- #1
Marioweee
- 18
- 5
- Homework Statement
- We can define a coherent state of scalar fields as ## \ket{\eta_k}=Ne^{A}\ket{0}## with ##A=\int \dfrac{d^3 k}{(2\pi)^3 \sqrt{2E_K}}\eta_k a^{\dagger}_{k} ## where the ##a_k## are the destruction operators of a real ##\phi## scalar field and ## \eta_k## functions of momentum k with sufficiently convergent behavior at infinity.
a) Compute N such that ##\braket{\eta_k | \eta_k}=1##.
- Relevant Equations
- ##[a_p, a_{q}^{\dagger}]=(2\pi)^3 \delta^{(3)}(\vec{p}-\vec{q})##
##e^{X}=\sum_{i}^{\infty}\dfrac{X^i}{i!} ##
What I have done is the following:
\begin{equation}
\braket{\eta_k | \eta_k}=|N|^2\sum_{n=0}^{\infty}\dfrac{1}{n!}\bra{0}(A^{\dagger})^nA^n\ket{0}=|N|^2\sum_{n=0}^{\infty}\dfrac{1}{n!}\int \dfrac{d^{3n}kd^{3n}p}{(2\pi)^{3n}(2\pi)^{3n}\sqrt{(2E_k)^{n}}\sqrt{(2E_p)^{n}}}\eta^{n}_{k}\eta_{p}^{n}\bra{0}(a_{k}a^{\dagger}_{p})^{n}\ket{0}
\end{equation}
We know that ##\bra{0}a_{k}a^{\dagger}_{p}\ket{0}=\bra{0}[a_{k},a^{\dagger}_{p}]\ket{0}+\bra{0}a^{\dagger}_{p}a_{k}\ket{0}=(2\pi)^3 \delta^{(3)}(\vec{p}-\vec{k})\bra{0}\ket{0}=(2\pi)^3 \delta^{(3)}(\vec{p}-\vec{k})## since the action of the destruction operator on the void returns us a zero. But we want to calculate ## \bra{0}(a_{k}a^{\dagger}_{p})^n\ket{0}## , however, this is nothing more than ## \bra{0}(a_{k}a^{\dagger}_{p})(a_{k}a^{\dagger}_{p})(a_{k}a^{\dagger}_{p})...\ket{0}=(2\pi)^{3n} \delta^{(3n)}(\vec{p}-\vec{k})##
Therefore
\begin{equation}
\braket{\eta_k | \eta_k}=|N|^2\sum_{n=0}^{\infty}\dfrac{1}{n!}\int \dfrac{d^{3n}kd^{3n}p}{(2\pi)^{3n}(2\pi)^{3n}\sqrt{(2E_k)^{n}}\sqrt{(2E_p)^{n}}}\eta^{n}_{k}\eta_{p}^{n}\bra{0}(a_{k}a^{\dagger}_{p})^{n}\ket{0}=|N|^2\sum_{n=0}^{\infty}\dfrac{1}{n!}\int \dfrac{d^{3n}kd^{3n}p}{(2\pi)^{3n}(2\pi)^{3n}\sqrt{(2E_k)^{n}}\sqrt{(2E_p)^{n}}}\eta^{n}_{k}\eta_{p}^{n}(2\pi)^{3n} \delta^{(3n)}(\vec{p}-\vec{k})=|N|^2\sum_{n=0}^{\infty}\dfrac{1}{n!}\int \dfrac{d^{3n}k}{(2\pi)^{3n}(2E_k)^{n}}\eta^{2n}_{k}=|N|^2\sum_{n=0}^{\infty}\dfrac{1}{n!}\left( \int \dfrac{d^{3}k}{(2\pi)^{3}(2E_k)}\eta^{2}_{k}\right)^n=|N|^2e^{\int \dfrac{d^{3}k}{(2\pi)^{3}(2E_k)}\eta^{2}_{k}}=1 \rightarrow N=e^{-\dfrac{1}{2}\int \dfrac{d^{3}k}{(2\pi)^{3}(2E_k)}\eta^{2}_{k}}
\end{equation}
Would this result be correct? If so, I have a doubt, since I have tried to make it more general since in principle we would need two sums for each exponential. I mean
\begin{equation}
\braket{\eta_k | \eta_k}=|N|^2\sum_{n=0}^{\infty}\sum_{m=0}^{\infty}\dfrac{1}{n!}\dfrac{1}{m!}\bra{0}(A^{\dagger})^nA^m\ket{0}=|N|^2\sum_{n=0}^{\infty}\sum_{m=0}^{\infty}\dfrac{1}{n!}\dfrac{1}{m!}\int \dfrac{d^{3n}kd^{3m}p}{(2\pi)^{3n}(2\pi)^{3m}\sqrt{(2E_k)^{n}}\sqrt{(2E_p)^{m}}}\eta^{n}_{k}\eta_{p}^{m}\bra{0}a_{k}^{n}(a^{\dagger}_{p})^{m}\ket{0}
\end{equation}
It is clear that for ##\bra{0}(a_{k}^{n}a^{\dagger}_{p})^{m}\ket{0}\neq 0 ## m and n must be equal and must appear in the equation ## \delta_{m,n}##. Now, if m is equal to n, we do the steps above. However, I find a problem with the factorial of n that appears twice, which does not allow to have the expression of the exponential. What am I doing wrong?
