# Quantisation definition

1. Apr 21, 2007

### logearav

2. Apr 21, 2007

### da_willem

Quantization refers to the discreteness of certain quantities (energy levels, angular momentum levels,..) in quantum theories. Mathematically this is obtained by imposing that, x, the position (operator) and. p, momentum (operator) do not commute, i.e.

$$xp-px = [x,p] = i \hbar$$

Or in field theory to quantize a system you impose a similar relation between a field [itex]\psi(\vec{x},t)[/tex] and the (canonical) momentum density [itex]\Pi (\vec{y}, t)[/tex], i.e.

$$[\psi(\vec{x},t) ,\Pi (\vec{y}, t)]= i \hbar \delta ^3 (\vec{x}-\vec{y})$$

3. Apr 24, 2007

### Demystifier

Not really. For example, for a free quantum particle neither position nor energy or momentum takes a discrete spectrum. In fact, the word "quantization" is misleading and is used for historical reasons. Instead of talking about "quantum" theory perhaps it would be more correct to talk about "operator" theory, or "hilbert-space" theory, or "uncertainty" theory, as these expressions better (though not perfectly) express the main point of QM.

4. Apr 30, 2007

### logearav

thanks folks for the definition

5. May 1, 2007

### masudr

Demystifier: While everything you say in your post is correct, I have a slight problem with the following:

There is no such thing as a free quantum particle; and furthermore, (as far as I know) the free planar wave does not exist in the space of square integrable functions.

6. May 1, 2007

### Haelfix

Neither do most wavefunctions you can think off if you really want to be mathematically exact. Thats ok though, people have learned how to deal with that. Technically you have to expand the hilbert space (the trick to pass to its dual) such that various nice mathematical properties linking distributions and hilbert spaces are satisfied. We call this a Gelfland triple.

7. May 1, 2007

### masudr

That sounds neat. But is it not unphysical?

8. May 2, 2007

### Demystifier

First, even if there are no free particles, there are at least particles for which (for practical purposes) the interaction can be neglected.
Second, you can work with wave packets that are square integrable superpositions of plane waves.

9. May 2, 2007

### masudr

I know both of those things. But your example of a free particle having a continuous spectrum still doesn't give any more weight to not calling it "quantum" as it is.

10. May 3, 2007

### jambaugh

In reply to the original question: Quantization is the procedure of taking a classical description of a physical system and generating the analogous quantum description. Hence we "quantize the simple harmonic oscillator" and we "quantize strings" we "quantize the classical electron/proton/...".

There are various schemes, the most often used being called canonical quantization.

Quantizing basically involves promoting the representation of observables from simple variables on sets of numerical values to linear operators for which those values are eigen-values in a linear associative algebra. This recognizes the fact that the values of an observable are obtained through execution of a physical action, the act of observation.

In the process the values of some but not necessarily all physical observables may become descrete, hence the term "quantize".

Now in the literature there is something historically mis-labeled "second quantization" . Second quantization is not technically quantization since one already begins with the quantum mechanical description. Second quantization is better described as quantification.

Quantification is the transition from the description of one object/system to the description of a variable number (including zero). The act of quantification is one of embedding a given quantum (or classical for that matter) system within a larger system. Quantum Field Theory typically takes a classical field, models it as an ensemble of physical systems at each point of space, quantizes those systems in such a way that their energies are discrete, and then treats the excitations as physical particles. This then is a quantification then quantization scheme. One may also begin with the single particle description of the field quantum (e.g. photon) and then quantify.

Here too are schemes from which one may choose:
Bose-Einstein, Fermi-Dirac, Maxwell-Boltzman quantification and some more exotic forms such as Anyonic quantification including a few I invented while a grad student.

The quantification scheme is refered to as the statistics of the particle/field/system being quantified. This is because the major effect, and first physical predictions of alternative quantification schemes occurs in the statistical mechanical description of the many particle system.

Regards,
James Baugh

11. May 15, 2007

### hyp

hello james
pls explain me scattering matrix and t-matrix

12. May 15, 2007

### jambaugh

Here is a rough idea of what the S matrix means. In practice one is usually only determining certain components of it.

Begin with the idea of the Hamiltonian of a quantum system. The Hamiltonian defines the propagation of modes over time. To get the unitary transition matrix we integrate the Hamiltonian Matrix:

$$U(t_2,t_1) = \exp\left( \int_{t_1}^{t_2} H(t)dt\right)$$

This allows us to determine the probability for an observed transition over a finite interval of time:
$$Pr(\psi_1(t_1) \to \psi_2(t_2)) = |\psi^\dag_2 U \psi_1|^2$$

Now for a scattering process we take the limiting cases:
$$t_1 \to -\infty, \quad t_2 \to +\infty$$
And this unitary matrix is the S-matrix.
$$S \equiv U(\infty,-\infty)$$

Given most scattering processes have a very small effect most of the time i.e. given that the probability of two particles colliding is quite small then the S-matrix should be very close to the identity operator. Thus it is convenient to write the S-matrix in the form:
$$S \equiv (\mathbf{1}+ikT)$$

Where k is a small constant and the imaginary factor is there because small variations of a unitary matrix from the identity should be anti-Hermitian and we want the T-matrix to be (roughly) Hermitian (self adjoint, having real eigen-values).

