Photo-Ionization

As a final example, let us investigate the photo-ionization of atomic hydrogen. In this phenomenon, a photon of angular frequency $\omega$ is absorbed by an electron that occupies the ground state of a hydrogen atom. The final energy of the electron,

$$E_f = \hbar\,\omega +E_i,$$ (8.221)

where $E_i=E_0$ is the (negative) hydrogen ground-state energy, is assumed to be positive, which corresponds to an unbound state. In other words, the absorption of the photon causes the hydrogen atom to dissociate.

Let ${\bf k}=(\omega/c)\,{\bf n}$ and $\epsilon$ be the wavevector and electric polarization vector of the photon, respectively. (Recall that ${\bf n}$ and $\epsilon$ are both unit vectors.) It follows from Equations (8.136) and (8.152) that the absorption cross-section of the hydrogen atom can be written

$$d\sigma_{i\rightarrow f}^{\rm abs} = \frac{4\pi^2\,\alpha}{m_e^{\... ...{\bf p}\,\right\vert i\right\rangle\right\vert^{\,2}\delta(\omega-\omega_{fi}),$$ (8.222)

where ${\bf p}$ is the electron momentum, and $\omega_{fi}=(E_f-E_i)/\hbar$ . Here, we have written $d\sigma_{i\rightarrow f}^{\rm abs}$ , rather than $\sigma_{i\rightarrow f}^{\rm abs}$ , because we are only considering final states in which the electron momentum, ${\bf p}_f$ , is directed into the range of solid angles $d{\mit\Omega}$ . As discussed in Section 8.6, we must operate on the previous expression with $\int dE_f\,(\cdots)\,\rho(E_f)$ , where $\rho(E_f)$ is the density of final states, to obtain the true absorption cross-section, which takes the form

$$d\sigma_{i\rightarrow f}^{\rm abs} =\left.\frac{4\pi^2\,\alpha\,\... ...t f\right\rangle\right\vert^{\,2}\rho(E_f)\right\vert _{E_f=E_i+\hbar\,\omega}.$$ (8.223)

Here, we have made use of the fact that ${\bf p}$ is an Hermitian operator, and have also employed the result $\delta(a\,x)=\delta(x)/a$ , where $a$ is a constant. (See Exercise 19.)

The electron is initially in the hydrogen ground state, whose wavefunction takes the form

$$\psi_i({\bf x})= \frac{1}{\sqrt{\pi}\,a_0^{\,3/2}}\,\exp\left(-\frac{r}{a_0}\right),$$ (8.224)

where $a_0$ is the Bohr radius. (See Chapter 4.) On the other hand, the final electron state is assumed to be a plane-wave state whose wavefunction is

$$\psi_f({\bf x})= \frac{1}{\sqrt{V}}\,\exp\left({\rm i}\,{\bf k}_f\cdot{\bf x}\right).$$ (8.225)

In writing this expression, we have conveniently assumed that the emitted electron is contained in a cubic box of dimension $L\gg a_0$ , and volume $V=L^{\,3}$ . Furthermore, the electron wavefunction is required to be periodic at the boundaries of the box. Of course, we can later take the limit $L\rightarrow\infty$ to obtain the general case. (However, it turns out that this is not necessary, because the final result does not depend on $L$ , provided that $L\gg a_0$ .) The wavefunction of the final electron state is normalized such that the probability of finding the electron in the box is unity: that is,

$$\int_V d^{\,3}{\bf x}\,\vert\psi_f({\bf x})\vert^{\,2} = 1.$$ (8.226)

Note that the final state is an eigenstate of ${\bf p}_f$ belonging to the eigenvalue

$${\bf p}_f = \hbar\,{\bf k}_f.$$ (8.227)

