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In words, $|{k}\rangle = \sqrt{2 \omega_k} |\vec k\rangle $ tells us that the three-momentum eigenstates $|\vec k\rangle $ are related to the corresponding four-momentum eigenstates $|{k}\rangle$ by a factor of $\sqrt{2 \omega_k}$. While I understand how to derive this starting from the definitions of the creation and annihilation operators ($a(k)$ vs. $a(\vec k)$ etc.), I'm lacking any real understanding why this relationship between three-momentum eigenstates and four-momentum eigenstates should hold.

(For a reference, see e.g. Eq. 2.64 in http://www.damtp.cam.ac.uk/user/tong/qft/qft.pdf. Moreover in Eq. 2.45 it is demonstrated that $|\vec k\rangle $ is indeed a three-momentum eigenstate.)

jak
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  • You are probably misreading your text, which follows a highly unconventional notation. It defines these states in (2.64) to yield the standard QFT (2.65) by contrast to the obsolete (2.54) of QM. I say "obsolete", because most mainstream QFT texts actually use $|\vec p\rangle$ for what you call $|p\rangle$ and never-ever huff and puff about (2.54). Caveat lector. – Cosmas Zachos Oct 25 '19 at 15:16

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