Thank you very much for any kind of help.
\begin{equation}
\braket{\eta_k | \eta_k}=|N|^2\sum_{n=0}^{\infty}\dfrac{1}{n!}\bra{0}(A^{\dagger})^nA^n\ket{0}=|N|^2\sum_{n=0}^{\infty}\dfrac{1}{n!}\int \dfrac{d^{3n}kd^{3n}p}{(2\pi)^{3n}(2\pi)^{3n}\sqrt{(2E_k)^{n}}\sqrt{(2E_p)^{n}}}\eta^{n}_{k}\eta_{p}^{n}\bra{0}(a_{k}a^{\dagger}_{p})^{n}\ket{0}
\end{equation}
We know that ##\bra{0}a_{k}a^{\dagger}_{p}\ket{0}=\bra{0}[a_{k},a^{\dagger}_{p}]\ket{0}+\bra{0}a^{\dagger}_{p}a_{k}\ket{0}=(2\pi)^3 \delta^{(3)}(\vec{p}-\vec{k})\bra{0}\ket{0}=(2\pi)^3 \delta^{(3)}(\vec{p}-\vec{k})## since the action of the destruction operator on the void returns us a zero. But we want to calculate ## \bra{0}(a_{k}a^{\dagger}_{p})^n\ket{0}## , however, this is nothing more than ## \bra{0}(a_{k}a^{\dagger}_{p})(a_{k}a^{\dagger}_{p})(a_{k}a^{\dagger}_{p})...\ket{0}=(2\pi)^{3n} \delta^{(3n)}(\vec{p}-\vec{k})##
Therefore
\begin{equation}
\braket{\eta_k | \eta_k}=|N|^2\sum_{n=0}^{\infty}\dfrac{1}{n!}\int \dfrac{d^{3n}kd^{3n}p}{(2\pi)^{3n}(2\pi)^{3n}\sqrt{(2E_k)^{n}}\sqrt{(2E_p)^{n}}}\eta^{n}_{k}\eta_{p}^{n}\bra{0}(a_{k}a^{\dagger}_{p})^{n}\ket{0}=|N|^2\sum_{n=0}^{\infty}\dfrac{1}{n!}\int \dfrac{d^{3n}kd^{3n}p}{(2\pi)^{3n}(2\pi)^{3n}\sqrt{(2E_k)^{n}}\sqrt{(2E_p)^{n}}}\eta^{n}_{k}\eta_{p}^{n}(2\pi)^{3n} \delta^{(3n)}(\vec{p}-\vec{k})=|N|^2\sum_{n=0}^{\infty}\dfrac{1}{n!}\int \dfrac{d^{3n}k}{(2\pi)^{3n}(2E_k)^{n}}\eta^{2n}_{k}=|N|^2\sum_{n=0}^{\infty}\dfrac{1}{n!}\left( \int \dfrac{d^{3}k}{(2\pi)^{3}(2E_k)}\eta^{2}_{k}\right)^n=|N|^2e^{\int \dfrac{d^{3}k}{(2\pi)^{3}(2E_k)}\eta^{2}_{k}}=1 \rightarrow N=e^{-\dfrac{1}{2}\int \dfrac{d^{3}k}{(2\pi)^{3}(2E_k)}\eta^{2}_{k}}
\end{equation}
Would this result be correct? If so, I have a doubt, since I have tried to make it more general since in principle we would need two sums for each exponential. I mean
\begin{equation}
\braket{\eta_k | \eta_k}=|N|^2\sum_{n=0}^{\infty}\sum_{m=0}^{\infty}\dfrac{1}{n!}\dfrac{1}{m!}\bra{0}(A^{\dagger})^nA^m\ket{0}=|N|^2\sum_{n=0}^{\infty}\sum_{m=0}^{\infty}\dfrac{1}{n!}\dfrac{1}{m!}\int \dfrac{d^{3n}kd^{3m}p}{(2\pi)^{3n}(2\pi)^{3m}\sqrt{(2E_k)^{n}}\sqrt{(2E_p)^{m}}}\eta^{n}_{k}\eta_{p}^{m}\bra{0}a_{k}^{n}(a^{\dagger}_{p})^{m}\ket{0}
\end{equation}
It is clear that for ##\bra{0}(a_{k}^{n}a^{\dagger}_{p})^{m}\ket{0}\neq 0 ## m and n must be equal and must appear in the equation ## \delta_{m,n}##. Now, if m is equal to n, we do the steps above. However, I find a problem with the factorial of n that appears twice, which does not allow to have the expression of the exponential. What am I doing wrong?
Thank you very much for any kind of help.
Last edited: