Ciro Santilli @cirosantilli 34

 Incoming links: Hilbert space

Bra-ket notation  Updated 2024-12-15  +Created 1970-01-01

Ket is just a vector. Though generally in the context of quantum mechanics, this is an infinite dimensional vector in a Hilbert space like

L^{2}

Bra is just the dual vector corresponding to a ket, or in other words projection linear operator, i.e. a linear function which can act on a given vector and returns a single complex number. Also known as... dot product.

For example:

(a x) ∣ y ⟩

(1)

is basically a fancy way of saying:

x \cdot y

(2)

that is: we are taking the projection of

y

along the

x

direction. Note that in the ordinary dot product notation however, we don't differentiate as clearly what is a vector and what is an operator, while the bra-ket notation makes it clear.

The projection operator is completely specified by the vector that we are projecting it on. This is why the bracket notation makes sense.

It also has the merit of clearly differentiating vectors from operators. E.g. it is not very clear in

x \cdot y

that

x

is an operator and

y

is a vector, except due to the relative position to the dot. This is especially bad when we start manipulating operators by themselves without vectors.

This notation is widely used in quantum mechanics because calculating the probability of getting a certain outcome for an experiment is calculated by taking the projection of a state on one an eigenvalue basis vector as explained at: Section "Mathematical formulation of quantum mechanics".

Making the projection operator "look like a thing" (the bra) is nice because we can add and multiply them much like we can for vectors (they also form a vector space), e.g.:

a x + a y

(3)

just means taking the projection along the

x + y

direction.

Ciro Santilli thinks that this notation is a bit over-engineered. Notably the bra's are just vectors, which we should just write as usual with

v

... the bra thing makes it look scarier than it needs to be. And then we should just find a different notation for the projection part.

Maybe Dirac chose it because of the appeal of the women's piece of clothing: bra, in an irresistible call from British humour.

But in any case, alas, we are now stuck with it.

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L^{2}

 Updated 2024-12-15  +Created 1970-01-01

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L^{p}

for

p = = 2

L^{2}

is by far the most important of

L^{p}

because it is quantum mechanics states live, because the total probability of being in any state has to be 1!

L^{2}

has some crucially important properties that other

L^{p}

don't (TODO confirm and make those more precise):

it is the only $L^{p}$ that is Hilbert space because it is the only one where an inner product compatible with the metric can be defined:
- math.stackexchange.com/questions/2005632/l2-is-the-only-hilbert-space-parallelogram-law-and-particular-ft-gt
- www.quora.com/Why-is-L2-a-Hilbert-space-but-not-Lp-or-higher-where-p-2
Fourier basis is complete for $L^{2}$ , which is great for solving differential equation

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Linear map  Updated 2024-12-15  +Created 1970-01-01

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A linear map is a function

f : V_{1} (F) \to V_{2} (F)

where

V_{1} (F)

and

V_{2} (F)

are two vector spaces over underlying fields

F

such that:

\forall v_{1}, v_{2} \in V_{1}, c_{1}, c_{2} \in F f (c_{1} v_{1} + c_{2} v_{2}) = c_{1} f (v_{1}) + c_{2} f (v_{2})

(1)

A common case is

F = R

V_{1} = R_{m}

and

V_{2} = R_{n}

One thing that makes such functions particularly simple is that they can be fully specified by specifyin how they act on all possible combinations of input basis vectors: they are therefore specified by only a finite number of elements of

F

Every linear map in finite dimension can be represented by a matrix, the points of the domain being represented as vectors.

As such, when we say "linear map", we can think of a generalization of matrix multiplication that makes sense in infinite dimensional spaces like Hilbert spaces, since calling such infinite dimensional maps "matrices" is stretching it a bit, since we would need to specify infinitely many rows and columns.

The prototypical building block of infinite dimensional linear map is the derivative. In that case, the vectors being operated upon are functions, which cannot therefore be specified by a finite number of parameters, e.g.

For example, the left side of the time-independent Schrödinger equation is a linear map. And the time-independent Schrödinger equation can be seen as a eigenvalue problem.

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Mathematics  Updated 2024-12-15  +Created 1970-01-01

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The proper precise definition of mathematics can be found at: Section "Formalization of mathematics".

The most beautiful things in mathematics are described at: Section "The beauty of mathematics".

Figure 1.
Study Hilbert spaces desert dilemma meme
. Source. Applies to almost all of mathematics of course. But we don't care, do we!

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Real world applications of the Lebesgue integral  Updated 2024-12-15  +Created 1970-01-01

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In "practice" it is likely "useless", because the functions that it can integrate that Riemann can't are just too funky to appear in practice :-)

Its value is much more indirect and subtle, as in "it serves as a solid basis of quantum mechanics" due to the definition of Hilbert spaces.

