Question

Has anyone here taken "Quantum Field Theory" Class??? I need help [Reward: Medal and Fan]

Answer 1

What's your question

Answer 2

wait a minute . . .

Answer 3

btw, i'm confused how to write dho (partial differential sysmbol) using equation here??

Answer 4

rho? \[\partial\]

Answer 5

`\partial`

Answer 6

ahh ok

Answer 7

Scalar charged particle is described by lagrangian
\[L = \frac{ 1 }{ 2 } (\partial_\mu \phi^* \partial^\mu \phi -m^2 \phi^* \phi)\] where m is particle mass and \(\phi(x)\) is complex scalar field which satisfies an equal time commutation relation
\[[\phi(x),\pi(y)] = [\phi^*(x),\pi^*(y)] = i\delta^3(\bar{x}-\bar{y})\] where x(y) is canonical momentum field

Answer 8

a) Show that the lagrangian above gives
\[\partial_\mu \partial^\mu \phi + m^2 \phi = 0\]

Answer 9

b) Show that
\[\phi(x) = \int \frac{ d^3k }{ (2\pi)^3} \frac{ 1}{ 2 \omega_k } (a_ke^{-ik.x}+ b^\dagger_{k} e^{ik.x})\]
where k is four-momentum and \(\omega_k\) is particle energy, is solution of \(\partial_\mu \partial^\mu \phi + m^2 \phi = 0\)

Answer 10

@iambatman

Answer 11

c) Show that both \(a_k\) and \(b^\dagger_k\) operator satisfy commutation relation :
\[[a_k,a^\dagger_q] = [b_k,b^{\dagger}_q] = (2\pi)^3 2\omega_k \delta^3 (\bar{k}-\bar{q})\]

Answer 12

d) Calculate the Hamiltonian for the system:
\(H = \int\limits d^3x : H : \) with \(H = \sum_i \pi_i \dot{\phi_i} - L\), where L is lagrangian

Answer 13

@ParthKohli, @Compassionate, @radar, @uri, @geerky42, @ganeshie8, @Abhisar

Answer 14

@iambatman

Answer 15

Sure, so what have you done?

Answer 16

i have finished part a)

Answer 17

have idea @Pompeii00 ?

Answer 18

I've never learned this stuff, but I've read about it on my own before. Have you tried to approach the first part (a)? I'm thinking it's probably just applying the Euler-Lagrange relation:\[\frac{\partial L}{\partial (\partial_\mu \phi)}=\frac{\partial L}{\partial_\mu \phi}\]

Answer 19

i have finished part a),

Answer 20

\[\frac{ \partial L }{ \partial \phi^*(x) } = \frac{ \partial }{ \partial \phi^*(x) }\left( \frac{ 1 }{ 2 } \partial_\mu \phi^* \partial^\mu \phi - \frac{ m^2 }{ 2 } \phi^* \phi \right)\]

Answer 21

then

Answer 22

\[\frac{ \partial L }{ \partial \phi^* (x) } = -\frac{ m^2 }{ 2 } \phi\]

Answer 23

and
\[\frac{ \partial L }{ \partial (\partial_\mu \phi^*(x)) } = \frac{ 1 }{ 2 } \partial^\mu \phi\]

Answer 24

based on euler-lagrange equation
\[\frac{ \partial L }{ \partial \phi^*(x) } - \partial_\mu \frac{ \partial L }{ \partial (\partial_\mu \phi^*(x)) } =0\]
\[-\frac{ m^2 }{ 2 } \phi - \partial_\mu (\frac{ 1 }{ 2 } \partial^\mu \phi ) = 0\]
\[\partial_\mu \partial^\mu \phi + m^2 \phi = 0\]

Answer 25

how about part b, c, and d?? i really need help

Answer 26

I agree. That's a nice derivation. I'm not sure about (b). I notice a complex exponential term, which reminds me of a Fourier Transform.

Answer 27

nice hat

Answer 28

for answer b), please you have to substitute the solution \[\phi \] into your equation:
\[(\partial ^{\mu} \partial _{\nu} -m ^{2})\phi=0\]
so you will get this:
\[k ^{\mu}k _{\mu}-m ^{2}=0\]
Please note that your solution is in the form of Fourier transform, so it is a tipical practice, substitute it into the equation which we attempt to solve in order to find some constraint

Answer 29

so the length of 4-vector k is equal to m

Answer 30

ok.., i'll try

Answer 31

oops...your equation is:
\[(\partial _{\mu} \partial ^{\mu} -m ^{2})=0\]

Answer 32

\[(\partial_\mu \partial^\mu + m^2) \phi(x) \]
\[= (\partial_\mu \partial^\mu + m^2) \left( \int\limits \frac{ d^3k }{ (2\pi)^3 } \frac{ 1 }{ 2 \omega_k } \left( a_k e^{-ik.x} + b^\dagger_k e^{ik.x} \right)\right)\]
\[=\int\limits \frac{ d^3k }{ (2\pi)^3} (\partial_\mu \partial^\mu + m^2) \frac{ 1 }{ 2 \omega-k } (a_k e^{-ik.x} + b_k^\dagger e^ik.x)\]
\[= \int\limits \frac{ d^3k }{ (2\pi)^3 } \frac{ 1 }{ 2 \omega_k } \left( (-k^2 + m^2) a_k e^{-ik.x} + (-k^2+ m^2) b_k^\dagger e^{ik.x} \right)\]

\[a_k = a(k)\]
\[b^\dagger_k = a^\dagger(k)\]
then
\[= \int\limits \frac{ d^3k }{ (2\pi)^3 } \frac{ 1 }{ 2 \omega_k } \left( (-k^2 + m^2) a(k) e^{-ik.x} + (-k^2+ m^2) a^\dagger (k) e^{ik.x} \right)\]

Answer 33

I used another method, namely the tensorial calculus, but the answer is the same!

