《量子力学教程》周世勋-课后答案.pdf

1 量子力学课后习题详解 第一章量子理论基础 11 由黑体辐射公式导出维恩位移定律：能量密度极大值所对应的波长 m 与温度 T 成反比，即 m T=b（常量） ； 并近似计算 b 的数值，准确到二位有效数字。 解根据普朗克的黑体辐射公式 dv e c hv d kT hv vv 1 18 3 3 = ，（1） 以及cv=，（2） ddv vv =，（3） 有 , 1 18 )( )( 5 = = = = kT hc v v e hc c d c d d dv 这里的 的物理意义是黑体内波长介于与+d之间的辐射能量密度。 本题关注的是取何值时， 取得极大值，因此，就得要求 对的一 阶导数为零， 由此可求得相应的的值， 记作 m 。 但要注意的是， 还需要验证 对的二阶导数在 m 处的取值是否小于零，如果小于零，那么前面求得的 m 就 是要求的，具体如下： 0 1 1 5 1 18 6 = + = kT hc kT hc e kT hc e hc 2 0 1 1 5= + kT hc e kT hc kT hc e kT hc = )1(5 如果令 x= kT hc ，则上述方程为 xe x = )1(5 这是一个超越方程。首先，易知此方程有解：x=0，但经过验证，此解是平庸的； 另外的一个解可以通过逐步近似法或者数值计算法获得：x=4.97，经过验证，此 解正是所要求的，这样则有 xk hc T m = 把 x 以及三个物理常量代入到上式便知 KmT m = 3 109 . 2 这便是维恩位移定律。据此，我们知识物体温度升高的话，辐射的能量分布的峰 值向较短波长方面移动，这样便会根据热物体（如遥远星体）的发光颜色来判定 温度的高低。 12 在 0K 附近，钠的价电子能量约为 3eV，求其德布罗意波长。 解根据德布罗意波粒二象性的关系，可知 E=hv， h P= 如果所考虑的粒子是非相对论性的电子（ 2 cE e 由于(1)、(3)方程中，由于=)(xU，要等式成立，必须 0)( 1 =x 0)( 2 =x 即粒子不能运动到势阱以外的地方去。 方程(2)可变为0)( 2)( 2 22 2 2 =+x mE dx xd 令 2 2 2 mE k=，得 0)( )( 2 2 2 2 2 =+xk dx xd 其解为kxBkxAxcossin)( 2 += 根据波函数的标准条件确定系数 A，B，由连续性条件，得 )0()0( 12 = )()( 32 aa= 0=B 0sin=kaA ), 3 , 2 , 1( 0sin 0 = = nnka ka A x a n Ax sin)( 2 = 11 由归一化条件 1)( 2 = dxx 得1sin 0 22 = a xdx a n A 由 mn a b a xdx a n x a m = 2 sinsin x a n a x a A sin 2 )( 2 2 = = 2 2 2 mE k= ), 3 , 2 , 1( 2 2 2 22 =nn ma En 可见 E 是量子化的。 对应于 n E的归一化的定态波函数为 = ax axU xU , 0 , 0 )( 0 运动，求束缚态( 0 0UE ，求： (1)粒子动量的几率分布函数； (2)粒子的平均动量。 解：(1)先求归一化常数，由 = = = = = = = 0 0 0 0 2 2 2 22 2 2 22 2 2 2 2 2 2 2 ) ) ) )( ( ( (1 1 1 1dxdxdxdxe e e ex x x xA A A Adxdxdxdxx x x x x x x x 2 2 2 2 3 3 3 3 4 4 4 4 1 1 1 1 A A A A = = = = 2 2 2 2/ / / /3 3 3 3 2 2 2 2 = = = =A A A A x x x x xexexexex x x x 2 2 2 22 2 2 2/ / / /3 3 3 3 2 2 2 2) ) ) )( ( ( ( = = = =) ) ) )0 0 0 0( ( ( ( x x x x 0 0 0 0) ) ) )( ( ( (= = = =x x x x ) ) ) )0 0 0 0( ( ( ( U U U U，求电子在均匀场外电场作用下穿过金属表面的透射系数。 解：设电场强度为 ，方向沿轴负向，则总势能为 ) ) ) )0 0 0 0( ( ( ( ) ) ) )( ( ( ( = = = =x x x xx x x xe e e ex x x xV V V V ， 0 0 0 0 V V V V 0 0 0 0 ）（）（ = = = =x x x xx x x xe e e eU U U Ux x x x 势能曲线如图所示。则透射系数为 ) ) ) ) ( ( ( (2 2 2 2 2 2 2 2 expexpexpexp 1 1 1 1 2 2 2 2 0 0 0 0 x x x x x x x x dxdxdxdxE E E Ex x x xe e e eU U U UD D D D 式中E E E E为电子能量。