Any EM-field in terms of photon

  • #1
olgerm
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I know formula ## p=\frac{h}{λ} ##
  • p is photon momentum
  • h is plankc constant
  • λ is EM-wave wavelenght
but it is only valid for one wave.

How to describe most general EM- field in terms of photons? Is there always discrete number of photons? If EM field is given in terms of
a)EM-vectorfield:
Ex(T;x;y;z) ; Ey(T;x;y;z) ; Ez(T;x;y;z) ; Bx(T;x;y;z) ; By(T;x;y;z) ; Bz(T;x;y;z)
for any x,y,z and one T.

b) potentsials and loren(t?)z gauge


then what are momentums of photons?
 

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  • #2
vanhees71
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The only way to describe "photons" right is quantum electrodynamics, a relativistic quantum field theory. The free electromagnetic field isn't so easily quantized, because of gauge invariance, but one way to do that for the free photon field is to start from classical (Maxwell) electrodynamics and fix the gauge completely. A convenient choice is the socalled radiation gauge, which uses the gauge constraints
$$A^0=\Phi=0, \quad \vec{\nabla} \cdot \vec{A}=0.$$
The vector potential is then decomposed into transverse plane-wave solutions via [corrected in view of #7]
$$\vec{A}(t,\vec{x})=\sum_{\lambda=\pm 1} \int_{\mathbb{R}^3} \frac{\mathrm{d}^3 \vec{k}}{(2 \pi)^{3/2} \sqrt{2 \omega_{\vec{k}}}} \left [A(\vec{k},\lambda) \vec{\epsilon}(\vec{k},\lambda) \exp(-\mathrm{i} \omega_{\vec{k}} t+\mathrm{i} \vec{k} \cdot \vec{x}) + \text{c.c.} \right ].
$$
Here ##\vec{\epsilon}(\vec{k},\vec{\lambda})## are the two transverse, i.e., ##\vec{k} \cdot \vec{\epsilon}(\vec{k},\vec{\lambda})=0##, left- and right-circular polarization vectors (leading to helicity states for the photons) and ##\omega_{\vec{k}}=|\vec{k}|## (setting ##\hbar=c=1## for convenience).

Then you can quantize the theory by assuming the canonical commutation relations for the field, i.e., substituting ##A(\vec{k},\lambda) \rightarrow \hat{a}(\vec{k},\lambda)##. Then the operators fulfill
$$[\hat{a}(\vec{k},\lambda),\hat{a}(\vec{k}',\lambda')]=0, \quad [\hat{a}(\vec{k},\lambda),\hat{a}^{\dagger}(\vec{k}',\lambda')]=\delta^{(3)}(\vec{k}-\vec{k}'),$$
and are thus are annihilation and creation operators for massless particles with momentum ##\vec{k}## and helicity ##\lambda##, because the commutation relations are those of an continuous number of idependent harmonic oscillators.

A convenient complete set of orthonormal basis vectors are the simultaneous eigenvectors of the number operators
$$\hat{N}(\vec{k},\lambda)=\hat{a}^{\dagger}(\vec{k},\lambda) \hat{a}(\vec{k},\lambda).$$
The ground state (aka "the vacuum") is given by the state, where all eigenvalues of these operators ##N(\vec{k},\vec{\lambda})=0##, fulfilling
$$\hat{a}(\vec{k},\lambda)|\Omega \rangle=0.$$
Then any basis state is given by
$$|\{N(\vec{k},\lambda) \} \rangle =\prod_{\vec{k},\lambda} \frac{1}{\sqrt{N(\vec{k},\lambda)!}} \hat{a}^{\dagger N(\vec{k},\lambda)}|\Omega \rangle,$$
where the product runs over any finite set of momenta and helicities and ##N(\vec{k},\lambda) \in \{0,1,2,\ldots \}##. These are then states with a definite photon number, the socalled Fock states.

It's also clear that due to the use of commutation relations for the field operators, photons are bosons, i.e., interchanging the annihilation and creation operators within the product to build the Fock states, doesn't change the state in any way, i.e., the occupation number basis consists of totally symmetrized products of single-photon states.

A general state in this Hilbert space consists of arbitrary superpositions of such Fock states. It can be a state with a definite photon number or some superposition of such states. E.g., the states most close to a classical electromagnetic field (like the field produced by a laser) are the socalled coherent states, where the photon number is not determined but follows a Poisson distribution.

