# Coherent state of second quantized hamiltonian

In my preparation for the exam I tried to solve the exercise 2.4 in Coleman's Introduction to Many-Body Physics. I like diagonalizing Hamiltonian, so I picked this problem. Also to learn more about coherent states, which we just accidentally discussed.

In this problem one has the bosonic Hamiltonian \begin{align} H = \omega \left( a ^ { \dagger } a + \frac { 1 } { 2 } \right) + \frac { 1 } { 2 } \Delta \left( a ^ { \dagger } a ^ { \dagger } + a a \right) \end{align} and transforms it with the Bogoliubov transformation \begin{align} \begin{aligned} b & = u a + v a ^ { \dagger } \\ b ^ { \dagger } & = u a ^ { \dagger } + v a \end{aligned} \end{align} to \begin{align} H = \tilde { \omega } \left( b ^ { \dagger } b + \frac { 1 } { 2 } \right) \end{align} This is done by $$\tilde \omega = \frac{1}{2uv} \Delta$$. Now the coherent state comes into play:

The Hamiltonian has a boson pairing term. Show that the ground state of $$H$$ can be written as a coherent condensate of paired bosons, given by \begin{align} | \tilde { 0 } \rangle = e ^ { - \alpha \left( a ^ { \dagger } a ^ { \dagger } \right) } | 0 \rangle \end{align} Calculate the value of $$\alpha$$ in terms of $$u$$ and $$v$$. (Hint: $$| \tilde { 0 } \rangle$$ is the vacuum for $$b$$, i.e. $$b | \tilde { 0 } \rangle = \left( u a + v a ^ { \dagger } \right) | \tilde { 0 } \rangle = 0$$. Calculate the commutator of $$\left[ a , e ^ { - \alpha a ^ { \dagger } a ^ { \dagger } } \right]$$ by expanding the exponential as a power series. Find a value of $$\alpha$$ that guarantees that $$b$$ annihilates the vacuum $$| \tilde { 0 } \rangle$$.)

1. Why is this a coherent state? - I don't know much about coherent states, but basically it means that $$\hat { a } | \alpha \rangle = \alpha | \alpha \rangle$$. So I don't see this condition being fulfilled. They state it's the ground state because $$b$$ annihilates the vacuum and simultaneously it should be the coherent state. But Wikipedia says:

Physically, this formula ($$\hat { a } | \alpha \rangle = \alpha | \alpha \rangle$$) means that a coherent state remains unchanged by the annihilation of field excitation or, say, a particle.

As far as I understood this, the coherent state shouldn't be zero after application of $$b$$.

1. How to calculate the commutator? - I tried the following: $$\left[ a , \mathrm { e } ^ { - \alpha a ^ { \dagger } a ^ { \dagger } } \right] = \left[ a , \sum _ { k = 0 } ^ { \infty } \frac { \left( - \alpha a ^ { \dagger } a ^ { \dagger k } \right) } { k ! } \right] = \left[ a , - \alpha a ^ { \dagger } a ^ { \dagger } + \frac { \alpha ^ { 2 } } { 2 } \left( a ^ { \dagger } a ^ { \dagger } \right) ^ { 2 } + \ldots \right] = - 2 \alpha a ^ { \dagger } - 2 \alpha ^ { 2 } \left( a ^ { \dagger } \right) ^ { 3 } - \ldots$$ After this I cannot compute the next element in the sum and didn't know how to continue.
• You do know how to calculate the next term in the sum, by $[a,f(a^\dagger)]=f'(a^\dagger)$, no? For the rest, see this answer. – Cosmas Zachos Feb 6 '19 at 21:35
• @CosmasZachos: Yes this is true. $b | \tilde { 0 } \rangle = u a \mathrm { e } ^ { - \alpha a ^ { \dagger } a ^ { \dagger } } | 0 \rangle + v a ^ { \dagger } \mathrm { e } ^ { - \alpha a ^ { \dagger } a ^ { \dagger } } | 0 \rangle = \left[ a , e ^ { - \alpha a ^ { \dagger } a ^ { \dagger } } \right] + \mathrm { e } ^ { - \alpha a ^ { \dagger } a ^ { \dagger } } | 0 \rangle - \frac { v } { 2 \alpha } \left[ a , e ^ { - \alpha a ^ { \dagger } a ^ { \dagger } } \right] | 0 \rangle$ .... – Leviathan Feb 7 '19 at 18:02
• \begin{aligned} b | \tilde{0} \rangle & = \left[ a , e ^ { - \alpha a ^ { \dagger } a ^ { \dagger } } \right] \left( u - \frac { v } { 2 \alpha } \right) | 0 \rangle \\ \Rightarrow \alpha & = \frac { u } { 2 v } \end{aligned} – Leviathan Feb 7 '19 at 18:03
• Right. You can do it, after all. Except in the bottom line you flipped u with v. – Cosmas Zachos Feb 7 '19 at 19:17
• Yes, thanks for that. It was too late to change it. Also in $$+ \mathrm { e } ^ { - \alpha a ^ { \dagger } a ^ { \dagger } } \underline{a} | 0 \rangle -$$ is missing. – Leviathan Feb 7 '19 at 19:55

