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My question is the following :

I am studying a 1D harmonic oscillator chain.

My classical hamiltonian contains terms such as $U_n$ where $U_n=x_n-x_n^0$, it represents the position away from equilibrium of my $n$'th atom. I call $a$ the spacing of my lattice (so we have $x_n^0=n.a$)

By using discrete Fourier transform, we can write it as : $ U_n=\frac{1}{\sqrt{N}}\sum_{k} U_k e^{-i(k.n.a)} $ ($N$ is the number of atoms on my lattice).

I know how to quantize $U_n$ (I use the correspondence principle, I know that I will pass from $U_n$ to $\widehat{U_n}$ which is an operator acting on Hilbert space such as it is diagonal in the position basis and has $x_n-x_n^0$ eigenvalue for a ket $|x_n>$. Indeed $U_n$ just represents a position "shifted" from the origin so we know how to quantize it.

But how to quantize the right side of the equality ? I mean, I would have an operator : $$\widehat{\frac{1}{\sqrt{N}}\sum_{k} U_k e^{-i(k.n.a)}}$$ (the hat has to go over my whole expression).

How to know if $\frac{1}{\sqrt{N}}$ would become an operator for example ? Indeed $N$ is the number of atoms on my lattice so do I need to quantize it ?

I think the answer here is I just have to quantize $U_k$ but I would like to understand well how I can know what I have to quantize.

[edit] : do we decide to quantize a physical quantity if we want to be more precise in the calculation? For example the $N$ here can be quantized if I want to say that the number of particles on my chain has a certain probability to be equal to a certain value? So to decide to quantize it or not is just a question of "accuracy" of the model?

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The description of your physical system necessarily involves some variables that "encode" the configuration of your system -- define the configuration space. Those are usually called the "generalized coordinates". In your case you can choose $q_n=x_n$ or $q_n=U_n$ or even $q_n={\cal F}[U_n]$ as the set of generalized coordinates, since any of them uniquely define the configuration of your system.

Using the generalized coordinates you then define the dynamics of you system by specifying its Lagrangian ${\cal L}(q_n, \dot{q}_n, t)$. From it you can get the generalized momenta, and derive its Hamiltonian: $$p_n = \frac{\partial {\cal L}}{\partial \dot{q}_n}$$ $${\cal H}(p_n, q_n,t) = \left(\sum_n \dot{q}_n p_n \right) - {\cal L}(q_n, \dot{q}_n, t)$$

Quantization (well, it is actually called "the canonical quantization") is a procedure of taking the generalized coordinates $q_n$ and generalized momenta $p_n$ and saying that they are operators now. With commutation relations: $$[\hat{q}_i, \hat{p}_j] = i\hbar \delta_{i,j}$$

Now lets go through your questions:

Should $N$ be an operator?

No, since, $N$ is not a part of your configuration space.

Do we decide to quantize a physical quantity if we want to be more precised in the calculation ?

N-no... First of all we do not quantize quantities. We quantize the description of system. And we do it when values of $q_n$ and $p_n$ are small enough for $[\hat{q}_n, \hat{p}_n] = i\hbar$ to be something we cannot neglect.

For example the "N" here can be quantized if I want to say that the number of particles on my chain has a certain probability to be equal to a certain value ?

Not really. You can consider $N$ to be a random variable. You can do averaging over it or whatever. That'll be like statistical mechanics. Nothing necessarily quantum there.

So to decide to quantize it or not is just a question of "accuracy" of the model ?

In a sense. Classical description of the world is the approximation when we consider $\hbar$ to be too small for us to have any visible effect. When this approximation is not valid, we must use the "accurate" description of the reality where $q_n$ and $p_n$ are non-commuting operators.

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  • $\begingroup$ The thing that I don't fully understand is the following. I have : $ U_n=\frac{1}{\sqrt{N}}\sum_{k} U_k e^{-i(k.n.a)} $ How can I know the $U_k$ will become an operator here ? Because if I follow what you said, I have to change my generalized coordinates to operators. But here I have a fourier transform so how can I link it with the generalized coordinates ? I have to use the fact that $ U_k=\frac{1}{\sqrt{N}}\sum_{n} U_n e^{+i(k.n.a)} $, so it depends on $x_n$ : it has to be transformed in operator ? $\endgroup$ – StarBucK Sep 27 '16 at 15:54
  • $\begingroup$ Any set of values that uniquely define a configuration of your system can serve as the generalized coordinates. The values of the Fourier transform of $U_n$ uniquely define a configuration of your system. Thus the Fourier transform of $U_n$ can serve as the generalized coordinates. $\endgroup$ – Kostya Sep 27 '16 at 16:00
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I agree with Kostya's answer, but would suggest an alternative way of arriving at your quantiziation. In these types of systems, quantization can be derived from boundary conditions imposed on the system. In this instance, you would take a finite loop of N atoms. This gives the boundary condition

$ U_n = U_{n + N} $

or

$ \frac{1}{\sqrt{N}} \sum_k U_k e^{-kna} = \frac{1}{\sqrt{N}} \sum_k U_k e^{-i(kna + kNa)} $

which is only true when $ kNa = 2m\pi $

implying that

$ k = \frac{2m\pi}{Na} $

where $m$ is an integer. This in turn is your quantization for the system. If you are working with an infinite system, you take $N \to \infty$ and arrive at a continuous k.

You can apply this argument to other boundary conditions as well and it will change the type of discretization for finite systems. However the continuum limit stays the same.

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  • $\begingroup$ Thank you for your answer. In fact I'm very interested by what you said "quantization can be derived from boundary conditions imposed on the system". I never really understood the link between applying boundary conditions (and then we have a set of possible k like you wrote), and the ""quantum quantization"" where we replace the configuration variables "q" and "p" by their corresponding operators. What I mean is that we call both things quantization but for me these two things are very different. Is there a link between applying B.C on wavevectors and changing (q,p) to observables ? $\endgroup$ – StarBucK Sep 27 '16 at 20:18
  • $\begingroup$ Yes, the boundary conditions limit the possible solutions to your equation but can also be introduced, as Kostya pointed out, in your configuration space. You can either start with a continuous variable and end up with a quantized reciprocal variable as in the case of the quantum well where the hard-wall boundaries quantizes the energy, or you can start with a discretized system as above and arrive at either a continuous or discrete reciprocal variable depending on boundary conditions. $\endgroup$ – Greg Petersen Sep 27 '16 at 21:12

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