For a very nice introduction to quantum optics, see

Scully, M. O., Zubairy, M. S.: Quantum Optics, Cambridge University Press, 1997
 
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  • #3
olgerm
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Thank you for thorough answer. I am quite good in math, but do not understand many of notations, you have used.

fix the gauge completely. A convenient choice is the so called radiation gauge, which uses the gauge constraints
$$A^0=\Phi=0, \quad \vec{\nabla} \cdot \vec{A}=0.$$
Are both of these equations coulomb gauge constraints?
##\Phi=0## is true only at time T?
##\vec{\nabla} \cdot \vec{A}=0## is true any time?
##\sum_{\lambda=\pm 1}(f(λ))=f(λ)+f(-λ)##?
What does c.c. mean?
What do bracets ( [] ) mean?
What does † mean?
What does a^ mean?

assuming the canonical commutation relations for the field
What does that mean?

Sorry I do not know a lot about QM. Hopefully you can explain these basic things to me.
 
  • #4
f95toli
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Sorry I do not know a lot about QM. Hopefully you can explain these basic things to me.

Then you need to start by learning a bit more about QM. Photons are -unfortunately- very complicated animals, and the full formalism/theory is so complicated that it is not really covered until you get the graduate level QM and even then you will only cover the basics. You are unlikely to ever learn this properly unless you work in quantum optics (and even then only if you do theory).
(for the record: I do NOT understand all of the maths, only the bits and pieces necessary for my own research :sorry:)
 
  • #5
vanhees71
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Thank you for thorough answer. I am quite good in math, but do not understand many of notations, you have used.


Are both of these equations coulomb gauge constraints?
##\Phi=0## is true only at time T?
##\vec{\nabla} \cdot \vec{A}=0## is true any time?
##\sum_{\lambda=\pm 1}(f(λ))=f(λ)+f(-λ)##?
What does c.c. mean?
What do bracets ( [] ) mean?
What does † mean?
What does a^ mean?

What does that mean?

Sorry I do not know a lot about QM. Hopefully you can explain these basic things to me.

Ok, the quantum theory, we cannot treat in terms of forum discussions. You should use quantum theory textbooks to study this. My favorite to start with is

J.J. Sakurai, Modern Quantum Mechanics

Then you need some QFT to understand photons. For that, I recommend the above mentioned book by Scully rather than a usual relativistic QFT book for particle physicists (although QED of course comes originally from that field), because there other aspects are more emphasized than the nature of light itself.

What we can discuss, however is the basics of classical electrodynamics of free fields, which is crucial to understand before you turn to quantum theory anyway.

You start from Maxwell's equations. In the following I'm using Heaviside-Lorentz units with ##c=1##. First we consider only the homogeneous equations,
$$\vec{\nabla} \times \vec{E}+\partial_t \vec{B}=0, \quad \vec{\nabla} \cdot \vec{B}=0.$$
The 2nd equation tells you, using, Helmholtz's decomposition theorem that ##\vec{B}## is a pure solenoidal field, i.e., it is the curl of another vector field, ##\vec{A}##, the vector potential:
$$\vec{B}=\nabla \times \vec{A}.$$
Plugging this into the first equation (Faraday's Law) you get
$$\vec{\nabla} \times (\vec{E}+\partial_t \vec{A})=0,$$
and again from Helmholtz's decomposition theorem (or Poincare's lemma for that matter) you see that there must be a scalar potential such that
$$\vec{E}+\partial_t \vec{A}=-\vec{\nabla} \Phi.$$
Thus we have up to now
$$\vec{E}=-\partial_t \vec{A}-\vec{\nabla} \Phi, \quad \vec{B}=\vec{\nabla} \times \vec{A}.$$
Now you can forget about the homogeneous Maxwell equations for a moment, because writing the fields in terms of the scalar and vector potential, they are automatically fulfilled.

