How is a non-zero overlap of coherent states consistent with their overcompleteness?

Question

I am trying to understand how the overcompleteness of coherent states, $$ I = \int\frac{d\alpha}{\pi}|\alpha\rangle\langle\alpha|, $$ is consistent with a non-vanishing overlap of coherent states, $$ \langle\alpha|\beta\rangle = \exp(-(|\alpha|^2+|\beta|^2 -2\alpha^*\beta)/2). $$ If I consider $$ I^2 = \frac{1}{\pi^2}\left(\int d\alpha\,|\alpha\rangle\langle\alpha|\right)\left(\int d\beta\,|\beta\rangle\langle\beta|\right) $$ $$ =\frac{1}{\pi^2}\int d\alpha d\beta\, |\alpha\rangle\langle\beta| \langle\alpha|\beta\rangle=I. $$ How can this be true if $$ \langle\alpha|\beta\rangle\neq \pi\delta(\alpha-\beta)? $$

Answer 1

1. Remember that $\alpha$ (and $\beta$) in your expressions are complex numbers, so the completeness relation for coherent states should be understood as $$ \mathbb{I}= \frac{1}{\pi} \int\limits_{-\infty}^\infty d ({\rm Re \, \alpha}) \int\limits_{-\infty}^\infty d({\rm Im \, \alpha})\; |\alpha \rangle \langle \alpha |. \tag{1} \label{1} $$

2. There is a sign mistake in your expression for the scalar product $\langle \alpha | \beta \rangle$. It should read instead $$ \langle \alpha | \beta \rangle = e^{-\frac{1}{2}(|\alpha|^2+|\beta|^2-2\alpha^\ast \beta)}. \tag{2} \label{2} $$

3. In view of my first remark, it is questionable what your delta function with complex arguments should even mean.

4. The easiest way in dealing with coherent states is by using their definition $$ |\alpha \rangle := e^{\alpha a^\dagger -\alpha^\ast a} |0\rangle= e^{-|\alpha|^2/2}e^{\alpha a^\dagger} |0\rangle =e^{-|\alpha|^2/2}\sum\limits_{n=0}^\infty \frac{\alpha^n}{\sqrt{n!}} |n \rangle, \tag{3} \label{3} $$ where $a, a^\dagger$ with $[a, a^\dagger]= \mathbb{I}$ are the ladder operators of a harmonic oscillator with groundstate $|0\rangle$ and $a^\dagger a |n\rangle = n | n\rangle$ for $n=0,1,2,\ldots$ with $\langle m | n\rangle=\delta_{mn}$ and $\sum\limits_{n=0}^\infty|n \rangle \langle n | = \mathbb{I}$.

It is now an easy exercise to compute the double integral $$ \int\! d^2 \alpha \, |\alpha \rangle \langle \alpha | = \!\int\limits_{-\infty}^\infty \!\!dx \!\!\int\limits_{-\infty}^\infty \!\!dy \,e^{-x^2} e^{-y^2} \sum\limits_{n=0}^\infty \sum\limits_{m=0}^\infty \frac{(x+iy)^n}{\sqrt{n!}} \frac{(x-iy)^m}{\sqrt{m!}} |n \rangle \langle m |, \tag{4} \label{4} $$ where $\alpha = x+iy$ with $x,y \in \mathbb{R}$. Using polar coordinates $x= r \cos \varphi$, $y= r \sin \varphi$ you should convince yourself that \eqref{1} indeed holds. With the same technique, you should be able to prove \eqref{2} and at the same time demystify your "puzzle".