(This follows from the standard Schrödinger representation ${\bf p} = -{\rm i}\,\hbar\,\nabla$ .) Furthermore, in writing the wavefunction (8.228), we have neglected any interaction between the emitted electron and the hydrogen nucleus. This neglect is only reasonable provided the final electron energy,

$$E_f = \frac{p_f^{\,2}}{2\,m_e}= \frac{\hbar^{\,2}\,k_f^{\,2}}{2\,m_e}$$ (8.228)

is much larger than the ionization energy (i.e., the binding energy) of the hydrogen atom,

$$I= -E_0 = \frac{\hbar^{\,2}}{2\,m_e\,a_0^{\,2}}.$$ (8.229)

The condition $E_f\gg I$ yields

$$k_f\,a_0\gg 1.$$ (8.230)

In other words, the de Broglie wavelength of the emitted electron must be much larger than the typical dimension of the hydrogen atom.

Let us calculate $\rho(E_f)$ . Suppose that ${\bf k}_f=(k_x,\, k_y,\, k_z)$ . The periodicity constraint on $\psi_f({\bf x})$ at the boundaries of the box imply that

$$k_x =\frac{2\pi\,n_x}{L},$$ (8.231)

$$k_y =\frac{2\pi\,n_y}{L},$$ (8.232)

$$k_z =\frac{2\pi\,n_z}{L},$$ (8.233)

where $n_x$ , $n_y$ , and $n_z$ are integers. It follows that

$$dn_x\,dn_y\,dn_z = \frac{V}{(2\pi\,\hbar)^{3}}\,dp_x\,dp_y\,dp_z,$$ (8.234)

where $p_x=\hbar\,k_x$ , et cetera. Thus, the number of final electron states contained in a volume $d^{\,3} {\bf p}_f=dp_x\,dp_y\,dp_z$ of momentum space is $\rho({\bf p}_f)\,d^{\,3}{\bf p}_f$ , where

$$\rho({\bf p}_f) = \frac{V}{(2\pi\,\hbar)^{\,3}}.$$ (8.235)

Note that $\rho({\bf p}_f)$ is constant. Hence, we deduce that the number of final electron states for which $p_f=\vert{\bf p}_f\vert$ lies between $p_f$ and $p_f+dp_f$ , and ${\bf p}_f$ is directed into the range of solid angles $d{\mit\Omega}$ , is $\rho(p_f)\,dp_f= \rho({\bf p}_f)\,d^{\,3}{\bf p}_f$ , where $d^{\,3}{\bf p}_f = p_f^{\,2}\,dp_f\,d{\mit\Omega}$ . In other words,

$$\rho(p_f) = \frac{V\,p_f^{\,2}}{(2\pi\,\hbar)^{\,2}}\,d{\mit\Omega}.$$ (8.236)

Finally, the number of final electron states whose energies lie between $E_f$ and $E_f+dE_f$ is $\rho(E_f)\,dE_f = \rho(p_f)\,(dp_f/dE_f)\,dE_f$ , which yields

$$\rho(E_f) =\frac{V\,m_e\,k_f}{(2\pi)^{\,3}\,\hbar^{\,2}}\,d{\mit\Omega},$$ (8.237)

where use has been made of Equations (8.230) and (8.231).

Equations (8.226) and (8.240) imply that the differential cross-section for the photo-ionization of atomic hydrogen takes the form

$$\frac{d\sigma_{i\rightarrow f}^{\,{\rm abs}}}{d{\mit\Omega}} = \f... ...}^{-{\rm i}\,{\bf k}\cdot{\bf x}}\,\right\vert f\right\rangle\right\vert^{\,2}.$$ (8.238)

Furthermore, making use of Equations (8.227) and (8.228), we can write

$$\left\langle i\left\vert\,\mbox{\boldmath$\epsilon$}\cdot{\bf p}\... ...mbox{\boldmath$\epsilon$}\cdot{\bf p}\,{\rm e}^{-{\rm i}\,{\bf q}\cdot{\bf x}},$$ (8.239)

where

$${\bf q} = {\bf k}-{\bf k}_f.$$ (8.240)

Note, by momentum conservation, that $\hbar\,{\bf q}$ is the recoil momentum of the hydrogen nucleus after ionization. The usual Schrödinger representation ${\bf p} =-{\rm i}\,\nabla$ reveals that

$$\left\langle i\left\vert\,\mbox{\boldmath$\epsilon$}\cdot{\bf p}\... ...t_V d^{\,3}{\bf x} \,{\rm e}^{-r/a_0}\,{\rm e}^{-{\rm i}\,{\bf q}\cdot{\bf x}},$$ (8.241)

where we have made use of the standard electromagnetic result $\epsilon$ $\cdot{\bf k} = 0$ [49].