Now let's turn to the inhomogeneous Maxwell equations, i.e.,
$$\vec{\nabla} \times \vec{B}-\partial_t \vec{E}=\vec{j} , \quad \vec{\nabla} \cdot \vec{E}=\rho,$$
where ##\rho## and ##\vec{j}## are the electric-charge and electric-current density. For free fields no charges or currents are present, and we'll set them to 0 in a moment, but first let's plug in the potentials into these equations.
$$\vec{\nabla} \times (\vec{\nabla} \times \vec{A})-\partial_t(-\partial_t \vec{A}-\vec{\nabla} \Phi)=\vec{j}.$$
Using Cartesian coordinates you can simplify them by using the identity ##\vec{\nabla} \times (\vec{\nabla} \times \vec{A})=\vec{\nabla} (\vec{\nabla} \cdot \vec{A})-\Delta \vec{A}##, where the Laplacian is defined as ##\Delta=\vec{\nabla} \cdot \vec{\nabla}##. Using this, the above equation becomes
$$\partial_t^2 \vec{A}-\Delta \vec{A} +\vec{\nabla}(\vec{\nabla} \cdot \vec{A}+\partial_t \Phi)=\vec{j}.$$
Now we can simplify the solution of these equations tremendously by recognizing that the potentials are not uniquely defined by the physically observable fields, ##\vec{E}## and ##\vec{B}##. Indeed you can add the gradient of an arbitrary scalar field, ##\chi##, to the vector potential, without changing ##\vec{B}##, i.e.,
$$\vec{A}'=\vec{A}-\vec{\nabla} \chi$$
is as good as ##\vec{A}## to get ##\vec{B}=\vec{\nabla} \times \vec{A}=\vec{\nabla} \times \vec{A}'##. At the same time we have to make sure that
$$\vec{E}=-\partial_t \vec{A}-\vec{\nabla} \Phi=-\partial_t (\vec{A}'+\vec{\nabla} \chi)-\vec{\nabla \Phi} = -\partial_t \vec{A}' - \vec{\nabla}(\Phi+\partial_t).$$
Thus we set
$$\Phi'=Phi+\partial_t \chi$$
so that we have the electromagnic field in the same way given by the primed as by the unprimed potentials. The transformation
$$\vec{A}'=\vec{A}-\vec{\nabla} \chi, \quad \Phi'=\Phi+\partial_t \chi$$
is called gauge transformation, and what we just derived tells us that electrodynamics is invariant under this gauge transformation for the potential.

Now because of this arbitrariness, we can put an extra condition on the potentials, and it's convenient to use the Lorenz-gauge condition
$$\partial_t \Phi+\vec{\nabla} \cdot \vec{A}=0,$$
because then our above equation for the potential from the one inhomogeneous Maxwell equation simplifies to
$$(\partial_t^2-\Delta) \vec{A}=\vec{j},$$
i.e., the three Cartesian components decouple from each other, and each other fulfills just an inhomogeneous wave equation with the current density as source.

Now we turn to the last equation, Gauss's Law. Plugging in the electric field in terms of the potentials we get
$$\vec{\nabla} \cdot \vec{E} = -\vec{\nabla} \cdot (\partial_t \vec{A}+\vec{\nabla} \Phi)=\rho,$$
but using the Lorenz gauge condition we can as well eliminate ##\vec{A}## from this equation, because we can put
$$\vec{\nabla} \cdot \vec{A}=-\partial_t \Phi,$$
which finally also gives a wave equation for ##\Phi##, decoupled from all components of the vector potential:
$$(\partial_t^2-\Delta)\Phi=\rho.$$
Now we specialize to free electromagnetic fields, i.e., we assume that there are no charges and currents present anywhere, i.e., ##\rho=0## and ##\vec{j}=0##. Then we have
$$(\partial_t^2-\Delta) \Phi=0, \quad (\partial_t^2-\Delta) \vec{A}=0.$$
Now the Lorenz gauge condition does not completely fix the potentials together with these equations, because we can still take a gauge transformation, for which the gauge field ##\chi## itself also satisfies the homogeneous wave equation, i.e., we set
$$\Phi'=\Phi+\partial_t \chi, \quad \vec{A}'=\vec{A}-\vec{\nabla} \chi$$
for a function ##\chi## satisfying
$$(\partial_t^2-\Delta \chi)=0.$$
Then also the new potentials fulfill the Lorenz gauge condition as do the old ones, because we have
$$\partial_t \Phi'+\vec{\nabla} \cdot \vec{A}'=\partial_t \Phi+\vec{\nabla} \cdot \vec{A} + (\partial_t^2-\Delta)\chi=0.$$
Now we can choose this remaining restricted gauge freedom to envoke the additional constraint that
$$\Phi'=\Phi+\partial_t \chi=0.$$
Since we have ##(\partial_t^2-\Delta \Phi)=0##, this is consistent with ##(\partial_t^2-\Delta) \chi=0##, and that's only the case for free fields, i.e., for ##\rho=0##. Thus now we have the simplified equations
$$(\partial_t^2-\Delta) \vec{A}'=0, \quad \vec{\nabla} \cdot \vec{A}'=0,$$
where the latter equation follows from the Lorenz gauge condition for ##(\Phi',\vec{A}')## and our additional condition ##\Phi'=0##. This is called the radiation gauge, and as stressed above, it can be used only for free fields (or for fields with ##\rho=0##). The above derivation shows that the gauge constraints ##\Phi'=0##, ##\vec{\nabla} \cdot \vec{A}=0## must hold for all times.

