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`CHAPTER V One-Dimensional Potential Wells and Barriers Part I. The Delta Function Potential Introduction So far, we have encountered two very different types of solution to the time independent Schrödinger equation. In the case of the infinite square well the Schrödinger equation s L &lt;8 ÐBÑ oe I8 &lt;8 ÐBÑ (5.1)gives us a set of normalizable, orthogonal wave functions (or eigenfunctions) &lt;8 ÐBÑ, each associated with a specific, discrete energy value (or eigenvalue) I8 . The most general time-dependent solution to the Schrödinger equation in this case is a linear combination of these eigenfunctions of the form G Bß &gt; oe8-8 /3=8 &gt;&lt;8 ÐBÑ(5.2)where =8 oe I8 Îh , and where the expansion coefficients, -8 , are determined from the wave function at time &gt; oe ! from the equation -8 oe (  * G Bß ! &lt;8 ÐBÑ .B(5.3)These coefficients must also satisfy the normalization condition l-8 l# oe &quot;8(5.4)The expectation value of energy of this system is given by s ØLÙ oe8l-8 l# I8(5.5)In the case of the free particle, however, we encountered a solution to Schrödinger's time-independent equation which was associated with a continuous range of possible energy values (for I !). The most general normalizable solution to the Schrödinger equation in this case is the integral GÐBß &gt;Ñ oe where 9Ð5Ñ oe &quot; È #1 (&quot; È #1 ( 9Ð5Ñ/3 5B=&gt;.5(5.6)GÐBß !Ñ/35B.B(5.7)One-Dimensional Potential Wells and Barriers2Although the &quot;eigenfunctions&quot; of the momentum operator are not properly normalizable wavefunctions, the expectation value of the energy for a particle described by our free-particle wave function can be written in terms of an energy distribution dependent upon the function 9Ð5Ñ. By analogy, we would expect the expectation value of the energy to be given by something like s ØLÙ oe ( l 9 5 l#h5 # .5 #7(5.8)You should perform this operation starting with the free-particle wave functions s :# GÐBß &gt;Ñ, and the Hamiltonian operator L oe s Î#7 expressed in configuration space (where s oe : 3h `Î`BÑ and show that you obtain the result above. Notice that this expression is just the expectation value of the energy (expressed in terms of the momentum : oe h5 ) in momentum space. The Delta-Function Potential Well Problem We now want to consider a particularly simple (though somewhat unrealistic) problem which exhibits both of these types of solutions - the delta-function potential problem. In this problem, we assume that the potential energy is of the form Z ÐBÑ oe  \$ ÐBÑ (5.9)which acts something like an infinitely narrow, infinitely deep potential well in an otherwise constant potential background. We will approach the solution to this problem much like we did in the case of the infinite square well. Schrödinger's time-independent equation for this problem is h# ` 2 &lt;ÐBÑ #7 `B# Z B &lt;ÐBÑ oe I &lt;ÐBÑ (5.10)where Z B oe  \$ B . This potential function is somewhat unique and looks something like what is shown in the diagram below, with the effective depth of the well going to negative infinity, but with the width of the well being so narrow that the area of the well is unity! This means that for a \$ -function potential, region II in the diagram below has zero width. We simply draw this potential with a finite width to be able to think clearly about what we are doing. Because region II has zero width, we only need to be concerned about the wave function in regions I and III.One-Dimensional Potential Wells and BarriersV(x)3x+-Figure 4.1 The delta-function Potential Well We must now find the solution to Schrödinger's equation in regions I and III. In each of these regions we must consider two different cases, one where I ! and one where I !. We will first examine the case where I !. The Delta-Function Potential Well Solution for I In region I and III, the Schrödinger equation is h# ` 2 &lt;ÐBÑ oe I &lt;ÐBÑ #7 `B# or `2 &lt;ÐBÑ oe `B# which has a solution of the form &lt;ÐBÑ oe E/-B Plugging this back into the Schrödinger equation reveals that - oe ,, 35 In the case where I !, 5 is imaginary, so we write 5oeÊ and the solution becomes &lt;M ÐBÑ oe E/ &lt;MMM ÐBÑ oe J /,B!