V Further examples: bound states, extended states, and tunneling

Consider that the particle moves in a constant potential $V(x)=V_I$. It may be a good idea to sketch the potential on a piece of paper.

Classically, the total energy of the particle would be given by $E=p^2/2m+V_I$. Part of the total energy is taken up by the potential energy $V_I$. The remaining kinetic energy is $p^2/2m=(E-V_I)$. We can solve for $p$ and find that the classical particle has momentum

$$p=\pm \sqrt{2m(E-V_I)}.$$

(56)

Classically, no energies are allowed, since the momentum has to be real.

Quantum-mechanically, the particle is described by a wave function $\psi (x)$ which solves the Schrödinger equation

$$E\psi (x)=-\frac{{\mathrm{\hslash }}^2}{2m}{\psi }^{\prime \prime }(x)+V_I\psi (x).$$

(57)

It is useful to rearrange this equation by subtracting on both sides $V_I\psi (x)$. Then we obtain

$$(E-V_I)\psi (x)=-\frac{{\mathrm{\hslash }}^2}{2m}{\psi }^{\prime \prime }(x).$$

(58)

This is a homogenous linear ordinary differential equation of second order with constant coefficients. Its general solution is of the form

$$\psi (x)=A\exp (ikx)+B\exp (-ikx),$$

(59)

where $A$ and $B$ are arbitrary complex numbers. The wave number $k$ is also a constant, and is found by inserting $\psi (x)$ into the Schrödinger equation:

	$(E-V_I)[A\exp (ikx)+B\exp (-ikx)]$
	$={\displaystyle \frac{{\mathrm{\hslash }}^2{k}^2}{2m}}[A\exp (ikx)+B\exp (-ikx)].$		(60)

This gives the condition $(E-V_I)=\frac{{\mathrm{\hslash }}^2{k}^2}{2m}$, which is satisfied for

$$k=\frac{\sqrt{2m(E-V_I)}}{\mathrm{\hslash }}.$$

(61)

Note that this is of the form $k=|p|/\mathrm{\hslash }$, where $p$ is the classical momentum (56). (This is just the de Broglie relation for momentum.) A second solution is $k=-|p|/\mathrm{\hslash }$, but we have already taken care of it by the second exponential in (59). For $E>V_I$ (the classically allowed case), $k$ is positive, and the solution $\psi (x)$ is the superposition of a plane wave that moves to the right (amplitude $A$) and a plane wave that moves to the left (amplitude $B$). If we would make an experiment that measures the direction of the particle, we would find the probabilities for motion into each direction as

	$P(\text{particle moves right})$	$=$	${\displaystyle \frac{{|A|}^2}{{|A|}^2+{|B|}^2}},$		(62)
	$P(\text{particle moves left})$	$=$	${\displaystyle \frac{{|B|}^2}{{|A|}^2+{|B|}^2}}.$		(63)

(We could normalise the wave function to ${|A|}^2+{|B|}^2=1$ to obtain simpler expressions, but here it is useful to keep the discussion more general.)

We can also find solutions for . The wave number $k$ is then imaginary,

$$k=\frac{\sqrt{2m(E-V_I)}}{\mathrm{\hslash }}=i\frac{\sqrt{2m(V_I-E)}}{\mathrm{\hslash }}\equiv i\kappa ,$$

(64)

where we introduced the real number $\kappa =\frac{\sqrt{2m(V_I-E)}}{\mathrm{\hslash }}$. The wave function is then of the form

$$\psi (x)=A\exp (-\kappa x)+B\exp (\kappa x).$$

(65)

The first term increases rapidly for $x\to -\mathrm{\infty }$, and the second term increases rapidly for $x\to \mathrm{\infty }$. According to the mathematical property III listed in Sec. III.1, we have to discard these solutions. Our only choice is to set $A=B=0$. Hence, just as in the classical case, the particle cannot have energy .
Points to remember

• •

  Probabilities of momentum can be read off by decomposing the wave function into a superposition of plane waves.

• •

  The energy of a quantum particle is bounded from below by the minimum of the potential energy of a system.

A hard wall is a region in space where $V(x)=\mathrm{\infty }$. Let us consider the potential

(66)

which is constant for and represents a hard wall for $x>0$.
For simplicity, let us assume that $V_I=0$.

Classically, the particle moves with momentum $p=\sqrt{2mE}$ to the right, bounces from the hard wall at $x=0$, and then moves with momentum $p=-\sqrt{2mE}$ to the left.

For $x>0$, the Schrödinger equation is of the form $E\psi (x)=\mathrm{\infty }\psi (x)$, which only can be fulfilled for $\psi (x)=0$. Hence, just as in the classical case, the particle is never found in the hard wall.

For , the Schrödinger equation is just of the same form as in the previous section, Eq. (58), with $V_I=0$. The general solution is again of the form Eq. (59), but now we have to take care of the mathematical property II listed in Sec. III.1: the wave function has to be continuous at $x=0$. Since in the wall $\psi (x)=0$, we have to require that

$$\psi (0)=A+B=0.$$

(67)

Hence, $B=-A$, and the wave function becomes

(68)

Compared to the previous section, we hence have ‘lost’ one free coefficient in the wave function. This is the consequence of the boundary condition $\psi (x)=0$ at the hard wall. Since ${|B|}^2={|A|}^2$, the probability to find the particle moving to the right is equal to the probability to find the particle moving to the left.

A soft wall is a region in space where $V(x)=V_{II}$ is constant and larger than the potential in the bordering regions. Let us consider the potential

(69)

which represents a soft wall in the region $x>0$.

Classically, a soft wall serves the same purpose (namely, being a wall) as a hard wall as long as the total energy . Just as in case of the hard wall, the particle moves with momentum $p=\sqrt{2mE}$ to the right, bounces from the soft wall at $x=0$, and then moves with momentum $p=-\sqrt{2mE}$ to the left.

We call the ‘region I’ and $x>0$ the ‘region II’, and denote the wave function in each region by ${\psi }_I$ and ${\psi }_{II}$, respectively.

In region II ($x>0$), the Schrödinger equation is given by

$$E{\psi }_{II}(x)=-({\mathrm{\hslash }}^2/2m){\psi }_{II}^{\prime \prime }(x)+V_{II}{\psi }_{II}(x),$$

(70)

which is of the same form as Eq. (58) but with $V_I$ replaced by $V_{II}$. Since we assume , the solutions are of the form given in Eq. (65),

$${\psi }_{II}(x)=A_{II}\exp (-{\kappa }_{II}x)+B_{II}\exp ({\kappa }_{II}x),$$

(71)

where ${\kappa }_{II}=\frac{1}{\mathrm{\hslash }}\sqrt{2m(V_{II}-E)}$. When we first encountered this solution, we ruled it out completely, since it was increasing over all bounds for $x\to \pm \mathrm{\infty }$. Now, however, we are only concerned with the behaviour for $x>0$. Hence, we only can conclude that $B_{II}=0$, since the other part of the solution (with coefficient) decays rapidly to 0 for $x\to \mathrm{\infty }$. Thus, we have

$${\psi }_{II}(x)=A_{II}\exp (-\kappa x)\mathit{\hspace{1em}}\text{for }x>0.$$

(72)

For , the Schrödinger equation is also of the form of Eq. (58), but with $V_I=0$. The general solution is given in Eq. (59), so that

$${\psi }_I(x)=A_I\exp (i{k}_{I}x)+B\exp (-i{k}_{I}x),k_I=\frac{1}{\mathrm{\hslash }}\sqrt{2mE}.$$

(73)

Now we have to join the two regions I and II together. This procedure is called wave matching, and again is based on the continuity requirements (mathematical property II above). First of all, the wave function has to be continuous at $x=0$. This gives

