- #1

nrqed

Science Advisor

Homework Helper

Gold Member

- 3,569

- 190

## Main Question or Discussion Point

Consider the SUSY charge

[tex] Q= \int d^3y~ \sigma^\mu \chi~ ~\partial_\mu \phi^\dagger~ [/tex]

The SUSY transformation of fields, let's say of the scalar field, can be found using the commutator

[tex] i [ \epsilon \cdot Q, \phi(x)] = \delta \phi(x) [/tex]

using the equal time commutator

[tex] [\phi(\vec{x},t), \dot{\phi}^\dagger(\vec{y},t) ] = \delta^3(\vec{x} - \vec{y}) [/tex]

Everything works fine

However, I don't see why we need to assume this. The charge is time independent so we should be able to use whatever time we like to calculate the commutator. But if we pick a different time than the time of [tex] \phi(x) [/tex], we get zero for the transformation of the scalar field, which is incorrect.

So why do we need to set the two times equal?

Thanks in advance,

Patrick

[tex] Q= \int d^3y~ \sigma^\mu \chi~ ~\partial_\mu \phi^\dagger~ [/tex]

The SUSY transformation of fields, let's say of the scalar field, can be found using the commutator

[tex] i [ \epsilon \cdot Q, \phi(x)] = \delta \phi(x) [/tex]

using the equal time commutator

[tex] [\phi(\vec{x},t), \dot{\phi}^\dagger(\vec{y},t) ] = \delta^3(\vec{x} - \vec{y}) [/tex]

Everything works fine

**at the condition of assuming that the time in [tex] \phi^\dagger(y) [/tex] is equal to the time in in [tex] \phi(x) [/tex]**.However, I don't see why we need to assume this. The charge is time independent so we should be able to use whatever time we like to calculate the commutator. But if we pick a different time than the time of [tex] \phi(x) [/tex], we get zero for the transformation of the scalar field, which is incorrect.

So why do we need to set the two times equal?

Thanks in advance,

Patrick