Answer 2

We have $$ |\alpha\rangle=e^{-\frac{|\alpha|^2}{2}}\sum_{n=0}^\infty\frac{\alpha^n}{\sqrt{n!}}|n\rangle. $$ Thus $$ \langle m|\alpha\rangle=e^{-\frac{|\alpha|^2}{2}}\frac{\alpha^m}{\sqrt{m!}}. $$ This yields $$ I=\int\frac{d^2\alpha}{\pi}e^{-|\alpha|^2}\sum_{m,n}\frac{\alpha^{*m}\alpha^n}{\sqrt{m!}\sqrt{n!}}|n\rangle\langle m|. $$ So, $$ I^2=\int\frac{d^2\alpha}{\pi}\frac{d^2\beta}{\pi} e^{-|\alpha|^2-|\beta|^2} \sum_{m,n}\frac{\alpha^{*m}\alpha^n}{\sqrt{m!}\sqrt{n!}}\frac{\beta^{m}}{\sqrt{m!}} \sum_{m_1}\frac{\beta^{*m_1}}{\sqrt{m_1!}}|n\rangle\langle m_1|. $$ Thus, $$ \langle k_1|I^2|k_2\rangle=\int\frac{d^2\alpha}{\pi}\frac{d^2\beta}{\pi} e^{-|\alpha|^2-|\beta|^2} \sum_{m}\frac{\alpha^{*m}\alpha^{k_1}}{\sqrt{m!}\sqrt{k_1!}}\frac{\beta^{m}}{\sqrt{m!}} \frac{\beta^{*k_2}}{\sqrt{k_2!}} $$ Choose $\alpha=|\alpha|e^{i\theta_1}$ and $\beta=|\beta|e^{i\theta_2}$. Thus, $d^2\alpha=|\alpha|d|\alpha|d\theta_1$ and $d^2\beta=|\beta|d|\beta|d\theta_2$. The integral will yield \begin{align} \langle k_1|I^2|k_2\rangle&=\int\frac{|\alpha|d|\alpha|d\theta_1}{\pi}\frac{|\beta|d|\beta|d\theta_2}{\pi} e^{-|\alpha|^2-|\beta|^2}\sum_m\frac{|\alpha|^{m+k_1}e^{i(k_1-m)\theta_1)}}{\sqrt{m!}\sqrt{k_1!}} \frac{|\beta|^{m+k_2}e^{i(-k_2+m)\theta_2)}}{\sqrt{m!}\sqrt{k_1!}} \\&= 4\int d|\alpha|d|\beta|e^{-|\alpha|^2-|\beta|^2}\frac{|\alpha|^{2k_1+1}|\beta|^{2k_1+1}}{(k_1!)^2}\delta_{k_1k_2}=\delta_{k_1k_2}. \end{align}

Answer 3

After many helpful answers I have managed the solve this integral and hence my question. Here is my answer:

Inserting the definition of the coherent states into the double integral gives $$ \int d\alpha d\beta\, e^{-(|\alpha|^2+|\beta|^2)/2}\sum_{n,m}\frac{\alpha^n(\beta^*)^m}{\sqrt{n!m!}}|n\rangle\langle m|\langle\alpha|\beta\rangle. $$ The matrix elements of this integral, $A_{ij}$, in the Fock basis are $$ A_{ij} =\langle i|\left(\int d\alpha d\beta\, e^{-(|\alpha|^2+|\beta|^2)/2}\sum_{n,m}\frac{\alpha^n(\beta^*)^m}{\sqrt{n!m!}}|n\rangle\langle m|\langle\alpha|\beta\rangle\right)|j\rangle $$ $$ \to\quad A_{ij} =\int d\alpha d\beta\, e^{-(|\alpha|^2+|\beta|^2)/2}\frac{\alpha^i(\beta^*)^j}{\sqrt{i!j!}}\langle\alpha|\beta\rangle. $$ Now, using the overlap of the coherent states, $\langle \alpha |\beta\rangle$, $$ A_{ij} =\int d\alpha d\beta\, e^{-(|\alpha|^2+|\beta|^2)/2}\frac{\alpha^i(\beta^*)^j}{\sqrt{i!j!}}e^{-(|\alpha|^2+|\beta|^2-2\alpha^*\beta)/2} $$ $$ \to\quad A_{ij} =\int d\alpha d\beta\, e^{-(|\alpha|^2+|\beta|^2)+\alpha^*\beta/2}\frac{\alpha^i(\beta^*)^j}{\sqrt{i!j!}}. $$ Next, we expand the $\exp(\alpha^*\beta/2)$ term, $$ A_{ij} =\int d\alpha d\beta\, e^{-(|\alpha|^2+|\beta|^2)}\sum_k\frac{(\alpha^*)^k\beta^k}{k!}\frac{\alpha^i(\beta^*)^j}{\sqrt{i!j!}}. $$ We perform the integration in complex polar coordinates (changing the index $i\to\ell$ to avoid confusion), $$ A_{\ell j} = \int d|\alpha|d|\beta|d\theta d\phi |\alpha||\beta|\, e^{-(|\alpha|^2+|\beta|^2)}\sum_k\frac{|\alpha|^k|\beta|^k}{k!}\frac{|\alpha|^\ell e^{i(\ell-k)\theta}|\beta|^je^{i(k-j)\phi}}{\sqrt{\ell!j!}}. $$ Integrating over the angular components give Kronecker deltas and two factors of $2\pi$, $$ A_{\ell j} = 4\pi^2\int d|\alpha|d|\beta| |\alpha||\beta|\, e^{-(|\alpha|^2+|\beta|^2)}\sum_k\frac{|\alpha|^k|\beta|^k}{k!}\frac{|\alpha|^\ell \delta_{\ell k}|\beta|^j\delta_{kj}}{\sqrt{\ell!j!}}. $$ Collapsing the summation over $k$, $$ A_{\ell j} = 4\pi^2\int d|\alpha|d|\beta|\, e^{-(|\alpha|^2+|\beta|^2)}\frac{|\alpha|^{2\ell+1}|\beta|^{\ell+1}}{\ell!}\frac{|\beta|^j}{\sqrt{\ell!j!}}\delta_{\ell j}. $$ The integral over $|\alpha|$ is a standard Gaussian integral that evaluates to $\ell!/2$, $$ A_{\ell j} = 4\pi^2\int_0^\infty d|\beta|\, e^{-|\beta|^2}\frac{|\beta|^{\ell+j+1}}{2\sqrt{\ell!j!}}\delta_{\ell j}. $$ We need only consider the case $j=\ell$ since the element vanishes otherwise. In this case, $$ A_{\ell \ell} = 2\pi^2\int_0^\infty d|\beta|\, e^{-|\beta|^2}\frac{|\beta|^{2\ell+1}}{\ell!}. $$ Which is the same as the $|\alpha|$ integral and so, $$ A_{\ell \ell}=\pi^2. $$ Therefore, $$ A_{ij} = \pi^2\delta_{ij} $$ So, $I^2$ does indeed equal $I$ and the overcompletness relation is consistant with the non-orthogonality of the coherent states.

