Suppose I have a two-dimensional harmonic oscillator, $H= \hbar\omega(a_x^{\dagger}a_x+a_y^{\dagger}a_y)$. We define the operator $b=\frac{1}{\sqrt{2}}(a_x+ia_y)$.

If eigenkets of the hamiltonian are $|n_x, m_y \rangle = |n_x \rangle |m_y \rangle$, and if

$$|\psi(0) \rangle = \exp(\alpha b^{\dagger}-\alpha^* b)|0_x \rangle |0_y \rangle$$

Find $\psi(x,y,0) = \langle x, y | \psi(0) \rangle$.

My attempt

I attempt to solve the question firstly computating the case when $| \psi \rangle = | 0_x \rangle |0_y \rangle$.

It must be: $$\langle x,y | b |0_x \rangle |0_y \rangle=0$$

So, we have $\langle x,y | b |0_x \rangle |0_y \rangle = \frac{1}{\sqrt{2}}\sqrt{\frac{m\omega}{2\hbar}}\langle x,y | (a_x+ia_y) | 0_x \rangle |0_y \rangle$;

and, if my calculations are correct, we obtain:

$$x\psi-\frac{\hbar}{m\omega}\frac{\partial \psi}{\partial x}-i(y\psi-\frac{\hbar}{m\omega}\frac{\partial \psi}{\partial y})=0.$$

This implies that $$x\psi-\frac{\hbar}{m\omega}\frac{\partial \psi}{\partial x}=0$$


$$y\psi-\frac{\hbar}{m\omega}\frac{\partial \psi}{\partial y}=0$$

but now there is some mathematical issue, because $\psi$ is function of both $x$ and $y$. How I can solve these differential equations? And is this the main strategy for solving the question above?


3 Answers 3


@Abezhiko has done the bulk of the crucial combinatorics, up to the first line of his multi-level formula. Set $m\omega/\hbar= 1$, nondimensionalizing, and reinstate it only in the end if you must, but why?

You then have, in a straightforward definition, $$ \langle x,y|e^{\alpha b^\dagger - \alpha^*b} |0_x,0_y\rangle = e^{-\frac{1}{2}|\alpha|^2} \langle x,y| e^{\alpha b^\dagger} |0_x,0_y\rangle \\ = e^{-\frac{1}{2}|\alpha|^2}\langle x| e^{{\alpha\over \sqrt{2}} (x-\partial_x)} |0_x\rangle \langle y| e^{-i{\alpha\over \sqrt{2}} (y-\partial_y) } |0_y\rangle \\ = {e^{-\frac{1}{2}|\alpha|^2}\over \sqrt{\pi}}e^{{\alpha\over \sqrt{2}} (x-\partial_x)} e^{-x^2/2} ~~ e^{-i{\alpha\over \sqrt{2}} (y-\partial_y) } e^{-y^2/2}, $$ using this bridge, $\langle x |0_x\rangle= \exp(-x^2/2)/\sqrt[4]{\pi}=\psi_0(x)$, the ground Hermite function. No indirection or solving differential equations required!

Can you take it from there, mindful of the basic degenerate CBH identity? $$ e^{z (x-\partial_x)} e^{-x^2/2}= e^{-z^2/2}e^{zx}e^{-z\partial_x}e^{-x^2/2}\\ =e^{-z^2/2+zx}e^{-(x-z)^2 /2} =e^{-z^2/2-x^2/2+2zx}.$$

Response to comments $$\langle x,y | b |0_x \rangle |0_y \rangle=\tfrac{1}{2}(x+\partial_x +i(y+\partial_y))\langle x,y | 0_x, 0_y \rangle \\ =\tfrac{1}{2}(x+\partial_x +i(y+\partial_y)) \psi_0(x)\psi_0(y)=0,$$ as per the above bridge statement for the vacuum, trivially. But this is a basic identity, not related to any evaluation strategy...

  • $\begingroup$ I know that $|\psi(0) \rangle$ is a coherent state. And we can write $|\psi(0) \rangle = |\alpha_x \rangle |\alpha_y \rangle$. Otherwise, $|\psi(0) \rangle = D(\alpha_x)D(\alpha_y)|0_x \rangle |0_y \rangle$, and of course we can imagine that $D(\alpha_x)$ is a translation operator (since $\alpha_x$ is real) and $D(\alpha_y)$ is a translation on momentum space (infact $\alpha_y$ is imaginary). But I don't know how works $D(\alpha_y)$ and I don't understand the math in your answer. Is of course my ignorance. Could you say me how to improve my math about this matter? $\endgroup$
    – Damark
    Nov 5, 2023 at 0:41
  • $\begingroup$ Yes. I mean, I know that we can assume $a_x=x-\frac{i}{m\omega}\frac{\partial}{\partial x}$, but how it works? And how I have already asked: $D(\alpha_y)$ is of course a translation in momentum space, but how it works in this case? By the way, my attempt (see on question): why it is wrong? I can assume $\psi = \psi_x \psi_y$, and dividing all for $\psi_x \psi_y$, I have two differential equations and the solutions is a gaussian. But surely I made some mistakes. $\endgroup$
    – Damark
    Nov 5, 2023 at 7:44
  • $\begingroup$ I'm sorry, i'm messed up. I don't want absolutely that anyone solve for me this excercise, I want to solve by myself! By the way, I know of course Glauber's formula. Your calculation are clear. My point of view is: my attempt is to solve two differential equations. Is it correct to assume that the wavefunction is a multiplication of two function i.e. $\psi = \psi_x(x)\psi_y(y)$? This is my issue. Of couse $a_x$ and $a_y$ commute, and I assume this proved that $\psi = \psi_x(x)\psi_y(y)$. It is right? $\endgroup$
    – Damark
    Nov 5, 2023 at 18:35
  • $\begingroup$ Of course! They live in independent tensor factor spaces, so your instructor has given you two vacua. $\endgroup$ Nov 5, 2023 at 20:53