Assuming that $L\gg a_0$ , we can write

$$\int_V d^{\,3}{\bf x} \,{\rm e}^{-r/a_0}\,{\rm e}^{-{\rm i}\,{\bf... ...{\rm e}^{-r/a_0}\int_{-1}^1 d\mu\,{\rm e}^{-{\rm i}\,q\,r\,\mu}\oint d\varphi',$$ (8.242)

where $\mu=\cos\theta'$ . Here, $\theta'$ , $\varphi'$ are spherical angles whose polar axis is parallel to ${\bf q}$ . Thus, we obtain

$$\int_V d^{\,3}{\bf x} \,{\rm e}^{-r/a_0}\,{\rm e}^{-{\rm i}\,{\bf... ...}}=\frac{4\pi\,a_0^{\,2}}{q}\int_0^\infty dy\,y\,{\rm e}^{-y}\,\sin(q\,a_0\,y).$$ (8.243)

However (see Exercise 20),

$$\int_0^\infty dy \,y\,{\rm e}^{-y}\,\sin(b\,y) = \frac{2\,b}{(1+b^{\,2})^2},$$ (8.244)

which implies that

$$\int_V d^{\,3}{\bf x} \,{\rm e}^{-r/a_0}\,{\rm e}^{-{\rm i}\,{\bf q}\cdot{\bf x}}=\frac{8\pi\,a_0^{\,3}}{[1+(q\,a_0)^2]^{\,2}}.$$ (8.245)

Thus, Equations (8.241), (8.244), and (8.248) yield

$$\frac{d\sigma_{i\rightarrow f}^{\,{\rm abs}}}{d{\mit\Omega}} =\fr... ...ox{\boldmath$\epsilon$}\cdot{\bf k}_f)^{\,2}\,a_0^{\,3}}{[1+(q\,a_0)^2]^{\,4}},$$ (8.246)

where use has been made of the standard electromagnetic dispersion relation $\omega=k\,c$ [49].

It is convenient to orientate our coordinate system such that $\epsilon$ $={\bf e}_x$ and ${\bf n}={\bf e}_z$ . Thus, we can specify the direction of the emitted electron in terms of the conventional polar angles, $\theta$ and $\varphi$ . In fact,

$${\bf k}_f = k_f\,(\sin\theta\,\cos\varphi,\,\sin\theta\,\sin\varphi,\,\cos\theta),$$ (8.247)

and $\epsilon$ $\cdot{\bf k}_f = k_f\,\sin\theta\,\cos\varphi$ , with $d{\mit\Omega} = \sin\theta\,d\theta\,d\varphi$ . Hence, we obtain

$$\frac{d\sigma_{i\rightarrow f}^{\,{\rm abs}}}{d{\mit\Omega}} =2^{... ...f\,a_0)^3\,\frac{\sin^2\theta\,\cos^2\varphi}{[1+(q\,a_0)^2]^{\,4}}\,a_0^{\,2},$$ (8.248)

where

$$\hat{I}= \frac{I}{\hbar\,c\,k} = \frac{I}{\hbar\,\omega}$$ (8.249)

is the normalized ionization energy.

