In a set of notes it is stated that: Given coherent states of a harmonic oscillator $$| \alpha \rangle = \pi^{-\frac{1}{2}} \text{exp}(-\frac{1}{2}|\alpha|^2)\sum_{n = 0}^{\infty} \frac{\alpha^n}{(n!)^{\frac{1}{2}}}|n \rangle~~~~~~~~~~(1)$$where $|n \rangle$ is the nth energy eigenstate of the oscillator. We can define the completeness relation as the integral $$\int d^2 \alpha | \alpha \rangle \langle \alpha | = \sum_{n=0}^{\infty}| n \rangle \langle n| = 1~~~~~~~~~~~~~~~~(2)$$ But these states are not orthonormal. They do not span the Hilbert space in the two-dimensional $\alpha$ space.

Question: How is the "two dimensional $\alpha$ plane" defined? If $\{ | \alpha \rangle \}$ satisfies completeness relation (2) then what space does it span?

  • $\begingroup$ I believe your formula for the coherent state is missing a term $|\alpha|$ in the exponential. The "two dimensional alpha plane" seems to be a weird way to refer to the complex numbers (which are integrated over in your second equation). The coherent states span (in the generalized sense expressed by the second equation) the entire Hilbert space of the harmonic oscillator. For mathematical details, search for "Segal-Bargmann spaces", which make this somewhat more transparent. $\endgroup$ Jul 13, 2017 at 20:11
  • $\begingroup$ "They do not span" is some strong (also wrong) words. The (closure of) the linear span of the $|\alpha\rangle$ is indeed the full Hilbert space. A good way to analyze coherent states is as a vector space frame, and there's several good analogs in finite dimensions (like, say, $\{(\cos(2\pi k/3),\sin(2\pi k/3)):k=0,1,2\}$ over $\Bbb R^2$) along those lines. $\endgroup$ Jul 14, 2017 at 18:40

2 Answers 2


So let's start a step back because your coherent states are not normalized as I would normalize them.

Coherent states

The coherent states come from their response to the bosonic annihilator, $$\hat b |x , y\rangle = (x + i y) |x,y\rangle.$$From this one can derive that any particular one's representation among the number states must satisfy, $$\hat b~\sum_n c_n |n\rangle = \sum_n c_n \sqrt{n} |n-1\rangle=(x + i y) \sum_n c_n |n\rangle,$$giving the recursive relation that $c_n = \frac{x+iy}{\sqrt n}~c_{n-1}.$ Starting from $c_0$ we then find indeed the relation that $$|x,y\rangle = c_0~\sum_n \frac{(x + i y)^n}{\sqrt{n!}} |n\rangle.$$The remaining $c_0$ with the proper normalization gives $$\langle x,y|x,y\rangle = 1 = |c_0|^2 \sum_n \frac{(x-iy)^n(x+iy)^n}{n!} = |c_0|^2 \exp\big(x^2 + y^2\big).$$Choosing these to all have the same complex phase for their vaccum component finally yields,$$|x, y\rangle = \exp\left(-\frac12(x^2 + y^2)\right)\sum_n \frac{(x + i y)^n}{\sqrt{n!}}~|n\rangle.$$

So the question is, why does your expression have a leading $\pi^{-1/2}$ in it? That's because they resolve the identity in a somewhat weird way. What does that mean?

Resolving the identity

Suppose you have an expression for some average $\langle A \rangle.$ QM is very clear that this expression may be written based on its quantum state $|\psi\rangle$ as $\langle \psi|\hat A|\psi\rangle.$

But using the fact that $1 = \sum_n |n\rangle\langle n|,$ for example, we can insert these sums ad-hoc into that expression to find that in fact this expectation value also reads, $$\langle A \rangle = \sum_{mn} \langle\psi|m\rangle\langle m|\hat A|n\rangle\langle n|\psi\rangle = \sum_{mn} \psi^*_m~A_{mn}~\psi_n.$$ So that is the value of resolving the identity; it means that you can define this matrix $A_{mn}$ which fully specifies the action of $\hat A$ on the Hilbert space, recovering every single expectation value from the matrix.

