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June 19, 2009 02:21 PM
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When you take the dot product of two vectors in an ordinary (n-dimensional) Cartesian space, you are measuring the length of the projection of one vector along the direction of the other. This is equivalent to measuring the cosine of the angle between them. If the scalar product is zero, the two vectors are orthogonal. That is, the projection of one vector on the other has zero length.
In an abstract space like a Hilbert space, the scalar product is a measure of the "overlap" of two of the "vectors" in the space. Typically in a problem described by a Hilbert space, we decompose the space into an orthonormal basis, which is a set of functions (vectors in the space), that are orthogonal in the sense that their scalar product (however that's defined for the space) is zero. Given an arbitrary function, one can find it's representation in terms of the orthonormal basis by computing the scalar product with each of the basis functions in turn. Those scalar products are the coefficients in the representation of the function as a sum of terms over the basis.
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http://en.wikipedia.org/wiki/Hilbert_space
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What does the dot product in Hilbert Space suggest?
A dot product is a scalar value. Dot products can be used to determine angles between vectors. What does the scalar value in Hilbert space mean?
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| June 19, 2009 05:25 PM |
In an abstract space like a Hilbert space, the scalar product is a measure of the "overlap" of two of the "vectors" in the space. Typically in a problem described by a Hilbert space, we decompose the space into an orthonormal basis, which is a set of functions (vectors in the space), that are orthogonal in the sense that their scalar product (however that's defined for the space) is zero. Given an arbitrary function, one can find it's representation in terms of the orthonormal basis by computing the scalar product with each of the basis functions in turn. Those scalar products are the coefficients in the representation of the function as a sum of terms over the basis.
Source(s):
http://en.wikipedia.org/wiki/Hilbert_space
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One orthogothanal 4X4 matrix which I could apply a vector to and transform it to a new basis.
It sounds like if you take the Hamiltonian basis and multiple it by a polynomial vector that it will produce a scalar. What does that number mean?
IIRC, the eigenvectors of the Hamiltonian form an orthogonal basis for a Hilbert space. A simple example (and the one most used in introductory QM classes) is the "particle in a box". The time independent solutions to the
Schroedinger equation in the box are sine waves with nodes at the two ends of the box. Any function that satisfies the Hamiltonian--that is, functions that have nodes at the ends of the box--can be decomposed as a series whose terms are the various pure sine wave solutions multiplied by the scalar product of those eigensolutions with the function to be decomposed. Essentially, a superposition of discrete energy states. The square of a particular coefficient is interpreted as the probability of measuring the particle in that particular eigenstate.
This is an incredibly condensed version of two or three chapters of an intro QM textbook. Is it making any sense at all?
You've basically described the superposition characteristic of a quantum qubit equation and suggested that it is stable until observation. The adiabatic approach does not disrupt the superposition to get a result.
QC are so difficult to understand. I think understanding the Hamiltonian helps decode, how a QC is possible. You should claim a page on QC and explain the math and physic approach to computing and N Complete problems.
I've summarized your points and reference this page on my QC page. Post any comments you want me to add to the page.
Quantum computing is something I know almost nothing about, to be totally honest. I can help with understanding the most basic parts of the mathematics and physics, but my knowledge is about a foot wide and an inch deep!
I'll take a look at your page and see if I can make any suggestions that are within my basic knowledge, but I will be off-line for the next week or so, so I can't promise anything soon.