Elena & Fabrice's Web

(Part of the Wolverhampton Lectures of Physics's Quantum Physics Course)

We met with a paradox last time that although the system, including the potential, is rotationally invariant, with spherical-symmetry, the solutions for the 1st excited states are not, and break this symmetry. The resolution of this paradox is that we found solutions in the sense that we found a basis from which one can construct any other solutions, time-dependent or not. Indeed, one can construct time-independent rotationally-independent solutions from linear superpositions of the solutions that we got in the $x$, $y$ basis. Note that the ground state had the correct symmetry. The 1st excited states, on which we will focus now, were displaying a $x$ and $y$ asymmetry, which is not surprising, however, because we found the solutions by the separation of variables $\psi=XY$ and solving for the $x$ and $y$ variables independently. We remind the 1st excited states were:

\begin{align} \psi_{1,0}(x,y)&=\sqrt{2\over\pi}x\exp(-(x^2+y^2)/2)\\ \psi_{0,1}(x,y)&=\sqrt{2\over{\pi}}y\exp(-(x^2+y^2)/2) \end{align}

both with energy $2\hbar\omega$. Note that the exponential is rotationally symmetric, since

$$x^2+y^2=r$$

only depends on the distance, and this is why the ground state has the good symmetry. To make the 1st excited state rotational-invariant, since wavefunctions accommodate complex numbers, we can use this combination:

$$x+iy=r e^{i\theta}$$

which indeed only depends on the distance and the angle, meaning that now our 2D wavefunction picks up a phase. Indeed, it is clear from the above that defining $\psi_\pm\equiv{1/\sqrt2}(\psi_{1,0}\pm i\psi_{0,1})$ yields a function of $r$ and $\theta$ only:

$$\psi_{\pm}(r,\theta)=\sqrt{1\over\pi}re^{\pm i\theta}\exp(-r^2/2)\,.$$

Of course the energy is still the same, so this is also a pair of solutions for the 1st excited state (out of which one can reconstruct all the other 1-st excited states solutions). Interestingly, the modulus square indeed only depends on the distance (since $|e^{i\theta}|=1$) but the wavefunction itself depends on $\theta$. Is this physical? Yes. This describes circulation, or flow, similarly to a uniformly charged disk that rotates, producing a steady current. We can capture that by looking at the angular momentum, since we are in 2D (something we couldn't do last year). The classical angular momentum is $\vec L=\vec r\times\vec p$. In 2D, we can only rotate in the plane, so we are interested in the $z$-component (since rotation is described perpendicular to the plane of rotation, and our plane is $x$, $y$, so rotation goes in $z$). From the vector product, we have:

$$L_z=xp_y-yp_x$$

which, quantized (correspondence principle), yields

$$L_z=-i\hbar(x\partial_y-y\partial_x)$$

which is the operator in the $x$, $y$ basis, again, not surprisingly because that's the basis we used to define it. Let's try to express $L_z$ in the polar basis instead. This is a problem of change of variables. Changing variables when derivatives are involved relies on the chain rule. So we have a function $\psi(r,\theta)$ (eventually) with $r=\sqrt{x^2+y^2}$ and $\theta=\arctan(y/x)$ and we want to know how $\partial_{x,y}$ depend on $\partial_{r,\theta}$. The chain-rule tree for that scenario is:

\begin{align} {\partial \psi\over\partial x}&={\partial\psi\over\partial r}{\partial r\over\partial x}+{\partial\psi\over\partial\theta}{\partial\theta\over\partial x}\\ {\partial \psi\over\partial y}&={\partial\psi\over\partial r}{\partial r\over\partial y}+{\partial\psi\over\partial\theta}{\partial\theta\over\partial y} \end{align}

and since

\begin{align} {\partial r\over\partial x}&={x\over r}={\cos\theta}\\ {\partial r\over\partial y}&={y\over r}={\sin\theta}\\ {\partial\theta\over\partial x}&={-y\over r^2}={-\sin\theta\over r}\\ {\partial\theta\over\partial y}&={x\over r^2}={\cos\theta\over r} \end{align}

