Recent content by geoduck
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Graduate Improved energy-momentum tensor changing dilation operator
I'm trying to show that \int d^3x \,x^\mu \left(\partial_\mu \partial_0-g_{\mu 0} \partial^2 \right)\phi^2(x)=0 . This term represents an addition to a component of the energy-momentum tensor \theta_{\mu 0} of a scalar field and I want to show that this does not change the dilation operator...- geoduck
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- Dilation Energy-momentum Energy-momentum tensor Operator Tensor
- Replies: 2
- Forum: Quantum Physics
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Graduate Continuing to Euclidean Space Justified in Path Integral?
It seems to me that in a path integral, since you are integrating over all field configurations, that going into Euclidean space is not valid because some field configurations will give poles in the integrand of your action, and when the integrand has poles you can't make the rotations required...- geoduck
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- Euclidean Euclidean space Integral Path Path integral Space
- Replies: 2
- Forum: Quantum Physics
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Graduate What are matrix elements in QFT?
Without interaction, then in an expression like \langle p |\phi^4(x)|p\rangle , the bra and the ket are the same state: <p|p>=1. However, with interaction, <p|p>≠1, since the bra is an Out-state, and the ket is an In-state. It seems if you calculate \langle p |\phi^4(x)|p\rangle with...- geoduck
- Post #3
- Forum: Quantum Physics
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Graduate What are matrix elements in QFT?
Suppose you want the 1-particle matrix elements of an operator in QFT, e.g. \langle p' |\phi^4(x)|p\rangle It seems you would calculate this perturbatively by first Fourier transforming the x-variable to q, assuming an incoming particle with momentum p, an outgoing particle with momentum p'...- geoduck
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- Elements Matrix Qft
- Replies: 3
- Forum: Quantum Physics
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Graduate Large n-pt functions renormalized by small n-pt functions?
Suppose you have a λφ4 theory. Books only seem to calculate counter-terms for 2-pt and 4-pt functions. But what about 3 particles scattering into 3 particles? Do the counter-terms determined by renormalizing the 2-pt and 4-pt functions cancel divergences in 3x3 scattering? For example, take...- geoduck
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- Functions
- Replies: 1
- Forum: Quantum Physics
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Graduate Independence of Complex Fields?
Are a field and its complex conjugate independent? It seems like they're not, as one is the complex conjugate of the other, so if you have one, you know the other. However, it seems in path integrals, you integrate over the field and its conjugate, so they can take on values that are not the...- geoduck
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- Complex Fields Independence
- Replies: 1
- Forum: Quantum Physics
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Graduate Multiple Scattering in High Energy Physics Experiments
In a high energy physics experiment, do you ever scatter more than 2 particles at a time? Suppose you scatter 4 incoming particles and get 4 outgoing particles. Do you only look at connected diagrams with 8 external lines? Or do you also have to take into account the product of disconnected...- geoduck
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- Multiple Scattering
- Replies: 1
- Forum: High Energy, Nuclear, Particle Physics
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Graduate How to Determine Group from Commutation Relations?
Can Jacobi Identity and Leibniz derivation alone tell you what group it is? For example, take the SO(3) commutation relations. If you change [Jx,Jy]=iJz to [Jx,Jy]=-iJz and leave all other commutators the same, then I think you get something like SO(2,1) rather than SO(3). SO(2,1) would still...- geoduck
- Post #3
- Forum: Quantum Physics
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Graduate How to Determine Group from Commutation Relations?
Is there a way to determine the group from the commutation relations? For example, the commutation relations: [J_x,J_y]=i\sqrt{2} J_z [J_y,J_z]=\frac{i}{\sqrt{2}} J_x [J_z,J_x]=i\sqrt{2} J_y is actually SO(3), as can be seen by redefining J'_x =\frac{1}{\sqrt{2}} J_x : then J'_x , J_y and...- geoduck
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- Commutation Group Relations
- Replies: 2
- Forum: Quantum Physics
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Graduate What Defines a Local Operator in Position Space?
Thanks. If I have a correlation function, isn't it usually infinity even if all fields in the correlator are at different spacetime points? The exception seems to be a free field theory where everything is finite: therefore it seems infinitys come from structure of hamiltonian, and not the...- geoduck
- Post #3
- Forum: Quantum Physics
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Graduate What Defines a Local Operator in Position Space?
Is it okay to define a local operator as an operator whose matrix elements in position space is a finite sum of delta functions and derivatives of delta functions with constant coefficients? Suppose your operator is M, and the matrix element between two position states is <x|M|y>=M(x,y). It...- geoduck
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- Functions Local Operators
- Replies: 3
- Forum: Quantum Physics
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Graduate Counterterms in saddle point expansion
Don't counter-terms have one more power of the coupling in them than bare terms? For example, for λΦ4, I worked out the the coupling has dimensions [\lambda]=\frac{1}{\hbar c}. So if we write \lambda=k*\frac{1}{\hbar c}, where k is a pure number, then whether your higher order perturbative...- geoduck
- Post #4
- Forum: Quantum Physics
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Graduate Counterterms in saddle point expansion
In the saddle point evaluation of the path integral, at tree level, you plug in the classical solution of the field into the integrand. However, when determining the classical solution, we ignore counterterms. The counterterms only show up to renormalize divergences after a saddle point...- geoduck
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- Expansion Point
- Replies: 2
- Forum: Quantum Physics
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Graduate QFT & QM: Spinors, NRQM & Dirac Field Explained
Suppose you have a quantized Dirac theory. Can you get away with just describing the electron's degree of freedom by ignoring the creation and annihilation operators for the positron in your free fields (which are used in the interaction picture)? Basically set the positron creation and...- geoduck
- Post #22
- Forum: Quantum Physics
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Graduate Why does i-epsilon prescription not cure spin 1 propagator
Within the path integral framework, the reason you have to use Fadeev-Popov quantization for spin 1 is because the matrix of the coefficients of the quadratic part of the free-Lagrangian is non-invertible. But doesn't an i\epsilon prescription take care of that? The same thing happens with the...- geoduck
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- Propagator Spin
- Replies: 1
- Forum: Quantum Physics