0 0 0 0 1 1 1 1= = = = x x x x， 2 2 2 2 x x x x由下式确定 0 0 0 0) ) ) ) ( ( ( (2 2 2 2 0 0 0 0 = = = = = = = =E E E Ex x x xe e e eU U U Up p p p e e e e E E E EU U U U x x x x = = = = 0 0 0 0 2 2 2 2 令 2 2 2 20 0 0 0 sinsinsinsin e e e e E E E EU U U U x x x x = = = =，则有 45 ) ) ) )( ( ( (2 2 2 2 3 3 3 3 2 2 2 2 ) ) ) ) 3 3 3 3 coscoscoscos ( ( ( ( ) ) ) )( ( ( (2 2 2 22 2 2 2 sinsinsinsin2 2 2 2) ) ) )( ( ( (2 2 2 2) ) ) ) ( ( ( (2 2 2 2 0 0 0 0 0 0 0 0 2 2 2 2 0 0 0 0 3 3 3 3 0 0 0 0 0 0 0 0 2 2 2 2 0 0 0 0 2 2 2 20 0 0 0 0 0 0 00 0 0 0 1 1 1 1 2 2 2 2 E E E EU U U U e e e e E E E EU U U U E E E EU U U U e e e e E E E EU U U U d d d d e e e e E E E EU U U U E E E EU U U UdxdxdxdxE E E Ex x x xe e e eU U U U x x x x x x x x = = = = = = = = = = = = 透射系数 ) ) ) )( ( ( (2 2 2 2 3 3 3 3 2 2 2 2 expexpexpexp 0 0 0 0 0 0 0 0 E E E EU U U U e e e e E E E EU U U U D D D D 5指出下列算符哪个是线性的，说明其理由。 2 2 2 2 2 2 2 2 2 2 2 2 4 4 4 4 dxdxdxdx d d d d x x x x； 2 2 2 2 ； = = = = n n n n K K K K1 1 1 1 解： 2 2 2 2 2 2 2 2 2 2 2 2 4 4 4 4 dxdxdxdx d d d d x x x x是线性算符 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 21 1 1 1 2 2 2 2 2 2 2 2 2 2 2 2 1 1 1 1 2 2 2 22 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 1 1 1 11 1 1 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 22 2 2 21 1 1 11 1 1 1 2 2 2 2 2 2 2 2 2 2 2 2 4 4 4 44 4 4 4 ) ) ) )( ( ( (4 4 4 4) ) ) )( ( ( (4 4 4 4) ) ) )( ( ( (4 4 4 4 u u u u dxdxdxdx d d d d x x x xc c c cu u u u dxdxdxdx d d d d x x x xc c c c u u u uc c c c dxdxdxdx d d d d x x x xu u u uc c c c dxdxdxdx d d d d x x x xu u u uc c c cu u u uc c c c dxdxdxdx d d d d x x x x + + + + = = = = + + + += = = =+ + + + 2 2 2 2 不是线性算符 2 2 2 2 2 2 2 22 2 2 2 2 2 2 2 1 1 1 11 1 1 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 22 2 2 21 1 1 12 2 2 21 1 1 1 2 2 2 2 1 1 1 1 2 2 2 2 1 1 1 1 2 2 2 2 2 2 2 22 2 2 21 1 1 11 1 1 1 2 2 2 2 u u u uc c c cu u u uc c c c u u u uc c c cu u u uu u u uc c c cc c c cu u u uc c c cu u u uc c c cu u u uc c c c + + + + + + + + + + += = = =+ + + + = = = = n n n n K K K K1 1 1 1 是线性算符 = = = = = = = = = = = = = = = = + + + += = = =+ + + += = = =+ + + + N N N N K K K K N N N N K K K K N N N N K K K K N N N N K K K K n n n n K K K K u u u uc c c cu u u uc c c cu u u uc c c cu u u uc c c cu u u uc c c cu u u uc c c c 1 1 1 1 2 2 2 22 2 2 2 1 1 1 1 1 1 1 11 1 1 1 1 1 1 1 2 2 2 22 2 2 2 1 1 1 1 1 1 1 11 1 1 1 1 1 1 1 2 2 2 22 2 2 21 1 1 11 1 1 1 6指出下列算符哪个是厄米算符，说明其理由。 