- #1

- 282

- 12

- I
- Thread starter fxdung
- Start date

- #1

- 282

- 12

- #2

- 11,219

- 3,883

- #3

- 11,219

- 3,883

In non-relativistic QFT it's the same. In relativistic QFT one changes the normalization of ##\phi## so that it becomes Lorentz invariant, the consequence of which is that ##|1x\rangle## differs from ##|x\rangle##. See e.g. myAnd by the way what is the difference between /x> and /1x>=Phi(field operator)(x)/0>?

http://de.arxiv.org/abs/hep-th/0202204

Eqs. (55)-(60). In particular, the function (59) is Lorentz invariant, but is not a ##\delta##-function.

- Last Post

- Replies
- 27

- Views
- 3K

- Replies
- 3

- Views
- 915

- Last Post

- Replies
- 6

- Views
- 627

- Last Post

- Replies
- 61

- Views
- 3K

- Last Post
- Quantum Interpretations and Foundations

- Replies
- 14

- Views
- 1K

- Replies
- 1

- Views
- 2K

- Last Post

- Replies
- 3

- Views
- 2K

- Replies
- 1

- Views
- 1K

- Replies
- 3

- Views
- 1K

- Last Post

- Replies
- 5

- Views
- 2K