Answer 4

Here is a more straightforward way to deal with this integration on the complex plane. Actually, you can avoid all the fussy element-wise manipulations by working with operators instead of state representations. Starting from $$ I^2 = \int_{\Bbb{C}}\frac{\mathrm{d}^2\alpha}{\pi}\, |\alpha\rangle\langle\alpha| \int_{\Bbb{C}}\frac{\mathrm{d}^2\beta}{\pi}\, |\beta\rangle\langle\beta| = \int_{\Bbb{C}}\frac{\mathrm{d}^2\alpha}{\pi} \int_{\Bbb{C}}\frac{\mathrm{d}^2\beta}{\pi}\, e^{-\frac{1}{2}(|\alpha|^2 + |\beta|^2 - 2\alpha^*\beta)} |\alpha\rangle\langle\beta| $$ and using the fact that a coherent state $|\alpha\rangle$ is created from the vacuum with the help of the displacement operator $\hat{D}(\alpha)= e^{\alpha\hat{a}^\dagger - \alpha^*\hat{a}} = e^{\alpha\hat{a}^\dagger} e^{-\alpha^*\hat{a}} e^{-\frac{1}{2}|\alpha|^2}$, hence $|\alpha\rangle = \hat{D}(\alpha)|0\rangle = e^{-\frac{1}{2}|\alpha|^2} e^{\alpha\hat{a}^\dagger} |0\rangle$, we end up with $$ I^2 = \int_{\Bbb{C}}\frac{\mathrm{d}^2\beta}{\pi}\, e^{-\frac{1}{2}|\beta|^2} \int_{\Bbb{C}}\frac{\mathrm{d}^2\alpha}{\pi}\, e^{-|\alpha|^2 + \alpha^*\beta + \alpha\hat{a}^\dagger} |0\rangle\langle\beta|. $$ Note that the second integral is nothing else than a mere two-dimensional gaussian integral. Indeed, one has : $$ \begin{align} J &= \int_{\Bbb{C}}\frac{\mathrm{d}^2\alpha}{\pi}\, e^{-|\alpha|^2 + \alpha^*\beta + \alpha\hat{a}^\dagger} \;\;\text{with}\; \alpha = x+iy \\ &= \frac{1}{\pi} \int_{\Bbb{R}} e^{-x^2 + (\hat{a}^\dagger + \beta)x} \,\mathrm{d}x \int_{\Bbb{R}} e^{-y^2 + i(\hat{a}^\dagger - \beta)y} \,\mathrm{d}y \\ &= e^{\frac{1}{4}(\hat{a}^\dagger + \beta)^2} e^{-\frac{1}{4}(\hat{a}^\dagger - \beta)^2} \\ &= e^{\beta\hat{a}^\dagger} \end{align} $$ This result leads to $$ I^2 = \int_{\Bbb{C}}\frac{\mathrm{d}^2\beta}{\pi}\, e^{-\frac{1}{2}|\beta|^2} e^{\beta\hat{a}^\dagger} |0\rangle\langle\beta| = \int_{\Bbb{C}}\frac{\mathrm{d}^2\beta}{\pi}\, |\beta\rangle\langle\beta| = I $$ in the end. QED