2 2 2 2 2 2 2 2 4 4 4 4 dxdxdxdx d d d d dxdxdxdx d d d d i i i i dxdxdxdx d d d d ， 46 不是厄米算符不是厄米算符 ，当当 解：解： dxdxdxdx d d d d dxdxdxdx dxdxdxdx d d d d dxdxdxdx dxdxdxdx d d d d dxdxdxdx dxdxdxdx d d d d dxdxdxdx dxdxdxdx d d d d x x x x dxdxdxdx dxdxdxdx d d d d dxdxdxdx dxdxdxdx d d d d * * * *) ) ) )( ( ( ( * * * *) ) ) )( ( ( ( * * * * * * * * 0 0 0 00 0 0 0 * * * * * * * * * * * * - - - - = = = = = = = = = = = = 是厄米算符是厄米算符 dxdxdxdx d d d d i i i i dxdxdxdx dxdxdxdx d d d d i i i idxdxdxdx dxdxdxdx d d d d i i i i dxdxdxdx dxdxdxdx d d d d i i i ii i i idxdxdxdx dxdxdxdx d d d d i i i i = = = = = = = = = = = = * * * *) ) ) )( ( ( ( * * * *) ) ) )( ( ( ( * * * * * * * * * * * * - - - - 是厄米算符是厄米算符 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 - - - - 2 2 2 2 2 2 2 2 4 4 4 4 * * * *) ) ) )4 4 4 4( ( ( ( * * * *4 4 4 4 * * * * 4 4 4 4 * * * * 4 4 4 4 * * * * 4 4 4 4 * * * * 4 4 4 4 * * * *4 4 4 4 4 4 4 4* * * * dxdxdxdx d d d d dxdxdxdx dxdxdxdx d d d d dxdxdxdx dxdxdxdx d d d d dxdxdxdx dxdxdxdx d d d d dxdxdxdx d d d d dxdxdxdx dxdxdxdx d d d d dxdxdxdx d d d d dxdxdxdx dxdxdxdx d d d d dxdxdxdx d d d d dxdxdxdx d d d d dxdxdxdx dxdxdxdx d d d d = = = = = = = = + + + += = = = = = = = = = = = 7、下列函数哪些是算符 2 2 2 2 2 2 2 2 dxdxdxdx d d d d 的本征函数，其本征值是什么？ 2 2 2 2 x x x x， x x x x e e e e，x x x xsinsinsinsin ，x x x xcoscoscoscos3 3 3 3，x x x xx x x xcoscoscoscossinsinsinsin+ + + + 解：2 2 2 2) ) ) )( ( ( ( 2 2 2 2 2 2 2 2 2 2 2 2 = = = =x x x x dxdxdxdx d d d d 2 2 2 2 x x x x不是 2 2 2 2 2 2 2 2 dxdxdxdx d d d d 的本征函数。 x x x xx x x x e e e ee e e e dxdxdxdx d d d d = = = = 2 2 2 2 2 2 2 2 x x x x e e e e不是 2 2 2 2 2 2 2 2 dxdxdxdx d d d d 的本征函数，其对应的本征值为 1。 47 x x x xx x x x dxdxdxdx d d d d x x x x dxdxdxdx d d d d sinsinsinsin) ) ) )(cos(cos(cos(cos) ) ) )(sin(sin(sin(sin 2 2 2 2 2 2 2 2 = = = = = = = 可见，x x x xsinsinsinsin是 2 2 2 2 2 2 2 2 dxdxdxdx d d d d 的本征函数，其对应的本征值为1。 ) ) ) )coscoscoscos3 3 3 3( ( ( (coscoscoscos3 3 3 3) ) ) )sinsinsinsin3 3 3 3( ( ( () ) ) )coscoscoscos3 3 3 3( ( ( ( 2 2 2 2 2 2 2 2 x x x xx x x xx x x x dxdxdxdx d d d d x x x x dxdxdxdx d d d d = = = = = = = = x x x xcoscoscoscos3 3 3 3是 2 2 2 2 2 2 2 2 dxdxdxdx d d d d 的本征函数，其对应的本征值为1。 ) ) ) )coscoscoscos(sin(sin(sin(sin coscoscoscossinsinsinsinsinsinsinsin(cos(cos(cos(cos) ) ) )coscoscoscos(sin(sin(sin(sin 2 2 2 2 2 2 2 2 x x x xx x x x x x x xx x x xx x x xx x x x dxdxdxdx d d d d x x x xx x x x dxdxdxdx d d d d + + + + = = = = = = = = = = = =+ + + +） x x x xx x x xcoscoscoscossinsinsinsin+ + + +是 2 2 2 2 2 2 2 2 dxdxdxdx d d d d 的本征函数，其对应的本征值为1。 