Bibliography:

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Schrödinger picture  Updated 2024-12-15  +Created 1970-01-01

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To better understand the discussion below, the best thing to do is to read it in parallel with the simplest possible example: Schrödinger picture example: quantum harmonic oscillator.

The state of a quantum system is a unit vector in a Hilbert space.

"Making a measurement" for an observable means applying a self-adjoint operator to the state, and after a measurement is done:

the state collapses to an eigenvector of the self adjoint operator
the result of the measurement is the eigenvalue of the self adjoint operator
the probability of a given result happening when the spectrum is discrete is proportional to the modulus of the projection on that eigenvector.
For continuous spectra such as that of the position operator in most systems, e.g. Schrödinger equation for a free one dimensional particle, the projection on each individual eigenvalue is zero, i.e. the probability of one absolutely exact position is zero. To get a non-zero result, measurement has to be done on a continuous range of eigenvectors (e.g. for position: "is the particle present between x=0 and x=1?"), and you have to integrate the probability over the projection on a continuous range of eigenvalues.
In such continuous cases, the probability collapses to an uniform distribution on the range after measurement.
The continuous position operator case is well illustrated at: Video "Visualization of Quantum Physics (Quantum Mechanics) by udiprod (2017)"

Those last two rules are also known as the Born rule.

Self adjoint operators are chosen because they have the following key properties:

their eigenvalues form an orthonormal basis
they are diagonalizable

Perhaps the easiest case to understand this for is that of spin, which has only a finite number of eigenvalues. Although it is a shame that fully understanding that requires a relativistic quantum theory such as the Dirac equation.

The next steps are to look at simple 1D bound states such as particle in a box and quantum harmonic oscillator.

This naturally generalizes to Schrödinger equation solution for the hydrogen atom.

The solution to the Schrödinger equation for a free one dimensional particle is a bit harder since the possible energies do not make up a countable set.

This formulation was apparently called more precisely Dirac-von Neumann axioms, but it because so dominant we just call it "the" formulation.

Quantum Field Theory lecture notes by David Tong (2007) mentions that:

if you were to write the wavefunction in quantum field theory, it would be a functional, that is a function of every possible configuration of the field $ϕ$ .

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Uncertainty principle  Updated 2024-12-15  +Created 1970-01-01

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The wave equation contains the entire state of a particle.

From mathematical formulation of quantum mechanics remember that the wave equation is a vector in Hilbert space.

And a single vector can be represented in many different ways in different basis, and two of those ways happen to be the position and the momentum representations.

More importantly, position and momentum are first and foremost operators associated with observables: the position operator and the momentum operator. And both of their eigenvalue sets form a basis of the Hilbert space according to the spectral theorem.

When you represent a wave equation as a function, you have to say what the variable of the function means. And depending on weather you say "it means position" or "it means momentum", the position and momentum operators will be written differently.

This is well shown at: Video "Visualization of Quantum Physics (Quantum Mechanics) by udiprod (2017)".

Furthermore, the position and momentum representations are equivalent: one is the Fourier transform of the other: position and momentum space. Remember that notably we can always take the Fourier transform of a function in

L^{2}

due to Carleson's theorem.

Then the uncertainty principle follows immediately from a general property of the Fourier transform: en.wikipedia.org/w/index.php?title=Fourier_transform&oldid=961707157#Uncertainty_principle

In precise terms, the uncertainty principle talks about the standard deviation of two measures.

We can visualize the uncertainty principle more intuitively by thinking of a wave function that is a real flat top bump function with a flat top in 1D. We can then change the width of the support, but when we do that, the top goes higher to keep probability equal to 1. The momentum is 0 everywhere, except in the edges of the support. Then:

to localize the wave in space at position 0 to reduce the space uncertainty, we have to reduce the support. However, doing so makes the momentum variation on the edges more and more important, as the slope will go up and down faster (higher top, and less x space for descent), leading to a larger variance (note that average momentum is still 0, due to to symmetry of the bump function)
to localize the momentum as much as possible at 0, we can make the support wider and wider. This makes the bumps at the edges smaller and smaller. However, this also obviously delocalises the wave function more and more, increasing the variance of x

Bibliography:

www.youtube.com/watch?v=bIIjIZBKgtI&list=PL54DF0652B30D99A4&index=59 "K2. Heisenberg Uncertainty Relation" by doctorphys (2011)
physics.stackexchange.com/questions/132111/uncertainty-principle-intuition Uncertainty Principle Intuition on Physics Stack Exchange

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