Well we see something very similar when we look at the operator, $$\hat Q = \int_{-\infty}^\infty dx~\int_{-\infty}^\infty dy~|x,y\rangle\langle x, y| = \sum_{mn} \iint dx~dy~e^{-x^2-y^2}\frac{(x-iy)^m(x+iy)^n}{\sqrt{m!n!}} |m\rangle\langle n|.$$ At this point it is useful to shift to polar coordinates where $x + i y = r e^{i\theta},$ yielding $$\hat Q = \sum_{mn}\int_{0}^\infty dr~\int_0^{2\pi} r~d\theta~e^{-r^2}~\frac{r^{m+n} e^{i(n-m)\theta}}{\sqrt{m!n!}} |m\rangle\langle n|.$$ Note that the angle over $\theta$ integrates a sinusoid over one or more full periods and therefore vanishes if $m\ne n$; it is $2\pi$ if $m = n$, so we must get:$$\hat Q = \pi\sum_{n}\int_{0}^\infty dr~2r~e^{-r^2}~\frac{r^{2n} }{n!} |n\rangle\langle n|.$$Substituting $u=r^2, du=2r~dr$ we find that this is:$$\hat Q = \pi\sum_{n}\frac1{n!}~|n\rangle\langle n|~\int_{0}^\infty du~e^{-u}~u^n.$$If you've never seen the gamma function before, the integral on the right hand side is $n!$ and in fact it is the canonical way to extend the factorial function to non-integers to find e.g. that $(-1/2)! = \sqrt{\pi},$ though of course we only need the integers here. After cancelling that through we find out that in fact, $$\hat Q = \pi,$$ or in other words we recover this property of resolving the identity even though not all of these functions are orthogonal, because the way that they're non-orthogonal just comes down to a constant multiplicative factor. We can therefore state unequivocally, $$1 = \iint dx~dy~\frac1\pi~|x,y\rangle\langle x,y|.$$ Your expression absorbs a $1/\sqrt{\pi}$ term into each of these kets, and writes $\pi^{-1/2} |x, y\rangle = |\alpha\rangle$ (where $\alpha = x + i y$) for short, both of which help in writing these expansions. One then finds similarly to the above expression with $A_{mn}$, that $$\langle A \rangle = \iint d^2\alpha~d^2\beta~\psi^*(\alpha)~A(\alpha,\beta)~\psi(\beta).$$The only cost to this notation is that we then have to express the above integrals with the more clumsy $\int d^2\alpha$ which is short for something like $d\alpha_x~d\alpha_y$ where $\alpha = \alpha_x + i \alpha_y.$

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    $\begingroup$ Allow me please to add: Keep in mind that $\hat b$ is NOT hermitian so its eigenvectors need not have real eigenvalues. Thus $\vert x,y\rangle$ is just a convenient way to denote a state with eigenvalue $x+iy$ that lives in the Hilbert space of the 1d harmonic oscillator states $\endgroup$ Jul 13, 2017 at 21:36
  • $\begingroup$ @CRDrost Thanks for the comprehensive answer, I am relatviely new to QM so I'm still finding my feet (hence your answer is quite scary). Could I ask, how would you define the two dimensional $\alpha$ plane (in a possibly simple way)? $\endgroup$
    – user110903
    Jul 13, 2017 at 21:42
  • $\begingroup$ The simplest definition is "it's the dimensionless phase space of the harmonic oscillator" but that requires a bit more explanation. The harmonic oscillator Hamiltonian can be written as $\hbar\omega~\big(\hat x^2 + \hat y^2\big)$ for the two dimensionless "quadratures" $\hat x = \frac12(b^\dagger + b), \hat y = \frac i2(b^\dagger - b),$ at which point one sees that $\hat x |x,y\rangle = x|x,y\rangle$ and $\hat y|x,y\rangle = y|x,y\rangle.$ Since $[\hat x, \hat y] = i/2$ one sees that $\hat y$ is a dimensionless momentum operator conjugate to $\hat x$, a dimensionless position operator. $\endgroup$
    – CR Drost
    Jul 13, 2017 at 22:58
  • $\begingroup$ As an exercise, consider a harmonic oscillator of mass $m$ oscillating on a spring of with spring constant $k$, deviating from its rest position by a coordinate $q$. The Schrodinger eq'n is $i\hbar\dot\Psi=-(\hbar^2/2m)\partial_q^2\Psi + (k/2)q^2\Psi,$ define $x=q/q_0$ for some as yet unknown $q_0$ to find the first term has the form $(\hbar^2/2mq_0^2)$ while the second has the form $(k/2)q_0^2,$ choosing the "right" $q_0$ one gets $\hbar\omega=(\hbar^2/2mq_0^2)=(k/2)q_0^2$ or $q_0=(\hbar^2/mk)^{1/4}.$ This nondimensionalizes the equation from $(q,p)$ space to $(x, y)$ space. $\endgroup$
    – CR Drost
    Jul 13, 2017 at 23:09
  • $\begingroup$ Also if it helps, the evolution equation for each $|n\rangle$ under the harmonic oscillator Hamiltonian is $i\hbar\partial_t |n\rangle = \hbar\omega|n\rangle$ and consequently we get that for times not equal to 0, $|n\rangle\mapsto e^{-i\omega t~n}|n\rangle.$ Comparison with the above formulas for $|x,y\rangle$ in terms of $|n\rangle$, we see that it absorbs into $\big((x+iy)e^{-i\omega t}\big)^n,$ so we map $\alpha\mapsto \alpha e^{-i\omega t},$ the coherent states orbit this $(x,y)$ phase space plane in a circle. They are the "most classical" harmonic oscillator states in that sense. $\endgroup$
    – CR Drost
    Jul 13, 2017 at 23:23