Firstly, let's compute the following commutator : $$ [b,b^\dagger] = \frac{1}{2} [a_x + ia_y, a_x^\dagger - ia_y^\dagger] = \frac{1}{2} [a_x, a_x^\dagger] + \frac{1}{2} [a_y, a_y^\dagger] = 1 $$ Next, since $b$ and $b^\dagger$ commute with their constant commutator, Glauber's formula permits to write : $$ e^{\alpha b^\dagger - \alpha^*b} = e^{\alpha b^\dagger} e^{-\alpha^*b} e^{-\frac{1}{2}|\alpha|^2[b,b^\dagger]} $$ Moreover, the factor $e^{-\alpha^*b} = 1 - \alpha^*b + \mathcal{O}(b^2)$ contains only annihilation operators (apart from the constant term), so that it is equivalent to the identity operator when applied on the vacuum. In consequence, we are left with the following expression : $$ \begin{align} e^{\alpha b^\dagger - \alpha^*b} |0_x,0_y\rangle &= e^{-\frac{1}{2}|\alpha|^2} e^{\alpha b^\dagger} |0_x,0_y\rangle \\ &= e^{-\frac{1}{2}|\alpha|^2} \sum_{n=0}^\infty \frac{\alpha^n}{n!}(b^\dagger)^n |0_x,0_y\rangle \\ &= e^{-\frac{1}{2}|\alpha|^2} \sum_{n=0}^\infty \frac{\alpha^n}{2^nn!}(a_x^\dagger + ia_y^\dagger)^n |0_x,0_y\rangle \\ &= e^{-\frac{1}{2}|\alpha|^2} \sum_{n=0}^\infty \frac{\alpha^n}{2^nn!} \sum_{k=0}^n \binom{n}{k} (a_x^\dagger)^k(ia_y^\dagger)^{n-k} |0_x,0_y\rangle & (1) \\ &= e^{-\frac{1}{2}|\alpha|^2} \sum_{n=0}^\infty \sum_{k=0}^n \frac{\alpha^n}{2^nk!(n-k)!} (a_x^\dagger)^k(ia_y^\dagger)^{n-k} |0_x,0_y\rangle \\ &= e^{-\frac{1}{2}|\alpha|^2} \sum_{k=0}^\infty \sum_{n=k}^\infty \frac{\alpha^n}{2^nk!(n-k)!} (a_x^\dagger)^k(ia_y^\dagger)^{n-k} |0_x,0_y\rangle & (2) \\ &= e^{-\frac{1}{2}|\alpha|^2} \sum_{k=0}^\infty \sum_{l=0}^\infty \frac{\alpha^{k+l}}{2^{k+l}k!l!} (a_x^\dagger)^k(ia_y^\dagger)^l |0_x,0_y\rangle & (3) \\ &= e^{-\frac{1}{2}|\alpha|^2} \sum_{k=0}^\infty \frac{\alpha^n}{2^kk!}(a_x^\dagger)^k \sum_{l=0}^\infty \frac{(i\alpha)^l}{2^ll!}(a_y^\dagger)^l |0_x,0_y\rangle \\ &= e^{\frac{1}{2}|\alpha|^2} |\alpha_x,i\alpha_y\rangle \end{align} $$ where we used basically (1) the binomial theorem, (2) a summation swap and (3) an index shift, namely $l := n-k$.

In consequence, the wavefunction $\psi(x,y) = \langle x,y | \alpha_x,i\alpha \rangle = \psi_\alpha(x)\psi_{i\alpha}(y)$ is separable, since it is a separable state. This is where you get stuck with your differential equations.

If you want to rederive the wavefunction of a (single) coherent state, you may start from the eigenvalue problem defining it, namely $a|\alpha\rangle = \alpha|\alpha\rangle$, hence in the $x$-representation : $$ \sqrt{\frac{m\omega}{2\hbar}} \left(x + \frac{\hbar}{m\omega} \frac{\partial}{\partial x}\right) \psi_\alpha(x) = \alpha\psi_\alpha(x) $$ This is a linear first-order ODE you can solve by separation of the variables, whose solution is a complex gaussian, bearing in mind that the quantity $|\psi_\alpha(t)|^2$ will have to be normalized. It is also to be noted that the eigenvalues are time-dependent, such that $\alpha(t) = \alpha_0 e^{-i\omega t}$.