	${\psi }_I(0)$	$=$	${\psi }_{II}(0),$		(74)
	$A_I+B_I$	$=$	$A_{II}.$		(75)

Moreover, since the potential is finite, the first derivative of the wave function has to be continuous as well:

	${\psi }_I^{\prime }(0)$	$=$	${\psi }_{II}^{\prime }(0),$		(76)
	$ik(A_I-B_I)$	$=$	$-\kappa A_{II}.$		(77)

We have two conditions (75), (77), for three coefficients $A_I$, $B_I$, and $A_{II}$. So, we may choose $A_I$ as a given parameter and express the $B_I$ and $A_{II}$ in terms of it. This results in

$$ik(A_I-B_I)=-\kappa (A_I+B_I)\Rightarrow B_I=\frac{ik+\kappa }{ik-\kappa }{A}_I$$

(78)

and

$$A_{II}=A_I+B_I\Rightarrow A_{II}=\frac{2ik}{ik-\kappa }{A}_I.$$

(79)

Just as in the previous section, we only have one free coefficient in the wave function, $A_I$. In region $I$, the particle is described by the wave function ${\psi }_I$. Since ${|B_I|}^2={|A_I|}^2$ (please check!), the probability to find the particle moving to the right is still equal to the probability to find the particle moving to the left. In region $II$, the particle is described by the wave function ${\psi }_{II}$.

From our solution we find that $A_{II}$ is finite, hence, ${\psi }_{II}(x)$ does not vanish in the classically forbidden region II. This physical phenomenon is known as tunnelling into classically forbidden regions. A quantum particle can be found in regions of space which are energetically forbidden for a classical particle. One also says that the particle tunnels into the classically forbidden region.

In order to illustrate the dramatic consequences of tunnelling, consider the potential

(80)

which represents a potential barrier of height $V$ and length $L$. For , a classical particle arriving from the left (region I) would be reflected at the soft wall at $x=0$ and could never enter region III.

In order to find the quantum-mechanical solution of this problem, first observe that the potential is constant in each of the three regions, so that we can readily write down the separate solutions of the Schrödinger equation,

	${\psi }_I(x)=A_I{e}^{ikx}+B_I{e}^{-ikx}$		(81)
	${\psi }_{II}(x)=A_{II}{e}^{-\kappa x}+B_{II}{e}^{\kappa x}$		(82)
	${\psi }_{III}(x)=A_{III}{e}^{ikx}+B_{III}{e}^{-ikx},$		(83)

where $k=\frac{1}{\mathrm{\hslash }}\sqrt{2mE}$ and $\kappa =\frac{1}{\mathrm{\hslash }}\sqrt{2m(V_0-E)}$.

The amplitudes $A_I$ and $B_{III}$ describe incoming particles from the left and right, respectively. The amplitudes $B_I$ and $A_{III}$ describe outgoing particles. As mentioned above, we assume that the particle initially arrives from the left, described by the amplitude $A_I$, but does not arrive from the right. This requires us to set $B_{III}=0$. The amplitude $B_I$ describes the case that the particle is reflected from the barrier, while $A_{III}$ describes the case that the particle is transmitted. The probabilities $R$ of reflection and $T$ of transmission are given by

$$R=\frac{{|B_I|}^2}{{|A_I|}^2},T=\frac{{|A_{III}|}^2}{{|A_I|}^2}.$$

(84)

In order to calculate these probabilities, we again use continuity (the mathematical property II) and find four conditions for the five coefficients (setting $B_{III}=0$ as just discussed):

