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Question about a quantum field.

by cragar
Tags: field, quantum
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cragar
#1
Feb2-13, 09:05 PM
P: 2,466
What do they mean when they say a quantum field has an infinite amount of degrees of freedom? I hardly know anything about QFT so don't go to deep. Does this mean that the field
can move in an infinite amount of ways that are independent of each other.
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The_Duck
#2
Feb2-13, 10:14 PM
P: 864
Quote Quote by cragar View Post
What do they mean when they say a quantum field has an infinite amount of degrees of freedom? I hardly know anything about QFT so don't go to deep. Does this mean that the field
can move in an infinite amount of ways that are independent of each other.
Yes. But this is also true for a classical field.

A point particle moving in three dimensions has three degrees of freedom, because its state can be completely specified by three numbers (its spatial coordinates) and their time derivatives.

Now consider, say, the classical electromagnetic field. To specify the state of the electromagnetic field you need an infinity of numbers: the values of E and B throughout space. So the electromagnetic field has an infinite number of degrees of freedom. This continues to be true when you treat the electromagnetic field quantum mechanically, with quantum field theory.
cragar
#3
Feb2-13, 10:22 PM
P: 2,466
So in QFT you would just quantize the classical fields.

The_Duck
#4
Feb2-13, 10:56 PM
P: 864
Question about a quantum field.

Yes. You find, in the classical field theory of a field ##\phi##, the canonically conjugate momentum ##\pi(x)## to the field at ##\phi(x)## at each point in space x. Then you promote the field ##\phi## and conjugate momentum ##\pi## to operators ##\hat\phi## and ##\hat \pi## and impose the canonical commutation relation ##[\hat\phi(x), \hat\pi(y)] = i \hbar \delta(x - y)##. This is exactly analogous the to the equation ##[\hat{x}_i, \hat{p}_j] = i \hbar \delta_{ij}## for a single point particle in 3 dimensions. The procedure is exactly the same as quantizing a simple quantum mechanical system, except that since we have an infinite number of degrees of freedom we end up with an infinite number of commutation relations, one for each pair of spatial points x and y.
Sonderval
#5
Feb3-13, 05:33 AM
P: 130
Or you could use a different approach and use a wave function (I find this much more intuitive to understand, although it is much more difficult to calculate things this way):
For a point particle, you make the transition to QM by saaying: "The particle does not have a definitive position, but is in a superposition of all possible position states. For each state I have a probability amplitude, which I call the wave function."

Similarly, in QFT, you can say "For each classically possible field configuration, I have a probability amplitude for this configuration, which I call the wave function(al)" Of course this is an ugly beast, mathematically (IT is a function that assigns a number to a field configuration, which is another function (so it is a functional)), but it make many things very easy to understand.
For example, the vacuum state is not just a state where the field is zero (as in classical physics), but it is a superpositon of different field configurations - only the mean value (expectation value) is zero.
Similarly, you can see (with a bit of calculation) that a one-photon state is not a state where the electrical field has some definite value, but it is always a superposition of different field configurations where the expectation value of the electrical field is zero.
cragar
#6
Feb3-13, 09:24 AM
P: 2,466
Interesting, thanks for the responses. What is a good book that I should look at for intro
to QFT?
stevendaryl
#7
Feb3-13, 09:57 AM
Sci Advisor
P: 2,106
What's interesting to me about quantum field theory is the fact that two seemingly very different approaches lead to the same theory in the end.

The two approaches:
    1. Start with a classical field.
    2. Quantize it.
    3. Interpret excitations of the field as "particles".
    1. Start with classical theory of a single particle.
    2. Quantize it to get a wave function evolving according to Schrodinger's equation.
    3. Now, do a many-particle version.

It's strange to me that many-particle quantum mechanics is the same as quantizing a classical field.
Sonderval
#8
Feb3-13, 10:19 AM
P: 130
@cragar
Good books - that has been discussed very oftern hereabouts.
I like Zee, Stone, Ryder, also Srednicki (hope the spelling is right), and especially Hatfield (QFT of point particles and strings) which is the only one that treats all three approaches to QFT (canonical, path integrals and wave functionals).

@stevendaryl
Many-particle QM is not QFT - you have to add relativity to get there, otherwise there's no reason for spin or anti-particles.
stevendaryl
#9
Feb3-13, 10:21 AM
Sci Advisor
P: 2,106
Quote Quote by Sonderval View Post
Many-particle QM is not QFT - you have to add relativity to get there, otherwise there's no reason for spin or anti-particles.
You can have a nonrelativistic quantum field theory.
vanhees71
#10
Feb3-13, 10:38 AM
Sci Advisor
Thanks
P: 2,460
I add Peskin-Schroeder to begin with and Weinberg's books (Quantum Theory of Fields), which latter, I think, are the best books on the subject. I think Zee is more confusing than helping. All the others mentioned are good books.
stevendaryl
#11
Feb3-13, 10:39 AM
Sci Advisor
P: 2,106
Quote Quote by stevendaryl View Post
You can have a nonrelativistic quantum field theory.
Here's a comment from
http://www.fuw.edu.pl/~derezins/scat-ims.pdf