You are right concerning the sum over ##\lambda=\pm 1##; c.c. means "complex conjugate" and thus ##f+\text{c.c.}=f+f^*## for any complex valued function ##f##. The brackets are commutators of operators, ##\dagger## stands for the hermitean conjugate of a linear operator in Hilbert space, and ##\hat{a}## is such an operator in Hilbert space (see Quantum Mechanics textbooks).
 
  • #6
phyzguy
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Perhaps I can answer your specific notational questions:

Thank you for thorough answer. I am quite good in math, but do not understand many of notations, you have used.
Are both of these equations coulomb gauge constraints?
##\Phi=0## is true only at time T?
##\vec{\nabla} \cdot \vec{A}=0## is true any time?
These gauge constraints are true at all times.
##\sum_{\lambda=\pm 1}(f(λ))=f(λ)+f(-λ)##?
Yes.
What does c.c. mean?
c.c. means 'complex conjugate'. Basically A has to be real, so by adding the expression in the brackets to its complex conjugate, you guarantee the result is real.
What do bracets ( [] ) mean?
The square brackets denote commutator. So, for two operators A and B, [A,B] = AB - BA
What does † mean?
The dagger denotes the Hermitian conjugate of an operator, also called the adjoint or Hermitian adjoint. You can look up its definition. If the operator is written as a square matrix, then the Hermitian conjugate is the conjugate transpose of the matrix.
What does a^ mean?
It denotes that a is an operator.
 
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  • #7
olgerm
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The vector potential is then decomposed into transverse plane-wave solutions via
$$\vec{A}(t,\vec{x})=\sum_{\lambda=\pm 1} \int_{\mathbb{R}^3} \frac{\mathrm{d}^3 \vec{k}}{(2 \pi)^{3/2} \sqrt{2 \omega_{\vec{p}}}} \left [A(\vec{k},\lambda) \vec{\epsilon}(\vec{k},\lambda) \exp(-\mathrm{i} \omega_{\vec{k}} t+\mathrm{i} \vec{k} \cdot \vec{x}) + \text{c.c.} \right ].
$$
Here ##\vec{\epsilon}(\vec{k},\vec{\lambda})## are the two transverse, i.e., ##\vec{k} \cdot \vec{\epsilon}(\vec{k},\vec{\lambda})=0##, left- and right-circular polarization vectors (leading to helicity states for the photons) and ##\omega_{\vec{k}}=|\vec{k}|## (setting ##\hbar=c=1## for convenience).

Do I understand it correctly: You decompose magnetic potential fiels into fictional plane-waves with different frequencies and sum over the plane-waves parameters ##\omega_{\vec{p}}##, ##\omega_{\vec{k}}##, ##A(\vec{k},\lambda)##, ##\vec{\epsilon}(\vec{k},\lambda)##, ##\vec{k}## and ##\vec{x}## ?
 
  • #8
jtbell
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##\sum_{\lambda=\pm 1}(f(λ))=f(λ)+f(-λ)##?
Yes.
Shouldn't ##\sum_{\lambda=\pm 1}(f(λ))=f(+1)+f(-1)##?
 
  • #9
phyzguy
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Shouldn't ##\sum_{\lambda=\pm 1}(f(λ))=f(+1)+f(-1)##?

Yes, of course you're right. Sorry for the confusion.
 
  • #10
vanhees71
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Do I understand it correctly: You decompose magnetic potential fiels into fictional plane-waves with different frequencies and sum over the plane-waves parameters ##\omega_{\vec{p}}##, ##\omega_{\vec{k}}##, ##A(\vec{k},\lambda)##, ##\vec{\epsilon}(\vec{k},\lambda)##, ##\vec{k}## and ##\vec{x}## ?
Sorry, there's a typo. It must be ##\omega_{\vec{k}}=|\vec{k}|## everywhere. I don't know, what you mean bei "fictional". It's the usual Fourier decomposition of the electromagnetic field in plane-wave modes, i.e., momentum-helicity eigenstates. It's not fictional but just a mathematical tool to describe the electromagnetic field. An alternative basis is to use energy-angular-momentum eigenstates (partial waves, multipole expansion).
 