On a side note, Cosmas Zachos follows a mixed strategy in his answer. He starts from the ground state $\psi_0(x) = \langle x|0 \rangle \propto e^{-x^2/2}$. Then, one has : $$ \psi_\alpha(x) \propto \langle x | e^{\alpha a^\dagger} | 0 \rangle \propto \exp\left(\sqrt{\frac{m\omega}{2\hbar}} \left(x - \frac{\hbar}{m\omega} \frac{\partial}{\partial x}\right)\right) e^{-x^2/2} $$ The exponential can be split with the help of Glauber formula again, so that $\psi_\alpha(x) \propto e^{\mu x} e^{\nu\partial_x} e^{-x^2/2}$, where $e^{\nu\partial_x}$ will act on the gaussian on its right through the translation $x \mapsto x + \nu$.

  • $\begingroup$ But he wants $\langle x,y|$ thereof, which is so much simpler to evaluate... $\endgroup$ Nov 4, 2023 at 20:10
  • $\begingroup$ @CosmasZachos Hmm, my bad, I misread the question. $\endgroup$
    – Abezhiko
    Nov 4, 2023 at 21:02
  • $\begingroup$ But you did the crucial Glauber formula step... $\endgroup$ Nov 4, 2023 at 21:06
  • $\begingroup$ Of course you are right, and no issue there is in your answer. By the way, I already know that $|\psi(0) \rangle$ is a coherent state (see my comment below). $\endgroup$
    – Damark
    Nov 5, 2023 at 0:44
  • $\begingroup$ @Damark I've edited my answer in order to address your actual question. The key point lies on the fact that the wavefunction $\psi(x,y)$ is separable. $\endgroup$
    – Abezhiko
    Nov 5, 2023 at 18:05

I'll try to explain better my attempt. Thanks to Cosmas Zachos and Abezhiko for their patience.

We want to write $\langle x,y |\psi(0) \rangle$.

Firstly, is easy to show that $|\psi(0) \rangle$ is a coherent state:

$$|\psi(0) \rangle= exp(\alpha b^{\dagger}-\alpha^* b)|0_x,0_y \rangle$$

infact $exp(\alpha b^{\dagger}-\alpha^* b)=exp(\frac{\alpha}{\sqrt{2}}(a_x+ia_y)-\frac{\alpha^*}{\sqrt{2}}(a_x^{\dagger}-ia_y^{\dagger})= exp(\alpha_x a_x^{\dagger}-\alpha_x^* a_x +(\alpha_y a_y^{\dagger}+\alpha_y^*a_y))$

where $\alpha_x=\frac{\alpha}{\sqrt{2}}, \alpha_y=-\frac{i\alpha}{\sqrt{2}}$

But $a_x$ and $a_y$ commute, and we can use the Glauber's formula to obtain: $exp(\alpha_x a_x^{\dagger}-\alpha_x^* a_x) exp(\alpha_y a_y^{\dagger}+\alpha_y^*a_y)=D(\alpha_x)D(\alpha_y)$.

Then, we have $|\psi(0) \rangle = D(\alpha_x)D(\alpha_y)|0_x, 0_y \rangle = |\alpha_x, \alpha_y \rangle$.

Now, if: $\langle x,y | \psi(0) \rangle = \psi(x,y,0)=\psi(x)\psi(y)=\psi_x\psi_y$

it must be: $$\langle x,y|b |\psi(0) \rangle = \frac{\alpha_x}{\sqrt{2}} \psi_x \psi_y + i\frac{\alpha_y}{\sqrt{2}} \psi_x \psi_y$$

and, remembering that $a=\sqrt{\frac{m\omega}{2\hbar}}(x+\frac{i}{m\omega}p)$,

$$\langle x,y|b |\psi(0) \rangle =\sqrt{\frac{m\omega}{4\hbar}}\big(x\psi_x\psi_y+\frac{\hbar}{m\omega}\frac{d\psi_x}{dx}\psi_y+i(y\psi_x\psi_y+\frac{\hbar}{m\omega}\frac{d\psi_y}{dy}\psi_x)\big)$$

Now, the two results have to be equal: this implies that

$$\sqrt{\frac{m\omega}{4\hbar}}\big(x\psi_x\psi_y+\frac{\hbar}{m\omega}\frac{d\psi_x}{dx}\psi_y\big)= \frac{\alpha_x}{\sqrt{2}} \psi_x \psi_y$$

and mutatis mutandis the same for the imaginary part.

Dividing both equations for $\psi_x\psi_y$ we have a differential equations like this:

$$\sqrt{\frac{m\omega}{4\hbar}}\big(x+\frac{\hbar}{m\omega}\frac{1}{\psi_x}\frac{d\psi_x}{dx}\big)= \frac{\alpha_x}{\sqrt{2}}$$

easy to solve. After, it's enough to normalize both $\psi_x$ and $\psi_y$.


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