Scattering theory for N-body Schrodinger Hamiltonians is well understood,
thanks to the work of Enss, Sigal, Soer, Graf, the author and
others, see [9] and references therein. It can be interpreted as a rather
special example of a quantum field theory [6] for a class of Hamiltonians
preserving the number of particles.
Sonderval
#12
Feb3-13, 11:01 AM
P: 130
@stevendaryl
Yes, you are right (was thinking of "standard" QFT only).
mpv_plate
#13
Feb5-13, 07:54 AM
P: 87
Quote Quote by The_Duck View Post
...Then you promote the field ##\phi## and conjugate momentum ##\pi## to operators ##\hat\phi## and ##\hat \pi## ...
This might be a very silly question (so I apologize in advance), but how exactly is the field becoming an operator? A field is a function which takes a number and returns another number. An operator usually takes a function and returns another function. So in our case the field operator should take a field and return another field. And if the field is an eigenstate of the field operator, we get the same field multiplied by a constant.

But how does the field operator (function mapping numbers to numbers) operate on a field (also a function mapping numbers to numbers)? The field operator seems to have only been declared to be an operator, but nothing else really changed.

Maybe I just understand something in a really wrong way. Maybe there are no fields that are not operators at the same time, so the field operator does not operate on a field, but something more abstract. But I'm not sure how to grasp that.
Sonderval
#14
Feb5-13, 08:05 AM
P: 130
The field operator operates on a state, not on a field.
A state is not a function - for example, the vacuum state is a superposition of many different field configurations (as explained above). Luckily, you usually do not have to think too much about the meaning of a state since everything can be derived from a vacuum state using raisng/lowering operators.
I think that for the question of how fields become operators, Bob Klaubers qftinfo site is quite helpful. (Although he has non-standard views on many things).
stevendaryl
#15
Feb5-13, 08:11 PM
Sci Advisor
P: 2,106
Quote Quote by mpv_plate View Post
This might be a very silly question (so I apologize in advance), but how exactly is the field becoming an operator?
I'm not sure if this helps, or not, but it's actually the same sort of thing that happens in one-particle quantum mechanics. Let's look at the case of a 1-D harmonic oscillator.

Classically, you have the equations of motion:

[itex]m \dfrac{d^2 x}{dt^2} = -k x[/itex]

or alternatively, an energy equation:

[itex]E = \frac{1}{2m} p^2 + \frac{1}{2} k x^2[/itex]

where [itex]p = m \dfrac{dx}{dt}[/itex]

The quantity [itex]x[/itex] is just a function of time, just a number. But then you take the quantum version of this theory, and there are two different ways to think about it: The Schrodinger way is that instead of [itex]x[/itex] being a definite function of time, there is instead a probability amplitude of the particle being at position [itex]x[/itex] at time [itex]t[/itex]. The second way of thinking about it, the Heisenberg way, which is more closely related to quantum field theory, is that when you quantize the harmonic oscillator, [itex]x[/itex] goes from being simply a number to being an operator that acts on a quantum state.

The same thing happens in quantum field theory. The field [itex]\Phi[/itex] is classically just a number, a function of time (and space). But when you make a quantum version of the field, you can either do it the Schrodinger way, where there is a "wave functional" giving the probability amplitude of the number [itex]\Phi[/itex] having different values, or you can do it the Heisenberg way, where [itex]\Phi[/itex] becomes an operator, rather than a number.

It's actually the case that the 1-D harmonic oscillator can be thought of as a field theory in 1-D spacetime (which means, there is only time, and no spatial dimensions). So in this sense, quantum mechanics is a degenerate case of quantum field theory.
mpv_plate
#16
Feb6-13, 01:30 PM
P: 87
Quote Quote by Sonderval View Post
The field operator operates on a state, not on a field.
A state is not a function - for example, the vacuum state is a superposition of many different field configurations (as explained above). Luckily, you usually do not have to think too much about the meaning of a state since everything can be derived from a vacuum state using raisng/lowering operators.
I think that for the question of how fields become operators, Bob Klaubers qftinfo site is quite helpful. (Although he has non-standard views on many things).
Thank you. States are represented as vectors in a Fock space (if I'm right), so the field operator acts on vectors in the Fock space?

I checked Bob Klaubers web and he is introducing the annihilation and creation operators as coefficients of the Fourier transformation when solving the Klein-Gordon equation. The coefficient commutation relations are said to be the reason why these coefficients are not just numbers but operators. But I need to spend more time studying that, because from the first glance it is not clear to me what it means for a coefficient to be an operator, rather than a number.
mpv_plate
#17
Feb6-13, 01:46 PM
P: 87
Quote Quote by stevendaryl View Post
...when you quantize the harmonic oscillator, [itex]x[/itex] goes from being simply a number to being an operator that acts on a quantum state...

