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I'm reading Timo Weigand notes for Gupta-Bleuler quantization of free EM field.

On page 109, Author has made the following statements.

The Gupta-Bleuler condition for physical state is

$$|\vec{p},\zeta\rangle \in \cal{H}_\textrm{phys} \leftrightarrow p^\mu\zeta_\mu = 0 \tag{4.56}$$

since $p^2 = 0 $ for massless photon, such $\zeta^\mu$ can be decomposed as

$$\zeta^\mu = \zeta^\mu_T + \zeta^\mu_S \tag{4.57}$$

with $\zeta_S = c\cdot p, \zeta^2_S = 0$ and $\vec{\zeta}_T\cdot\vec{p} = 0, \zeta_T^2<0$. So $|\vec{p},\zeta\rangle \in \cal{H}_\textrm{phys}$ can be written as

$$|\vec{p},\zeta\rangle = |\vec{p},\zeta_T\rangle + |\vec{p},\zeta_S\rangle\tag{4.58}$$

where $|\vec{p},\zeta_T\rangle$ describes 2 transverse DOF of positive norm. and $|\vec{p},\zeta_S\rangle$ describes 1 combined timelike and longitudinal DOF of zero norm.

I have following doubts regarding this:

  1. How is such decomposition of $\zeta^\mu$ made?

  2. why do $|\vec{p},\zeta_T\rangle$ have positive norm and $|\vec{p},\zeta_S\rangle$ have zero norm?

  3. Why do the decomposed parts $\zeta^{\mu}_T$ and $\zeta^{\mu}_S$ follow the properties listed above the equation 4.58?

Please Help.

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1 Answer 1

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First, since the photon is massless, without loss of generality, and by use of the $SO(3)$ symmetry, you can take: $$p^{\mu}=|{\vec{p}}|(1,0,0,1)$$ Now there are only two types of independent 4-vectors $\xi^{\mu}$ satisfying $p^{\mu}\xi_{\mu}=0$: ($p^{\mu}\xi_{\mu}=p^{0}\xi^{0}-p^{j}\xi^{j}$)

$$\xi_{1}^{\mu}=\frac{\pm1}{\sqrt{a^2+b^2}}(0,a,b,0), \,\,\,\,\xi_{2\pm}^{\mu}=(1,0,0,1)$$ Any other $\xi^{\mu}$ satisfying $p^{\mu}\xi_{\mu}=0$ will be a linear combination of these 4-vectors. Obviously, $\xi_{1}^{\mu}\equiv\xi_{T}$ and $\xi_{2+}^{\mu}\equiv\frac{1}{c|\vec{p}|}\xi_{S}$. Now it is easy to check that the conditions above (4.58) are met.

To find the norm of state vectors, you can use eq.(4.54) in the notes you mentioned: $$<\vec{q},\xi | \vec{p},\xi'> =-(2\pi)^3 2 |\vec{p}|\delta(\vec{q}-\delta{p}) \xi.\xi'$$ For $|\vec{p},\xi_T>$ norm is positive because $\xi_T.\xi'_T<0$ and for $|\vec{p},\xi_S>$ norm is zero because $\xi_S$ is a null 4-vector.

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  • $\begingroup$ what is $\vec{c}$ here ? $\endgroup$
    – Abhinav
    Oct 14, 2023 at 11:07
  • $\begingroup$ @Abhinav Actually c is the speed of light. My apology! I edited the mistake in the answer. $\endgroup$
    – Navid
    Oct 15, 2023 at 3:41
  • $\begingroup$ thanks @Navid.. $\endgroup$
    – Abhinav
    Oct 15, 2023 at 5:49

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