  • #11
Demystifier
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loren(t?)z gauge
It's Lorenz gauge, but most books erroneously write Lorentz gauge.
 
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  • #12
olgerm
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I have been in university meanwhile, but have still not learnt anything related to this question and still do not understand most of the notation.
Is the basic idea, that (3D) Fourier transfourm of must be sum of finite number of diraci-delta functions?
What is the intuitive meaning of creation and anhilation operator? I borrowed the book J.J. Sakurai, Modern Quantum Mechanics, but did not get any answers from it.
 
  • #13
weirdoguy
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Because it belongs to the realm of quantum field theory, which is another step after quantum mechanics. But keep in mind that you can't skip the quantum mechanics part.
 
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  • #14
olgerm
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Because it belongs to the realm of quantum field theory, which is another step after quantum mechanics. But keep in mind that you can't skip the quantum mechanics part.
I just would like to have some basic knowledges about what these notations mean. It is hard to follow any textbook if I do not understand what are physical meanings of operators(creation and anhilation) and "vectors".
 
  • #15
olgerm
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Then you can quantize the theory by assuming the canonical commutation relations for the field, i.e., substituting ##A(\vec{k},\lambda) \rightarrow \hat{a}(\vec{k},\lambda)##. Then the operators fulfill
$$[\hat{a}(\vec{k},\lambda),\hat{a}(\vec{k}',\lambda')]=0, \quad [\hat{a}(\vec{k},\lambda),\hat{a}^{\dagger}(\vec{k}',\lambda')]=\delta^{(3)}(\vec{k}-\vec{k}'),$$
and are thus are annihilation and creation operators for massless particles with momentum ##\vec{k}## and helicity ##\lambda##, because the commutation relations are those of an continuous number of idependent harmonic oscillators.
Can you explain this part in more detail and explain all more complicated words you used. I think I could understand something of it if I just understood the notation. for example what is difference between ##A## and ##\hat{a}##.
 
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  • #16
Demystifier
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Sorry I do not know a lot about QM.
Then you should learn more QM first, and then after a couple of months of serious work you can come back here again.
 
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  • #19
weirdoguy
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Which page and chapter?
You still don't get it. You won't understand quantum field theory without understanding basics of quantum mechanics, and to learn those basics you need to delve through most of the book, not one or two chapters. There is no other way to understand physics.
 
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  • #21
ftr
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I have been in university meanwhile, but have still not learnt anything related to this question and still do not understand most of the notation.
Is the basic idea, that (3D) Fourier transfourm of must be sum of finite number of diraci-delta functions?
What is the intuitive meaning of creation and anhilation operator? I borrowed the book J.J. Sakurai, Modern Quantum Mechanics, but did not get any answers from it.
Can you explain this part in more detail and explain all more complicated words you used. I think I could understand something of it if I just understood the notation. for example what is difference between ##A## and ##\hat{a}##.
I think the teacher in this youtube is very good at explaining the basics if you do not have the patience and in a a hurry to understand the basics.
https://www.youtube.com/user/DrPhysicsAalso there are books that have the titles of introduction to theoretical physics (which you can get from your university library) where you will be introduced to the whole subject in one go. the level of rigor is not as horrifying as in standard textbook on the individual subjects. I feel the older ones are even better for starting physics enthusiasts. Also some popular physics books can give you a good feel of the subject but not for the purpose of understanding. this my general recommendation for people interested in modern physics.
 
  • #22
olgerm
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My problem with using all of these sources is that I cannot understand meaning(theyr relation with physical reality) of mathematical things(operators, brackets) used. In what part of the books are these things explained?
As I understand creation operator with argument ##\vec{k}## acting on a state equal to another state where at time and location specified by ##\vec{k}##will come another photon. And ##\vec{k}## has some limitations because otherwise any EM-field could be compoused that way and everything would be same as non-quantum physics, if there is nothing else than photons in the physical system.
And to find probability the new state I have to multiply bra of the new state with ket of the new state.
Could the EM-field be described by photons without using brackets and operators. For example Sounds something like: every EM-field must be inverse fourier transform of function that is composed of diracdeltafunctions that have limited degrees of freedom? For every delatafunction corresponds one photon.
 
  • #23
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My problem with using all of these sources is that I cannot understand meaning(theyr relation with physical reality) of mathematical things(operators, brackets) used. In what part of the books are these things explained?