## 1 Answer

A coherent state (in the Perelomov sense) is a displaced ground state. Here, the operators $$K_0=\frac{1}{2}\left(a^\dagger a+a a^\dagger\right)\, , K_+=a^\dagger a^\dagger\, ,\qquad K_-=a\,a$$ span an $$\mathfrak{su}(1,1)$$ algebra. The displacement operator for this is an $$SU(1,1)$$ transformation, which can be normal-ordered to give the (unnormalized) states $$\vert\xi\rangle = e^{\xi a^\dagger a^\dagger}\vert 0\rangle\, ,\qquad \hbox{or}\qquad \vert\xi\rangle = e^{\xi a^\dagger a^\dagger}\vert 1\rangle\, .$$

In the unitary form one can write $$e^{-i\alpha K_0}e^{-i\beta K_y}\vert 0\rangle$$ where $$2iK_y=K_+-K_-$$. The overlap $$\langle n\vert e^{-i\alpha K_0}e^{-i\beta K_y}\vert 0\rangle$$ is actually an $$SU(1,1)$$ group function; such functions have a closed form expression closely related the usual Wigner $$D$$-functions for $$SU(2)$$. For $$SU(1,1)$$ they can be found (along with other details) in

Ui, Haruo. "Clebsch-Gordan formulas of the SU (1, 1) group." Progress of Theoretical Physics 44.3 (1970): 689-702.

Searching for "su(1,1) coherent states" in Google will produce multiple helpful hits.

The key hint is that, if your write your $$H$$ in terms of $$K_0$$ and $$2K_x=K_++K_-$$, then the $$SU(1,1)$$ transformation $$e^{-i\beta K_y}$$ will digonalize $$H$$ for some $$\beta\in \mathbb{R}$$. This is similar to the way a Hamiltonian $$J_0+bJ_x$$ is diagonalised by a rotation about $$\hat y$$. So the key is to work through commutators like $$[K_\pm,K_y]$$ to unwrap the effect of the exponential on $$K_0$$ and $$K_x$$; if done correctly some magic will happen.

• I think you mean $| \xi \rangle = e ^ { \xi a a } | 1 \rangle$ instead of $| \xi \rangle = e ^ { \xi a ^ { \dagger } a ^ { \dagger } } | 1 \rangle$ or? – Leviathan Feb 6 '19 at 18:29
• No... the coherent state for $e^{\xi a^\dagger a^\dagger}\vert 1\rangle$ will contain $\vert 2p+1\rangle$, i.e. odd number states, whereas $e^{\xi a^\dagger a^\dagger}\vert 0\rangle$ will contain only even states. – ZeroTheHero Feb 6 '19 at 19:22
• I tried diagonalizing the Hamiltonian with the answer you gave. By using Baker-Hausdorff formula I came on $\mathrm { e } ^ { i \beta K _ { y } } H \mathrm { e } ^ { - i \beta K _ { y } } = \omega K _ { 0 } + \frac { 1 } { 2 } \Delta \left( K _ { + } + K _ { - } \right) + \frac { \omega \beta } { 2 } \left( K _ { + } + K _ { - } \right) - \frac { \omega ^ { 2 } \beta } { 4 } \left( K _ { + } - K _ { - } \right) + \ldots$ Maybe I did a mistake, therefore I'm not confident enough to post a answer. This looks to me that I cannot diagonalize it with one $\beta$. If the fourth term would cancel.. – Leviathan Feb 7 '19 at 18:10
• I could define $\beta=\frac{ \omega}{ \Delta}$... but this is not the case. – Leviathan Feb 7 '19 at 18:11
• You ought to get something like $\cosh(A) K_0\pm \sinh(A) K_x$ when rotating the $K_0$ terms, and something like $\cosh(A) K_x \pm \sinh(A) K_0$ when rotating the $K_x$ term. Then you'd pick the $\sinh(A)$ and $\cosh(A)$ to cancel the $\frac{1}{2}\Delta$ term. I have $t\to 0$ time now but can come back to this next week. – ZeroTheHero Feb 7 '19 at 18:13