(5.11)#7I &lt;ÐBÑ oe h#5# &lt;ÐBÑ(5.12)(5.13)(5.14)#7I #7lIl oeÊ oe 3, h# h#(5.15)F/ K/,Bfor B for B! !(5.16) (5.17),B,BOne-Dimensional Potential Wells and Barriers4Now, in region I, where B !, the solution blows up as B p , unless F oe !. Likewise, in region III, where B !, the solution blows up as B p , unless J oe !. This means that the solutions in region I and III are &lt;M ÐBÑ oe E/ &lt;MMM ÐBÑ oe K/,Bfor B for B! !(5.18) (5.19),BAs we have discussed earlier, the wave function must be continuous to have meaning as a probability amplitude. This means that for a \$ -function potential where region II in the diagram above has no real width, the wave function in region I must have the same value as the wave function in region III at the point where B oe !. This requires that E oe K, so the solution to Schrödinger's equation is &lt;M ÐBÑ oe E/ &lt;MMM ÐBÑ oe E/ or &lt;MßMMM ÐBÑ oe E/, lBl ,Bfor B for B! !(5.20) (5.21),B(5.22)The second condition on the wave function solution to Schrödinger's equation is that the first derivative of the wave function must be continuous for piecewise-continuous potentials. However, in the case of the delta-function potential, the potential in not piecewise continuous - it is infinite. Let's look again at the requirement imposed on the first derivative of the wave function by the Schrödinger equation. As you will remember, we integrated the Schrödinger equation with respect to B over a small interval ?% h # Bo % ` 2 &lt;ÐBÑ .B ( #7 Bo % `B# (Bo % Bo %Z B &lt;ÐBÑ .B oe I (Bo %&lt;ÐBÑ .BBo %(5.23)In the limit as ?% p ! the integral over the wave function must go to zero, since it is a continuous, single-valued function. The integral of the second derivative of the wave function is just the first derivative, so that we are left with ` &lt;ÐBÑ Bo lim º ?%Ä! `B Bo% %oe lim#7 Bo % Z B &lt;ÐBÑ .B ( ?%Ä! h # Bo %(5.24)As pointed out in the last chapter, the integral of the product of the potential energy function and the wave function can be represented by the diagram shown for piecewisecontinuous potential energy functions i.e., those having a finite number of finite steps .One-Dimensional Potential Wells and BarriersV(x)5(X)xHowever, for potential energy functions which have infinite steps, such as the infinite square well and the delta-function potential, the first derivative of the wave function may not be continuous at a boundary. In fact, using the definition of the deltafunction, we can evaluate the integral of the product of the wave function and the potential energy function to obtain ` &lt;ÐBÑ º ?%Ä! `B lim which we write as lim OE ` &lt;ÐBÑ º `B ` &lt;ÐBÑ º oe `B % #7 &lt;Ð!Ñ oe h# #7 E h#% %oe lim#7 ( ?%Ä! h #%\$ ÐBÑ&lt;ÐBÑ .B oe%#7 &lt;Ð!Ñ h#(5.25)?%Ä!(5.26)%Now the first partial derivative is evaluated in the region B is evaluated in the region B !, giving?%Ä!!, while the second partial #7 E h#lim OE ,E/ lim,Bº,E/%,Bº oe%(5.27)?%Ä!