Any QM textbook, I recommend Shankar.
 
  • #24
weirdoguy
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Could the EM-field be described by photons without using brackets and operators.

No.
 
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  • #25
olgerm
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Any QM textbook, I recommend Shankar.
Which page of this book explains this specific question?
 
  • #26
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  • #27
weirdoguy
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My problem with using all of these sources is that I cannot understand meaning(theyr relation with physical reality) of mathematical things(operators, brackets) used.

Huh? It's the very basic thing that you'll find in the begining of every quantum mechanics textbook that hermitean operators correspond to measurable quantities and their eigenvalues are the values of that quantity that you can measure in an experiment. The fact that you didn't find that means that you didn't even try to read any textbook. You are wasting your time being reluctant to our advices. There is no other way to learn quantum mechanics and quantum field theory than reading (not glancing through) a proper textbook, period.
 
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  • #28
Demystifier
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Which page of this book explains this specific question?
If he tells you the page, you will find an explanation that contains another concept that you don't understand. If you want to build a house to enjoy the view from the balcony, you must start with building the foundations of the house. The same is with building the understanding of quantum theory. There is no shortcut.
 
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  • #29
vanhees71
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You still don't get it. You won't understand quantum field theory without understanding basics of quantum mechanics, and to learn those basics you need to delve through most of the book, not one or two chapters. There is no other way to understand physics.
That said, maybe you also need some better understanding of classical physics (mechanics and electrodynamics) first. There's no other way to understand physics than to learn it systematically from scratch together with a good deal of math.

For a good understanding of non-relativistic QM you need classical mechanics (including analytical mechanics in the Hamilton version, including Poisson brackets) and some classical electrodynamics (as a first step electrostatics is the minimum in order to understand basic examples like the hydrogen atom). Mathwise you need for sure linear algebra, vector calculus, differential equations, and Fourier transformations.
 
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  • #30
olgerm
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classical electrodynamics (as a first step electrostatics is the minimum in order to understand basic examples like the hydrogen atom). Mathwise you need for sure linear algebra, vector calculus, differential equations, and Fourier transformations.
I know all these things except hamiltonian.
 
  • #31
olgerm
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I think I just understood something, that I wanted to have explained to me 3 years ago: bra and kets include numbers that describe the states that system can be in. What every number means is not specified, but can vary in every model(must be same everywhere in same model). Is it correct?

Maybe it is misconseption, but I have always thought, that QFT describes particles as fields so that probability of particle being a position is determined by the field value in that point. What ever are relations between these fields are should be describeable by equation between those fields. I cant think of any realtion between fields that can be described by hamilonian, but can not be described by (differencial) equations between the fields. Maybe brackets are more useful in systems that have constraint "forces", but for fields only it should be possible to write it without using hamiltonian.
 
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  • #32
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Quantum mechanics isn’t a change of notation, it’s a fundamentally different physical framework from classical mechanics. You don’t have to use the bra-ket notation but you must learn quantum mechanics to gleam any understanding on the subject.

The Hamiltonian is simply the generator of time-translation (evolving systems forwards/backwards in time) and appears in all frameworks of physics whether directly or indirectly. The classical theory of fields that you seem to be talking about can also be formulated in terms of Hamiltonians but this is not done because it is not as elegant or simple as the Lagrangian approach.
 
  • #33
olgerm
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Quantum mechanics isn’t a change of notation, it’s a fundamentally different physical framework from classical mechanics. You don’t have to use the bra-ket notation but you must learn quantum mechanics to gleam any understanding on the subject.
I know. I have been looking for sources that do not use bra-ket notation. Even better if it did not use hamiltonian ether.
 
  • #34
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While I question this approach, I guess you could learn QM from the path integral approach first. However it doesn’t seem apparent to me why you shouldn’t learn QM the way everyone else does, since it’s that way for good reason.
 
  • #35
olgerm
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why you shouldn’t learn QM the way everyone else does, since it’s that way for good reason.
It is not that I do want to learn it in specific way, but that I cant understand the sources that use brakets, operators, and commutivity. I have wanted someone to explain me what these mean or give me clear refence to look it up. I now finally understood something(post 31), but still not enougth.
I dont think any of my learning has been hinder by insufficient knowledge in QM, but by not understanding meaning of brakets, operators, and commutivity.

Is there really something about quantum field that can not be expressed without hamiltonian and brakets, but only by fields and differencial equations?
 
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