They do not span a 2-dimensional space; you can see this because $\vert\alpha\rangle$ is a linear combination of harmonic oscillator states. Thus state $\vert\alpha\rangle$ live in the $\infty$-dimensional space. They span that space because any state can be written as a linear combination of the $\vert \alpha\rangle$'s and the expansion coefficient is unique and completely fixed by $\vert \alpha\rangle$, just like you would expand over the set $\{\vert n\rangle,n=0,1,\ldots\}$.

Edit: It seems that part of the difficulty is connected to the observation that the eigenvalues of $\hat b$ are continuous and complex rather than real and discrete. Thus a sum of the type $$ \sum_n \vert n\rangle \langle n\vert = \mathbb{I} $$ that would hold for discrete eigenstates must be replaced by a continuous sum (i.e. an integral) in the continuous case. Since the eigenstates are labelled by complex numbers $x+iy$, and since the closure involves a continuous sum over all eigenstates, $$ \sum_n \vert n\rangle \langle n\vert = \mathbb{I}\propto \int dxdy \vert x,y\rangle\langle x,y\vert = \int d^2\alpha \vert\alpha\rangle\langle\alpha\vert $$ with $d^2\alpha=dxdy$ if $\alpha=x+iy$.

  • $\begingroup$ Okay thanks, but what exactly is the "two dimensional $\alpha$ plane", just a two dimensional vector space over a complex field and what is the relevance of this observation, just stating that the coherent states span the infinite dimensional Hilbert space but not a two-dimensional one, it semms a bit arbitrary? $\endgroup$
    – user110903
    Jul 13, 2017 at 20:28
  • $\begingroup$ The set of coherent stated is not a subspace of the Hilbert space of harmonic oscillator states : $|α\rangle + |β\rangle$ (renormalized) is NOT a coherent state, but it is a valid harmonic oscillator state. This allows the 2-dimensional set of coherent states to span the full $∞$ dimensional Hilbert space $\endgroup$ Jul 14, 2017 at 18:06
  • $\begingroup$ @JohnJack hopefully the added material helps. $\endgroup$ Jul 14, 2017 at 18:07
  • $\begingroup$ @ZeroTheHero It does help thanks, but also does not directly answer my question in the comments above about whether my definition of a 'two dimensional $\alpha$ plane' is correct or not? Similarly, what "2-dimensional space" are you referring to in your answer? $\endgroup$
    – user110903
    Jul 16, 2017 at 13:47
  • $\begingroup$ The 2d $\alpha$ plane is just the complex plane since the eigenvalues of $\hat b$ are continuous and complex. $\endgroup$ Jul 16, 2017 at 14:08

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