#,E/,%oe#7 E h#(5.28)which in the limit as % p 0 fixes the value of , (and therefore the energy I ) according to the equation ,oe 7 #7lIl oeÊ # h h# (5.29)One-Dimensional Potential Wells and Barriers6where  is the &quot;depth&quot; of the delta-function potential. We see from this equation that there is only one allowed energy for I ! given by Ioe 7 # # h #  oe h% #7 7 # #h # (5.30)The only thing remaining for the case where I determine E (! ! is to normalize the wave function andE# /#, B.B(E# /#, B.B oe &quot;(5.31)!E # / #, B ! º #,  E# / #, ­ E# c&quot; #,#, BE # / #, B  oe&quot; º #, ! / #, B(5.32)º!º!--oe&quot;(5.33)!!&quot; doeE# oe&quot; ,(5.34) (5.35)Ê E oe È,Thus, for the delta-function potential, we have only one eigenstate corresponding to an energy I ! given by &lt;I ÐBÑ oe È,/, lBlIoe7 # #h #(5.36)________________________________________________________________________ Problem 5.1 The Double Delta Function Potential Consider the double delta function potential energy function given by Z ÐBÑ oe (a) (b) c\$ ÐB +Ñ \$ ÐB +ÑdSketch this potential energy function. Show that there are at most two possible bound state energies (i.e., energies where I !) allowed. Skematically plot the wave function corresponding to each of these two possible solutions. Do both solutions always exist? (c) In particular, find the allowed energies for the two cases  oe hÎ7+ and  oe hÎ%7+, and sketch the corresponding wave functions. ________________________________________________________________________One-Dimensional Potential Wells and Barriers7The Delta-Function Potential Well Solution for I In region I and III, the Schrödinger equation is, again h# ` 2 &lt;ÐBÑ oe I &lt;ÐBÑ #7 `B# or `2 &lt;ÐBÑ oe `B# which has a solution of the form &lt;ÐBÑ oe E/-B Plugging this back into the Schrödinger equation reveals that - oe ,, 35 In the case where I !, 5 is real, given by 5oeÊ and the solution becomes &lt;M ÐBÑ oe E/ &lt;MMM ÐBÑ oe J /35B!(5.37)#7I &lt;ÐBÑ oe h#5# &lt;ÐBÑ(5.38)(5.39)(5.40)#7I h#(5.41)F/ K/35Bfor B for B! !(5.42) (5.43)35B35BThe time-dependent form of these two equations have the form of two traveling sinusoidal waves moving in opposite directions. Let's assume that we are dealing with a particle (or group of particles) that are originating in the negative half-plane (i.e., in the region where B !). The particles will move in from the left, encounter the deltafunction potential at B oe !, and will either continue moving in the B direction (i.e., they are transmitted through the region of potential change), or will be reflected and move back in a B direction. This means that in the region where B ! we must allow for the possibility of two opposite going waves (or particles), but in the region where B ! there is only one possibility - the particle (or wave) moves only in the B direction. This means that we must require that K oe !, based upon the initial conditions (the assumption that particles are originating in the negative half-plane). This leaves us with the equations &lt;M ÐBÑ oe E/35BF/35B35Bfor B for B! !(5.44) (5.45)&lt;MMM ÐBÑ oe J /One-Dimensional Potential Wells and Barriers8Since the wave function must be continuous, these two equations must be equal at B oe !, giving the condition E FoeJ (5.46)The second condition on the wave function solution to Schrödinger's equation is that the first derivative of the wave function must also be continuous for piecewisecontinuous potentials. But again, the potential in not piecewise continuous - it is infinite. As before, we integrate the Schrödinger equation with respect to B over a small interval ?% h # Bo % ` 2 &lt;ÐBÑ .B ( #7 Bo % `B# (Bo % Bo %Z B &lt;ÐBÑ .B oe I (Bo %&lt;ÐBÑ .BBo %(5.47)Again, in the limit as ?% p ! the integral over the wave function must go to zero, the integral of the second derivative of the wave function is just the first derivative, and so we are left with ` &lt;ÐBÑ Bo lim º ?%Ä! `B Bo% %#7 Bo % oe lim # ( Z B &lt;ÐBÑ .B ?