(here we have used $\arctan(x)'=1/(1+x)^2$ and the rules of derivatives for compositions of functions) then it is easily found, however lengthy the algebra, that:

\begin{align} {\partial \psi\over\partial x}&=\cos\theta{\partial\psi\over\partial r}-{\sin\theta\over r}{\partial\psi\over\partial\theta}\\ {\partial \psi\over\partial y}&={\sin\theta}{\partial\psi\over\partial r}+{\cos\theta\over r}{\partial\psi\over\partial\theta} \end{align}

or, removing $\psi$ (expressing it in operator form):

\begin{align} {\partial_x}&=\cos\theta{\partial_r}-{\sin\theta\over r}{\partial_\theta}\\ {\partial_y}&={\sin\theta}{\partial_r}+{\cos\theta\over r}{\partial_\theta} \end{align}

Finally, this gives us the main result (of this change-of-variable story for $L_z$):

$$x\partial_y-y\partial_x=\partial_\theta$$

since $x\sin\theta-y\cos\theta=0$ and $x\cos\theta+y\sin\theta=r$ (remember $x=r\cos\theta$ and $y=r\sin\theta$). Therefore:

$$L_z=-i\hbar\partial_\theta\,.$$

Indeed this shows how $L_z\psi_\pm=\pm\hbar\psi_\pm$ are states of a given, and fixed, angular momentum, since

$$L_z\psi_\pm=\pm\hbar\psi_\pm$$

meaning, according to the postulates of quantum mechanics, that such states are those with angular momentum (in the $z$-direction) $+\hbar$ (for $\psi_+$) and $-\hbar$ (for $\psi_-$). One state rotates anticlockwise, the other clockwise. Together they form a basis of 1st-excited state with rotational symmetry. The paradox is gone! in the most aesthetically possible way.

While we're at it, note that we have also obtained results such as nabla in the polar basis:

\begin{align} \nabla&=\hat\imath\partial_x+\hat\jmath\partial_y\\ &=\hat\imath(\cos\theta\partial_r-(\sin\theta/r)\partial_\theta)+\hat\jmath(\sin\theta\partial_r+(\cos\theta/r)\partial_\theta)\\ &=\partial_r(\hat\imath\cos\theta+\hat\jmath\sin\theta)+\partial_\theta(-\hat\imath(\sin\theta/r)+\hat\jmath(\cos\theta/r))\\ &=\hat e_r\partial_r+{\hat e_\theta\over r}\partial_\theta \end{align}

We'll also need the Laplacian, for the kinetic energy. We let you derive it yourself along these lines in the tutorials.

We now turn to the problem of finding the other (higher) excited states, in which case the method of building them from the solutions in the $x$, $y$ basis is not the most efficient/elegant. Instead, we are going to use the ladder-operator formalism, which simplifies considerably the algebra and brings in valuable physical insights. While we will work it out for the 2D oscillator, most of the concepts have been already introduced last year (for the 1D case) and so this will also serve as a reminder.

We introduce (following Dirac) the following operators:

\begin{align} a_{x}&={1\over\sqrt{2}}\left(\beta\hat x+{i\over\beta\hbar}\hat p_x\right)\\ a_{y}&={1\over\sqrt{2}}\left(\beta\hat y+{i\over\beta\hbar}\hat p_y\right) \end{align}

where we write explicitly the hat to remind ourselves that both $\hat x$ and $\hat p_x$ are operators. We don't do that on the $a$s although they too are, as a consequence, operators, namely, "ladder operators". We have also introduced as a shortcut notation:

$$\displaystyle\beta=\sqrt{\mu\omega\over\hbar}\,.$$

The operators which are observables are Hermitian, meaning they are equal to their complex-conjugate, i.e.,

\begin{align} \hat x^\dagger&=\hat x\,,\\ \hat p^\dagger&=\hat p\,. \end{align}

So taking the complex-conjugate of $a_x$ and $a_y$ we get:

\begin{align} a_{x}^\dagger&={1\over\sqrt{2}}\left(\beta\hat x-{i\over\beta\hbar}\hat p_x\right)\\ a_{y}^\dagger&={1\over\sqrt{2}}\left(\beta\hat y-{i\over\beta\hbar}\hat p_y\right) \end{align}

and from there, it is simple algebra to show that:

$$\displaystyle a_x^\dagger a_x={1\over2}\left(\beta^2x^2-{1\over\beta^2}\partial_x^2-1\right)$$