There is of course more detail to this such as how to represent relative phase shifts in taking the limits. The T matrix gives relative scattering probabilities since the absolute probabilities for non-trivial scattering processes generally have quite low values.

This should give you a general picture. To get the practical details take a course in quantum physics or at least find a good text.

Now you can define the S matrix in two ways. Given say an electron scattering off of a nucleus you can write down just the S matrix for the electron. But you can also define a vacuum'' S-matrix which has components for any multi-particle input and any multi-particle output, including e.g. for an electron and nucleus in and electron and nucleus out.

But one can also consider components consisting of an electron positron (or proton anti-proton) inputs and all kinds of strange particle composite outputs.

In principle all the physical laws are encapsulated in the S-matrix. The Feynman diagrams you see represent specific components (or perturbation terms for specific components) of the vacuum S-matrix.

And hence also all of those high-energy collision experiments are essentially explorations of the higher order components of the S-matrix.

Does that help?
Regards,
James

13. May 15, 2007

### hyp

thanks james,
yes, I m now searching about phase shift with t-matriy for hyperon nucleon interaction. but now I got the problem in my code with this s-matrix.if say, I do not know exact error ,what can be?Now I know it can be s-matrix unitary violation.So could u explain me again about unitary violation in s-matrix.thanks in advance

14. May 16, 2007

### jambaugh

There are a couple of issues. The first I'll bring up is purely mathematical.
If you consider a Unitary operator U it is expressible as an exponential of an anti-hermitian operator A which we can express as an imaginary multiple of a Hermitian operator H. Thus:

$$U = e^{i\theta H} = 1 + iH - H^2/2! - iH^3/3! \cdots$$

Thus in the case of the S-matrix we have:
$$S = e^{ik\Theta} = 1 + ik\Theta - k^2\Theta^2/2! \cdots$$
and hence the T-matrix is not actually hermitian:
$$T = (S-1)/ik = \Theta +ik\Theta^2/2! + O(k^2)$$
where $$\Theta$$ is Hermitian.

Rather the T-matrix is only "hermitian to order k" where k again is a small parameter.

Turning this around, if you insist on a Hermitian T-Matrix then the resulting S-matrix is no longer unitary:
$$S^\dag S = (1-ikT)(1+ikT) = 1 - k^2 T^2$$
where unitarity of S would require:
$$S^\dag S = 1$$
Thus assumption of a Hermitian T-matrix yields deviation of the S-matrix from unitarity of the order k.

To preserve unitarity to higher order you would have to introduce some anti-Hermitian component to the T matrix.

Now there are a number of physical and interpretational issues as well. Unitarity is necessary to conserve probabilities which in turn requires a correct distinction between physical and gauge degrees of freedom. Since all finite dimensional unitary groups are compact we cannot represent non-compact relativity transformations (e.g. Lorentz and Poincare transformations) without either
• invoking an infinite dimensional representation group,
• introducing gauge degrees of freedom,
• (as is usual) both.

So in a given gauge extended theory you may find the formal S-matrix fails to be unitary. One however introduces gauge constraints which project the formal S-matrix onto a physical S-matrix acting on a Hilbert sub-space of the gauge-extended (pseudo) Hilbert space. This physical S-matrix needs to be unitary to preserve transition probabilities. If one does not properly deal with the gauge conditions then this may not occur in the theoretical calculations.

An alternative to unitarity conditions is an alternative to the probability interpretation. One may try to formulate a theory wherein we replace normalized transition probabilities with parametric transition rates (which can still be meaningfully applied to single particle transitions). This is done informally and implicitly within Dirac theory when the quantity:
$$j_\mu = \overline{\psi}\gamma_\mu\psi$$
is interpreted as a probability current. (With negative currents interpreted as "electrons traveling backward in time a.k.a. positrons").

In this case then a loss of unitarity may be explained away via a rescaling of the rate parameter (e.g. time-dialations, length contractions, spatial volume or surface renormalization etc). What with worries of how to renormalize divergent calculations in standard QFT it is more critical to get finite answers rather than worry about so trivial a matter as whether the probabilities all add up to 1.

I say "may" here as I've yet to see a full and clear exposition on the subject. In the usual formulations many of these issues get swept under the rug via the use of infinite dimensional representations. Much of my PhD research (and current tentative explorations) investigate manifestly finite dimensional representations wherein "the rug is no longer big enough" for these issues to be "swept under" and ignored. I believe that "forcing them out in the open" is a necessary prerequisite to a sensible quantum theory of gravitation. But this is a topic for another thread so I'll say no more.

Pardon the long winded reply. I hope it was instructive.
Regards,
James Baugh

Last edited: May 16, 2007
15. May 16, 2007

### hyp

James,
yes, thanks very much for ur explaination. I appreciate it.I really need this.You are right my problem is in t-matrix.let say, imiginary part bcos of S=1-ikT.so my t matrix imiginary part is larger than 1.we use the formula for s=exp(ikQ),asin and acos function working only -1 to 1. that's why I got error.And I use formal s-matrix on the Hilbert sub space.that's why I need unitary condition. But I have a little bit confuse about that you said"if one does not properly deal with the gauge conditons then this may not occur in the theoretical calculations".what does it mean. if my question is not suit for under this title,pls tell me what title should I move bcos I m new user.