%Ä! h Bo %(5.48)Using the definition of the delta-function, we can evaluate the integral of the product of the wave function and the potential energy function to obtain ` &lt;ÐBÑ lim º ?%Ä! `B which we write as?%Ä! % %#7 oe lim # ( ?%Ä! h%\$ ÐBÑ&lt;ÐBÑ .B oe%#7 &lt;Ð!Ñ h#(5.49)lim OE` &lt;ÐBÑ º `B%` &lt;ÐBÑ º oe `B %#7 E h#F(5.50)Now the first partial derivative is evaluated in the region B is evaluated in the region B !, giving?%Ä!!, while the second partial #7 E h# Flim OE 35J /35Bº%^ 35E/ ^ 35E/35B35 F/35 % 35B º oe%F?%Ä!lim OE 35J /35 %35 %35F/35 % º oe%#7 E h# F?%Ä!lim 35 ^ JF /35 %E/oe#7 E h#(5.51)Now, in the limit this reduces to J F Eoe 3#7 E h#5 F oe #3&quot; E F (5.52)One-Dimensional Potential Wells and Barriers9where &quot; oe 7Îh # 5 . But from the condition on the continuity of the wave function we know that J oe E F , so this last equation reduces to F oe 3&quot; E or Foe Solving for J gives, since J oe E F, J oe &quot; &quot; 3&quot; E (5.55) 3&quot; E &quot; 3&quot; (5.54) F (5.53)It should be obvious that the amplitude E is related to the probability of measuring an incoming particle, while the amplitude F is related to the probability of measuring a reflected particle and the amplitude J to measuring a transmitted particle. To determine a precise relationship between these quantities we return to the concept of the probability current density defined by the equation 4ÐBß &gt;Ñ oe 3h ` GÐBß &gt;Ñ * &quot;G ÐBß &gt;Ñ #7 `B 3h t t t 'G* Ð&lt;ß &gt;ÑfGÐ&lt;ß &gt;Ñ #7 GÐBß &gt;Ñ ` G* ÐBß &gt;Ñ · `B (5.56)In three-dimensions this equation has the form t t &gt;Ñ oe 4Ð&lt;ß t t t GÐ&lt;ß &gt;ÑfG* Ð&lt;ß &gt;Ñ&quot; (5.57)which satisfies the continuity equation t ` 3Ð&lt;Ñ `&gt; t f4oe! (5.58)Recall that this equation states that the change in the probability density at some point is associated with the probability current flowing away from that point. If we integrate the continuity equation over some enclosed volume Z , we obtain ( `3 t 4 [email protected] oe ( f  t [email protected] oe * t  8 .= 4 ^ `&gt; Z W (5.59)Zwhere the last integral is due to Stokes' theorem. Thus, if the probability of finding a particle within an enclosed volume decreases in time, this corresponds to a probability flux, the sum of all probability currents flowing out of an enclosed volume. Thus in three-dimensions t corresponds to the relative number of particles per unit volume 4 flowing outward through the enclosing surface W , times the speed at which the particles leave. This has units of the number of particles passing through a given cross-sectional area per unit time. Thus, if we consider a stream of particles leaving an enclosed volume and incident upon some surface area .W , we would designate 438-3./8&gt; as the relative number of particles per unit volume that were approaching the surface, 4&lt;/06/-&gt;/. as theOne-Dimensional Potential Wells and Barriers10relative number of particles per unit volume that were reflected from that surface and 4&gt;&lt;+8=73&gt;&gt;/. as the relative number of particles per unit volume that were passing through the surface. We define the transmission and reflection coefficients in terms of these current densities according to the equations V´ 4&lt;/06/-&gt;/. 438-3./8&gt; and X ´ 4&gt;&lt;+8=73&gt;&gt;/. 