Now, energy conservation demands that

$$\hbar\,c\,k= \frac{\hbar^{\,2}\,k_f^{\,2}}{2\,m_e}+I.$$ (8.250)

[See Equation (8.224).] This expression can be rearranged to give

$$(k_f\,a_0)^{\,2} = \frac{1-\hat{I}}{\hat{I}},$$ (8.251)

and

$$\frac{k_f}{k} = \frac{2}{\beta}\,(1-\hat{I}),$$ (8.252)

where

$$\beta = \frac{\hbar\,k_f}{m_e\,c}= \frac{v}{c}.$$ (8.253)

Here, $v$ is the speed of the emitted electron. However, we have already seen, from Equation (8.233), that the approximations made in deriving Equation (8.251) are only accurate when $k_f\,a_0\gg 1$ . Hence, according to Equation (8.254), we also require that $\hat{I}\ll 1$ . Furthermore, because we are working in the non-relativistic limit, we need $\beta\ll 1$ . From Equation (8.255), this necessitates

$$k_f \gg k.$$ (8.254)

Finally, the inequality $\hat{I}\ll 1$ can be combined with the previous inequality to give

$$\frac{I}{\hbar\,c}\ll k\ll k_f$$ (8.255)

as the condition for the validity of Equation (8.251).

Now,

$$q^{\,2} = k_f^{\,2} - 2\,k\,{\bf k}_f\cdot{\bf n}+k^{\,2}=k_f^{\,2} - 2\,k\,k_f\,\cos\theta+k^{\,2},$$ (8.256)

which implies that

$$1+(q\,a_0)^{\,2} \simeq 1 + \hat{I}^{\,-1}(1-\hat{I}-\beta\,\cos\theta)= \hat{I}^{\,-1}\,(1-\beta\,\cos\theta),$$ (8.257)

where use has been made of Equation (8.255), and we have neglected terms of order $\beta^{\,2}$ . Equations (8.251), (8.254), and (8.260) can be combined to give the following expression for the differential photo-ionization cross-section of atomic hydrogen:

$$\frac{d\sigma_{i\rightarrow f}^{\,{\rm abs}}}{d{\mit\Omega}} \sim... ...3/2}a_0^{\,2}\,\frac{\sin^2\theta\,\cos^2\varphi}{(1-\beta\,\cos\theta)^{\,4}}.$$ (8.258)

Integrating over all solid angle, and neglecting terms of order $\beta^{\,2}$ , the total cross-section becomes

$$\sigma_{i\rightarrow f}^{\,{\rm abs}}\simeq \frac{2^{\,8}\pi}{3}\,\alpha\,\hat{I}^{\,7/2}\left(1-\hat{I}\right)^{\,3/2}a_0^{\,2}.$$ (8.259)

(See Exercise 21.)

Note that the previous two expressions are only accurate in the limits $\hat{I}\ll 1$ and $\beta\ll 1$ . Nevertheless, we have retained the factors $\left(1-\hat{I}\right)^{\,3/2}$ in these formulae because they emphasize that there is a threshold photon energy for photo-ionization: namely, $\hbar\,\omega = I$ . For $\hbar\,\omega< I$ (i.e., $\hat{I}>1$ ), the incident photons are not energetic enough to ionize the hydrogen atom, so there are no emitted photoelectrons. Consequently, the cross-section for photo-ionization tends to zero as $\hat{I}$ approaches unity from below. We have retained the factor involving $\beta$ in Equation (8.261) because it makes clear that the photoelectrons are emitted preferentially in the directions $\varphi=0$ , $\pi$ and $\theta=\cos^{-1}(2\,\beta)$ . Thus, in the non-relativistic limit, $\beta\rightarrow 0$ , the electrons are emitted preferentially along the $x$ -axis (i.e., parallel or anti-parallel to the incident photon's electric polarization vector.) On the other hand, as relativistic effects become important, and $\beta$ consequently increases, the directions of preferentially emission are beamed forward (i.e., they acquire a component parallel to the wavevector of the incident photon.) An accurate expression for the photo-ionization cross-section close to the ionization threshold (i.e., $\hat{I}\rightarrow 1$ ) can only be obtained using unbound positive energy hydrogen atom wavefunctions as the final electron states [106]. Likewise, an accurate expression for the cross-section in the finite-$\beta$ limit requires a relativistic treatment of the problem [96,97].