Answer 34

would you mind writing it here ??

Answer 35

or just take a photo

Answer 36

please, wait a moment....

Answer 37

ok thank you ;)

Answer 38

I'm sorry, I tried with my computer cam, but the image is bad quality!

Answer 39

never mind, I put into integral sign the two derivations successively

Answer 40

@gerryliyana sorry, I have a little question:
the operators a, namely a_k and a_q do commute?

Answer 41

yes...,

Answer 42

\[[a_k,a_q]=0\]

Answer 43

\[[a_k,a_q] = 0 = [a_k^\dagger, a_q^\dagger]\]

Answer 44

\[[a_k, a_q^\dagger] = \delta^3 (k-q)\]

Answer 45

ok! Now I try to insert two solution, namely:
\[\phi (x)=\int\limits \frac{ d ^{3} k}{ (2 \pi)^{3} }\frac{ 1 }{ 2 \omega _{k} }(a _{k}e ^{-ikx}+b _{k}^{*}e ^{ikx})\]
and:
\[\phi (x)=\int\limits \frac{ d ^{3} k}{ (2 \pi)^{3} }\frac{ 1 }{ 2 \omega _{k} }(a _{k}e ^{-iky}+b _{k}^{*}e ^{iky})\]
into your commutation relation

Answer 46

oops... the second solution is \[\pi (x)\]

Answer 47

oops.. in the second solution please read q in place of k

Answer 48

ok.., and then what's the next step??

Answer 49

wait a moment please, I'm trying...

Answer 50

ok ..., ;)

Answer 51

after that substitution, I get, as integrand function, the function below:

Answer 52

\[([a _{k},b _{q}^{*}]e ^{-ikx+iqy}+[b _{k}^{*},a _{q}]e ^{ikx-iqy})\frac{ 1 }{ (2 \pi)^{6} }*\frac{ 1 }{ 4\omega _{k}\omega _{q} }\]

Answer 53

where [,] is the commutator operator

Answer 54

now, we have to perform two integrations, namely \[d ^{3}q,d ^{3}k\]

Answer 55

ok

Answer 56

now, Isince we have:
\[b _{k}^{*}=a _{k}^{*}\]
I use your above commutation relation between a's and between a's and a*'s

Answer 57

the first term can be rewritten as below:
\[(2 \pi)^{3}\delta ^{3}(k-q)e ^{-ikx+iqy}\]

Answer 58

and the second term, can be rwritten as below:
\[(2 \pi)^{3}\delta ^{3}(k-q) e ^{ikx-iqy}\]

Answer 59

Sorry, I think to have made an error sign!

Answer 60

I got:
\[[\phi (x),\pi(y)]=0\]

Answer 61

great!

Answer 62

I try to evaluate this commutator:
\[[\phi (x),\pi ^{*}(y)]\]

Answer 63

ok

Answer 64

I got:
\[[\phi (x),\pi ^{*}(y)]=0\]

Answer 65

that's very strange, I think to have made an error, but I don't found it!

Answer 66

in other words, if I apply the commutation relation, namely:
\[[a _{k},a _{q}^{*}]=\delta ^{3}(k-q)\]
I lways get zero for all commutator between \[\phi (x),\pi(y)\]

Answer 67

and their complex conjugates

Answer 68

@gerryliyana what do you think about that, please?

Answer 69

i absolutely agree with you @Michele_Laino ...,

Answer 70

ok! Now I try to find your commutation relations, starting from your second equation of your problem

Answer 71

i'm ready . . .

Answer 72

I evaluated this commutator:
\[[\phi (x),\pi (y)]\]
and I got these four terms:
\[[a _{k},a _{q}]e ^{-ikx-iqy}\]

Answer 73

\[[a _{k},a _{q}^{*}]e ^{-ikx+iqy}\]

Answer 74

\[[a _{k}^{*},a _{q}]e ^{ikx-iqy}\]

Answer 75

and:
\[[a _{k}^{*},a _{q}^{*}]e ^{ikx+iqy}\]

Answer 76

now, we have to find necessary conditions under which that commutator is equal to:
\[i \delta ^{3}(x-y)\]

Answer 77

I think that those conditions are:
\[[a _{k},a _{q}]=[a _{k}^{*},a _{q}^{*}]=0\]
and:
\[[a _{k},a _{q}^{*}]=[a _{k}^{*},a _{q}]= constant*i \delta ^{3}(k-q)\]
where i is such that:
\[i ^{2}=-1\]

Answer 78

i found this

Answer 79

look at page 9., there is the same problem with ours

Answer 80

at that page there are solutions of some exercises, better it is if we know also texts of those exercise, I think!

Answer 81

have you seen equation 3.41 at that page?

Answer 82

btw thank you for helping me @Michele_Laino . I really appreciate it.

Answer 83

Thank you! @gerryliyana