438-3./8&gt; (5.60)The solutions we have obtained for our delta function potential are traveling plane waves of the form GÐBß &gt;Ñ oe E/3 5B =&gt; . The probability current density for a plane wave is given by toe 4 or toe 4 3h * &quot;E / #7 3h  E* / #73 5B =&gt;3h ` * &quot;G ÐBß &gt;Ñ G ÐBß &gt;Ñ #7 `BG ÐBß &gt;Ñ` * G ÐBß &gt;Ñ· `B · `(5.61)` E/3 5B `B 35E/3 5B=&gt;=&gt;E/3 5B=&gt;` * E/ `B 35 E* /3 5B =&gt;(5.62)toe 43 5B =&gt;E/3 5B=&gt;3 5B =&gt;(5.63)toe 43h 35lEl# #7 toe 435 lEl# `(5.64)3h #35lEl# ` #7 t oe h5 lEl# 4 7(5.65)(5.66)This has the form t oe : |E|# oe @|E|# 4 7 (5.67)and will be positive or negative depending upon the sign of 5 . This means that the incident, reflected and transmitted probability current densities for plane waves are givenß respectively, by t 38- oe h5 lEl# 4 7 t&lt;/06 oe h5 |F |# 4 7 (5.68)(5.69)One-Dimensional Potential Wells and Barriers11t&gt;&lt;+8= oe h5 |J |# 4 7 Thus, the reflection coefficient for our delta function potential is given by Voe &quot;# lFl# oe oe lEl# &quot; &quot;# &quot; &quot; &quot;Î&quot; #(5.70)(5.71)and the transmission coefficient is given by X oe lJ l# &quot; oe # lEl &quot; &quot;# (5.72)Notice that in this particular problem, the value of 5 oe È#7 I Z Îh # is the same in regions I and III of figure 4.1, but in other problems, the value of 5 may be different. In those cases, the transmission and reflection coefficients depend upon the value of 5 in the respective regions. You should notice that the sum of the reflection and transmission probabilities equal unity as it must. Now using the definition of &quot; and 5 , we find that &quot;# oe 7 # # 7 # # 7 # oe % oe # h % 5# h #7IÎh # #h I (5.73)So we can express the reflection and transmission coefficients for the delta-function potential well in terms of the energy and mass of the incident particle: &quot; &quot; #h # IÎ7# &quot; 7# Î#h # IVoe(5.74)X oe&quot;(5.75)Thus, for situations where the energy of the incident particle is small, there is a relatively high probability for reflection at B oe !, but as the energy of the incident particle increases, the probability for reflection goes down. However, there is one small problem we have not yet addressed. Plane wave solutions for the free particle are not normalizable, so are not really acceptable solutions! At first glance, this might seem a bit problematic, but we have shown that an acceptable solution can be formed from a combination of plane wave solutions using a Fourier integral. This means that each component of the Fourier integral will be transmitted or reflected with different probability, because each component has a slightly different energy. We can, however, obtain an estimate of the approximate transmission or reflection coefficient based upon the more prevalent component of the wave packet.One-Dimensional Potential Wells and Barriers12The Delta-Function Potential Barrier Problem For the case of a delta-function potential barrier, only the sign of the potential energy function is changed, giving the potential energy is of the form Z ÐBÑ oe  \$ ÐBÑ (5.76)which acts something like an infinitely narrow, infinitely tall potential barrier in an otherwise constant potential background. It should be obvious that there in no bound state solution; if I V everywhere, the wave function solution is the null solution. So we only have to consider the case where I !. For this case all we have to do to find the solution for the infinite barrier is to change the sign of . But the reflection and transmission coefficients are a function only of the square of , so that there is no change! Thus, whether we are dealing with a delta-function well or barrier, the transmission and scattering coefficients are identical. This means that a quantum mechanical particle can penetrate a potential barrier of infinite height - provided it is not too wide! This phenomena is called tunneling. We will see how this same situation plays out for a finite width barriers and wells in the next chapter.`

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