honest because it seemed to centre around (finite-dimensional) matrix algebra. Not a whiff of a Cauchy sequence.
 
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rmax
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Re: Quantum Question

September 28th, 2018, 7:40 am

Cuchulainn wrote:
rmax wrote:
Physicists invent all sorts of notations that are not used by mathematicians. Einstein has a different form of writing tensors than the usual mathematical notation. Bra-ket is just a short hand, and is easy to see the operator that is being applied.

On the Banach space thing, I am sure N was suggesting something along those lines. He didn't believe in HUP I seem to recall and suggested it was physicists not understanding maths. I still miss N after all these years.

I did a course on QM as undergrad. I was a bit underwhelmed to be honest because it seemed to centre around (finite-dimensional) matrix algebra. Not a whiff of a Cauchy sequence.

Right...but QM still gives good results and is probably a significant driver for the technological progress we have had. 
 
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Cuchulainn
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Re: Quantum Question

September 28th, 2018, 9:17 am

Right...but QM still gives good results and is probably a significant driver for the technological progress we have had. 

Of course. I'm not disputing that.It works in practice but it is useful to know how it works in theory.

//
Dirac's intuition was excellent but the maths was missing. Schwartz, Sobolev et al made it mathematically respectable; Sobolev spaces are all over the place these days. In a sense, it is a universal paranoia to move from the specific to the generic.


https://en.wikipedia.org/wiki/Distribution


The Finite Element Method (FEM) witnessed a similar transformation as engineering method to mathematical generalisation (long time ago I did FEM research (prove convergence in Sobolev spaces))  and then industry).

AFAIR it was a physicist JL Synge (father of the late great Cathleen Morawitz Synge) who invented Hypercircle method (protean FEM)..
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