	${\psi }_I(0)={\psi }_{II}(0)$		$\Rightarrow A_I+B_I=A_{II}+B_{II}$
	${\psi }_I^{\prime }(0)={\psi }_{II}^{\prime }(0)$		$\Rightarrow ik(A_I-B_I)=-\kappa (A_{II}-B_{II})$
	${\psi }_{II}(L)={\psi }_{III}(L)$		$\Rightarrow A_{II}{e}^{-\kappa L}+B_{II}{e}^{\kappa L}=A_{III}{e}^{ikL}$
	${\psi }_{II}^{\prime }(L)={\psi }_{III}^{\prime }(L)$				(85)
			$\Rightarrow -\kappa (A_{II}{e}^{-\kappa L}-B_{II}{e}^{\kappa L})=ik{A}_{III}{e}^{ikL}.$

This linear system of equations can be solved (e.g. by Gaussian elimination) in order to express all coefficients by the amplitude $A_I$ of the incoming particle. In particular, one finds the slightly unwieldy expression

$$A_{III}=A_I\frac{4ik\kappa \exp (ikL)}{{(\kappa +ik)}^2\exp (-\kappa L)-{(\kappa -ik)}^2\exp (\kappa L)}.$$

(86)

The transmission probability can be written as

	$T$	$=$	${\displaystyle \frac{{|A_{III}|}^2}{{|A_I|}^2}}$		(87)
		$=$	${\displaystyle \frac{4k^2{\kappa }^2}{4k^2{\kappa }^2+(k^2+{\kappa }^2){\sinh }^2(\kappa L)}}$		(88)
		$=$	${\displaystyle \frac{4E(V_0-E)}{4E(V_0-E)+V_0^2{\sinh }^2(\kappa L)}}.$		(89)

Here $\sinh (x)=\frac{1}{2}(e^x-e^{-x})$ is the hyperbolic sine function. Since $T>0$ we have to conclude that there is a finite probability that the particle tunnels through the wall.

For along barrier with $\kappa L\gg 1$, we can use that $\sinh x\sim e^x/2\gg 1$ for $x\to \mathrm{\infty }$, and simplify

$T={\displaystyle \frac{16E(V_0-E)}{V_0^2}}{e}^{-2\kappa L}.$

(90)

Hence, the tunnel probability becomes very small for long (macroscopic) barriers. The probability is also very small for a very high wall ($V_0\gg E$).

This physical phenomenon is known as tunnelling through a classically forbidden region. A quantum particle can penetrate walls, especially if they are short and energetically not too high.

For the problems discussed so far in this chapter, the particle can always escape to $x=-\mathrm{\infty }$, and in some cases also to $x=+\mathrm{\infty }$. Solutions of this type are called extended states. They cannot be normalised (the integral ${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{|\psi (x)|}^2𝑑x$ diverges), but still permit a probabilistic interpretation for the momentum (propagation direction) of the particle. The energy of extended states can be changed continuously. This is contrast to the energy of a particle in a square well, which we discussed in the preceding chapter and briefly revisit in the following section.

This is identical to the particle in the square well (see section IV). There, we encountered a type of solution which is generally known as bound states. This describes a quantum particle which is confined in a potential well (so that it cannot escape to $x=\pm \mathrm{\infty }$). The associated states are normalisable. Importantly, bound-state solutions of the Schrödinger equation can only be found for discrete (quantised) values of energy.

As mentioned before, the lowest bound state is called the ground state. Classically, the energy is minimised when the particle rests in the minimum of the potential. The quantum-mechanical ground-state energy, however, is elevated above this classical minimum by the zero-point energy. For the particle in the square well, the zero point energy is $E_1=\frac{{\pi }^2{\mathrm{\hslash }}^2}{2m{L}^2}$. Later we will see that this is due to the fact that a quantum particle can never be at rest at a fixed position, a restriction which is captured by Heisenberg’s uncertainty principle.

The distinction of bound states and extended states can be carried over to general one-dimensional potentials with arbitrary position dependence. Generally, one finds:

• •

  bound states at energies where the particle cannot escape to infinity. Only discrete values of the energy are allowed. The wavefunction can be normalised.

• •

  extended states at energies where the particle can escape to infinity (including escape by tunneling). A continuous range of energies is allowed.