... [itex]\Phi[/itex] becomes an operator, rather than a number.

It's actually the case that the 1-D harmonic oscillator can be thought of as a field theory in 1-D spacetime (which means, there is only time, and no spatial dimensions). So in this sense, quantum mechanics is a degenerate case of quantum field theory.
Thanks for this explanation. When the [itex]x[/itex] operates on a state, the state (vector in a Hilbert space) becomes sort of multiplied by [itex]x[/itex]. In the QFT, is a state (vector in a Fock space) also "multiplied" by the field operator?

I'm actually trying to understand what is mathematically changed in the field [itex]\Phi[/itex], to make it behave like an operator. Because multiplying a state by a field (function that assigns a number to each coordinate) does not seem to do the job of being an operator action.
stevendaryl
#18
Feb8-13, 10:21 AM
Sci Advisor
P: 2,106
Quote Quote by mpv_plate View Post
Thanks for this explanation. When the [itex]x[/itex] operates on a state, the state (vector in a Hilbert space) becomes sort of multiplied by [itex]x[/itex]. In the QFT, is a state (vector in a Fock space) also "multiplied" by the field operator?

I'm actually trying to understand what is mathematically changed in the field [itex]\Phi[/itex], to make it behave like an operator. Because multiplying a state by a field (function that assigns a number to each coordinate) does not seem to do the job of being an operator action.
Okay, here's what I hope is an enlightening discussion about the relationship between quantum field theory and quantum mechanics.


Suppose we have the following weird situation: You have a line of masses (labeled by consecutive integers) connected to each other by springs and also connected to a platform by springs. To simplify the motion (at the cost of a more complicated picture), I'm assuming that the masses are constrained to move vertically by rigid vertical rods running through them.

The energy associated with this system is given by:

[itex]E = \sum_n \lbrace \frac{m}{2} \dot{y_n}^2 + \frac{\lambda}{2} (y_n)^2 + \frac{\lambda}{2} (y_{n+1} - y_n)^2\rbrace[/itex]

where [itex]y_n[/itex] is the height of the [itex]n^{th}[/itex] mass above its equilibrium height, [itex]\dot{y_n}[/itex] is [itex]\dfrac{dy_n}{dt}[/itex], [itex]m[/itex] is the mass of each of the masses, and [itex]\lambda[/itex] is the spring constant.

Now, if we assume that the spring constant is large enough (a very stiff spring) that the masses don't move far from their equilibrium positions, then we can assume that [itex]y_n[/itex] can be approximated by a continuous function [itex]\Phi(x)[/itex] where [itex]x[/itex] measures the horizontal location of the mass. If [itex]L[/itex] is the distance between masses, then the horizontal location of [itex]n^{th}[/itex] mass is [itex]nL[/itex]. Then we can relate [itex]y_n[/itex] to [itex]y_{n+1}[/itex] by

[itex]y_{n+1} = \Phi((n+1)L) = \Phi(nL) + \dfrac{\partial \Phi}{\partial x} \delta x + \ldots = \Phi(nL) + \Phi' (nL) L + \ldots[/itex]

where [itex]\Phi' = \dfrac{\partial \Phi}{\partial x}[/itex]. In terms of [itex]\Phi[/itex], the energy can be written as

[itex]E = \sum_n \lbrace \frac{m}{2}\dot{\Phi}(nL)^2 + \frac{\lambda}{2} \Phi(nL)^2 + \frac{\lambda L^2}{2} \Phi'(nL)^2 \rbrace[/itex]

This discrete sum can be approximated by an integral:
[itex]E = \dfrac{1}{L} \int \lbrace \frac{m}{2}\dot{\Phi}(x)^2 + \frac{\lambda}{2} \Phi(x)^2 + \frac{\lambda L^2}{2} \Phi'(x)^2 \rbrace dx[/itex]

Now, the final bit of trickery is to define new constants:
[itex]\mu = \sqrt{\frac{\lambda}{m}}[/itex]
[itex]c = \sqrt{\frac{\lambda}{m}} L[/itex]

In terms of these constants, the energy for our system of oscillators looks like this:

[itex]E = \dfrac{m}{L} \int \lbrace \frac{1}{2}\dot{\Phi}(x)^2 + \frac{\mu^2}{2} \Phi(x)^2 + \frac{c^2}{2} \Phi'(x)^2 \rbrace dx[/itex]

Mathematically, this is the relativistic hamiltonian for a free real scalar field in a spacetime with only one spatial dimension, although the constant [itex]c[/itex] isn't actually the speed of light, but is the speed that vibrational waves travel through the system of harmonic oscillators.

Now, if you quantize the original system of harmonic oscillators, by replacing each of the masses by a quantum particle obeying Schrodinger's equation, that should be the same (given the approximations) as if you treated [itex]\Phi[/itex] as a quantum field.


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