(remember, these are operators) so that

$$\displaystyle (a_x^\dagger a_x+{1\over2})\hbar\omega=-{\hbar^2\over2\mu}\partial_x^2+{\mu\omega^2\over2}x^2$$

where we recognize the right-hand side as the 1D quantum harmonic oscillator Hamiltonian. Therefore, the 2D oscillator Hamiltonian, which we already know can be written has $H=H_x+H_y$, reads, in terms of ladder operators:

$$H=\left(a_x^\dagger a_x+a_y^\dagger a_y+1\right)\hbar\omega$$

The value of this is that ladder operators (whence the name) jump from one stationary solution to the next of higher energy by applying the so-called creation operators $a_x^\dagger$ (to create one excitation in $x$) and $a_y^\dagger$ (to create one excitation in $y$), or to one stationary solution to the next below by applying the adjoint (complex conjugate) $a_{x,y}$ operators. That is to say, labeling the states $\psi_{n_x,n_y}(x,y)$ as $|n_x,n_y\rangle$, we have:

\begin{align} a_x|n_x,n_y\rangle&=\sqrt{n_x}|n_x-1,n_y\rangle\\ a_x^\dagger|n_x,n_y\rangle&=\sqrt{n_x+1}|n_x+1,n_y\rangle\\ a_y|n_x,n_y\rangle&=\sqrt{n_y}|n_x,n_y-1\rangle\\ a_y^\dagger|n_x,n_y\rangle&=\sqrt{n_y+1}|n_x,n_y+1\rangle \end{align}

So one can construct the successive excited states from the ground state $|0,0\rangle$, e.g.:

$$|1,0\rangle=a_x^\dagger|0,0\rangle$$

To get this in the $x$, $y$ basis (or representation), we use:

$$\langle x,y|1,0\rangle=\langle x,y|a_x^\dagger|0,0\rangle$$

which, written in full, reads:

\begin{align} \psi_{1,0}(x,y)&={1\over\sqrt{2}}\iint\langle x,y|\left(\beta\hat x-{i\over\beta\hbar}\hat p_x\right)|\chi,\Upsilon\rangle\langle\chi,\Upsilon|0,0\rangle d\chi d\Upsilon\,,\\ &={1\over\sqrt{2}}\iint\left(\beta\hat\chi-{i\over\beta\hbar}\hat p_\chi\right)\psi(\chi,\Upsilon)\delta(x-\chi)\delta(y-\Upsilon)d\chi d\Upsilon\,,\\ &={1\over\sqrt{2}}\left(\beta\hat x-{i\over\beta\hbar}\hat p_x\right)\psi_{0,0}(x,y)\,, \end{align}

and since we know $\psi_{0,0}$ (it is $(\beta/\sqrt{\pi})\exp(-\beta^2(x^2+y^2)/2)$), we obtain in this way the first excited state in the x-direction. By iterations of this principle:

$$\displaystyle|n_x,n_y\rangle={1\over\sqrt{n_x!n_y!}}(a_x^\dagger)^{n_x}(a_y^\dagger)^{n_y}|0,0\rangle$$

which, from the properties of the Hermite polynomials, is how we get the general solution.

Now we turn to the problem of expressing the ladder operators in spherical coordinates. To do so, we introduce:

\begin{align} a_d&={1\over\sqrt{2}}\left(a_x-ia_y\right)\\ a_g&={1\over\sqrt{2}}\left(a_x+ia_y\right) \end{align}

from which one can compute:

\begin{align} a_d^\dagger a_d&={1\over2}(a_x^\dagger a_x+a_y^\dagger a_y-ia_x^\dagger a_y+ia_xa_y^\dagger)\,,\\ a_g^\dagger a_g&={1\over2}(a_x^\dagger a_x+a_y^\dagger a_y+ia_x^\dagger a_y-ia_xa_y^\dagger)\,, \end{align}

so that it is easy to see that:

$$H=(a_d^\dagger a_d+a_g^\dagger a_g+1)\hbar\omega$$

We can also reverse these relations:

\begin{align} a_{x}&={1\over\sqrt{2}}\left(\beta\hat x+{i\over\beta\hbar}\hat p_x\right)\\ a_{x}^\dagger&={1\over\sqrt{2}}\left(\beta\hat x-{i\over\beta\hbar}\hat p_x\right) \end{align}

to find (in matrix form for concision)

$$ \begin{pmatrix} \hat x\\ \hat p_x \end{pmatrix} = {i\hbar\over\sqrt{2}} \begin{pmatrix} -i/(\beta\hbar) & -i/(\beta\hbar)\\ -\beta & \beta \end{pmatrix} \begin{pmatrix} a\\ a^\dagger \end{pmatrix} $$

so that:

\begin{align} \hat x&={1\over\sqrt{2}\beta}(a_x+a_x^\dagger)\,,\\ \hat p_x&=-{i\hbar\beta\over\sqrt{2}}(a_x-a_x^\dagger)\,, \end{align}

From $L_z=-i\hbar(x\partial_y-y\partial_x)$ we find in ladder-operator form that $L_z=i\hbar(a_xa_y^\dagger-a_x^\dagger a_y)$, which, substituting in polar-coordinates operator forms, yields:

$$L_z=\hbar(a_d^\dagger a_d-a_g^\dagger a_g)\,.$$

The operators $ad_d$ and $a_g$ seem to bear a close relationship with the $a_x$ and $a_y$ operators, in fact, they follow the same algebra (so-called Bosonic algebra, cf. tutorial):

$$[a_d,a_d^\dagger]=[a_g,a_g^\dagger]=1\,.$$

This means that they are also linear creation and destruction operators for the excitation of a harmonic oscillator of some sort. In fact, they describe the creation of excitations in spherical coordinates. To see that, let us see how these get expressed in the spherical system:

$$a_d={1\over2}(\beta(x-iy)+{1\over\hbar}(\partial_x-i\partial_y))$$

which, using the expressions derived above:

\begin{align} {\partial_x}&=\cos\theta{\partial_r}-{\sin\theta\over r}{\partial_\theta}\\ {\partial_y}&={\sin\theta}{\partial_r}+{\cos\theta\over r}{\partial_\theta} \end{align}

allows us to write $a_d$ and $a_d^\dagger$ in terms of spherical coordinates:

\begin{align} a_d&={e^{-i\theta}\over2}\left(\beta r+{1\over\beta}\partial_r-{i\over\beta r}\partial_\theta\right)\,,\\ a_d^\dagger&={e^{i\theta}\over2}\left(\beta r-{1\over\beta}\partial_r-{i\over\beta r}\partial_\theta\right)\,. \end{align}

Observe that one does not go from $a_d$ to $a_d^\dagger$ merely by changing the sign of $i$ but that the transpose part of the $\dagger$ should be taken into account too, and, e.g., $\partial_r^\dagger=-\partial_r$ (just as $\partial_x^\dagger=-\partial_x$). Applying $a_d$ on a function that separates the angular and radial dependencies, i.E., of the type $e^{im\theta}F(r)$, yields:

$$\displaystyle a_d^\dagger(e^{im\theta}F(r))={e^{i(m+1)\theta}\over 2}\left[\left(\beta r+{m\over\beta r}\right)F(r)-{1\over\beta}{d\over dr}F(r)\right]\,.$$

In particular, if we apply this result to the case $F(r)=r^m e^{-\beta^2 r^2/2}$, we find:

$$\displaystyle a_d^\dagger\left[e^{im\theta}r^m e^{-\beta^2 r^2/2}\right]= \beta e^{i(m+1)\theta}r^{m+1}e^{-\beta^2r^2/2}$$

that is to say, it keeps the same structure. Indeed, this ladder operator adds one phase twist to the wavefunction and increments the polynomial push of the donut away from the center. Its iteration allows us to reconstruct all the $d$-excited states:

$$\displaystyle\langle r,\theta|n_d,0\rangle={\beta\over\sqrt{\pi n_d!}}e^{in_d\theta}(\beta r)^{n_d}e^{-\beta r^2/2}$$

We let you compute the effect of $a_g$ and the combined $a_d$, $a_g$ creations in the tutorial.