8、试求算符 dxdxdxdx d d d d ie ie ie ieF F F F ixixixix = = = = 的本征函数。 解：F F F F 的本征方程为 F F F FF F F F= = = = c c c c dxdxdxdx d d d d FeFeFeFe dxdxdxdx d d d d FeFeFeFed d d d dxdxdxdx d d d d FeFeFeFed d d ddxdxdxdxiFeiFeiFeiFe d d d d F F F F dxdxdxdx d d d d ie ie ie ie ixixixix ixixixixixixixixixixixix ixixixix ln ln ln lnln ln ln ln ) ) ) )( ( ( () ) ) )( ( ( ( + + + + = = = = = = = = = = = = = = = = = = = 即即 ix ix ix ix FeFeFeFe cececece = = = = （F F F FF F F F是是 的本征值） 9、如果把坐标原点取在一维无限深势阱的中心，求阱中粒子的波函数和能级的 表达式。 解： = = = = 2 2 2 2 , , , , 2 2 2 2 , , , , 0 0 0 0 ) ) ) )( ( ( ( a a a a x x x x a a a a x x x x x x x xU U U U 48 方程（分区域） ： ： = = = =) ) ) )( ( ( (x x x xU U U U0 0 0 0) ) ) )( ( ( (= = = =x x x x I I I I ) ) ) ) 2 2 2 2 ( ( ( ( a a a a x x x x ： = = = =) ) ) )( ( ( (x x x xU U U U0 0 0 0) ) ) )( ( ( (= = = =x x x x IIIIIIIIIIII ) ) ) ) 2 2 2 2 ( ( ( ( a a a a x x x x ： II II II II II II II II E E E E dxdxdxdx d d d d = = = = 2 2 2 2 2 2 2 22 2 2 2 2 2 2 2 0 0 0 0 2 2 2 2 2 2 2 22 2 2 2 2 2 2 2 = = = =+ + + + II II II II II II II II E E E E dxdxdxdx d d d d 令 2 2 2 2 2 2 2 2 2 2 2 2 E E E E k k k k = = = = 0 0 0 0 2 2 2 2 2 2 2 2 2 2 2 2 = = = =+ + + + II II II II II II II II k k k k dxdxdxdx d d d d ) ) ) )sin(sin(sin(sin( + + + += = = =kxkxkxkxA A A A II II II II 标准条件： = = = = = = = = ) ) ) ) 2 2 2 2 ( ( ( () ) ) ) 2 2 2 2 ( ( ( ( ) ) ) ) 2 2 2 2 ( ( ( () ) ) ) 2 2 2 2 ( ( ( ( a a a aa a a a a a a aa a a a IIIIIIIIIIIIII II II II II II II III I I I 0 0 0 0) ) ) )sin(sin(sin(sin(= = = =+ + + + kxkxkxkxA A A A 0 0 0 0 A A A A 0 0 0 0) ) ) )sin(sin(sin(sin(= = = =+ + + + kxkxkxkx 取0 0 0 0 2 2 2 2 = = = = a a a a k k k k ，即 2 2 2 2 a a a a k k k k= = = = ) ) ) ) 2 2 2 2 ( ( ( (sinsinsinsin) ) ) )( ( ( ( a a a a x x x xk k k kA A A Ax x x x II II II II + + + += = = = 0 0 0 0sinsinsinsin= = = =kakakakaA A A A 0 0 0 0sinsinsinsin = = = =kakakaka n n n nkakakaka= = = =) ) ) ) , , , ,2 2 2 2 , , , ,1 1 1 1( ( ( (= = = =n n n n n n n n a a a a k k k k = = = = 49 粒子的波函数为 + + + + = = = = 2 2 2 2 , , , , 0 0 0 0 2 2 2 2 ), ), ), ), 2 2 2 2 ( ( ( (sinsinsinsin ) ) ) )( ( ( ( a a a a x x x x a a a a x x x x a a a a x x x x a a a a n n n n A A A A x x x x 粒子的能级为 a a a a k k k kn n n n k k k kE E E E 2 2 2 22 2 2 2 2 2 2 22 2 2 22 2 2 2 2 2 2 2 2 2 2 2 = = = = = = = ) ) ) ) , , , , 3 3 3 3 , , , , 2 2 2 2 , , , , 1 1 1 1( ( ( (= = = =n n n n 由归一化条件，得 + + + += = = = = = = 2 2 2 2/ / / / 2 2 2 2/ / / / 2 2 2 22 2 2 2 2 2 2 2 ) ) ) ) 2 2 2 2 ( ( ( (sinsinsinsin) ) ) )( ( ( (1 1 1 1 a a a a a a a a dxdxdxdx a a a a x x x x a a a a n n n n A A A Ad d d dx x x x + + + + = = = = 2 2 2 2/ / / / 2 2 2 2/ / / / 2 2 2 2 ) ) ) ) 2 2 2 2 ( ( ( ( 2 2 2 2 coscoscoscos1 1 1 1 2 2 2 2 1 1 1 1 a a a a a a a a dxdxdxdx a a a a x x x x a a a a n n n n A A A A + + + + = = = = 2 2 2 2/ / / / 2 2 2 2/ / / / 2 2 2 22 2 2 2 ) ) ) ) 2 2 2 2 ( ( ( ( 2 2 2 2 coscoscoscos 2 2 2 2 a a a a a a a a dxdxdxdx a a a a x x x x a a a a n n n n A A A A a a a a A A A A 2 2 2 2 2 2 2 2 2 2 2 22 2 2 2 ) ) ) ) 2 2 2 2 ( ( ( ( 2 2 2 2 sinsinsinsin 2 2 2 22 2 2 2 a a a a a a a a a a a a x x x x a a a a n n n n n n n n a a a a A A A AA A A A a a a a + + + + = = = = 2 2 2 2 2 2 2 2 A A A A a a a a = = = = a a a a A A A A 2 2 2 2 = = = = 粒子的归一化波函数为 + + + + = = = = 2 2 2 2 , , , , 0 0 0 0 2 2 2 2 ), ), ), ), 2 2 2 2 ( ( ( (sinsinsinsin 2 2 2 2 ) ) ) )( ( ( ( a a a a x x x x a a a a x x x x a a a a x x x x a a a a n n n n a a a a x x x x 10、证明：处于 1s、2p 和 3d 态的氢原子中的电子，当它处于距原子核的距离分 别为 0 0 0 00 0 0 00 0 0 0 9 9 9 94 4 4 4a a a aa a a aa a a a、的球壳处的几率最（ 0 0 0 0 a a a a为第一玻尔轨道半径） 。 证：drdrdrdrr r r rR R R Rdrdrdrdrr r r rs s s s 2 2 2 2 2 2 2 2 1010101010101010 ) ) ) )( ( ( ( : : : :1 1 1 1= = = = drdrdrdrr r r re e e e a a a a a a a ar r r r 2 2 2 2 / / / /2 2 2 2 3 3 3 3 0 0 0 0 0 0 0 0 4 4 4 4) ) ) ) 1 1 1 1 ( ( ( ( = = = = 50 0 0 0 0 / / / /2 2 2 2 2 2 2 23 3 3 3 0 0 0 0 10101010 4 4 4 4) ) ) ) 1 1 1 1 ( ( ( () ) ) )( ( ( ( a a a ar r r r e e e er r r r a a a a r r r r = = = = 0 0 0 0 / / / /2 2 2 22 2 2 2 0 0 0 0 3 3 3 3 0 0 0 0 10101010 ) ) ) ) 2 2 2 2 2 2 2 2( ( ( () ) ) ) 1 1 1 1 ( ( ( (4 4 4 4 a a a ar r r r e e e er r r r a a a a r r r r a a a adrdrdrdr d d d d = = = = 0 0 0 0 / / / /2 2 2 2 0 0 0 0 3 3 3 3 0 0 0 0 ) ) ) ) 1 1 1 1 1 1 1 1( ( ( () ) ) ) 1 1 1 1 ( ( ( (8 8 8 8 a a a ar r r r rere rerer r r r a a a aa a a a = = = = 令0 0 0 0 10101010 = = = = drdrdrdr d d d d ，则得 0 0 0 0 11111111 = = = =r r r r 0 0 0 011111111 a a a ar r r r= = = = ) ) ) )1 1 1 1( ( ( () ) ) ) 2 2 2 2 1 1 1 1( ( ( () ) ) ) 1 1 1 1 ( ( ( (8 8 8 8 0 0 0 0 / / / /2 2 2 2 0 0 0 00 0 0 00 0 0 0 3 3 3 3 0 0 0 0 2 2 2 2 10101010 2 2 2 2 a a a ar r r r e e e e a a a a r r r r a a a a r r r r r r r r a a a aa a a adrdrdrdr d d d d = = = = ) ) ) ) 2 2 2 24 4 4 4 1 1 1 1( ( ( () ) ) ) 1 1 1 1 ( ( ( (8 8 8 8 0 0 0 0 / / / /2 2 2 2 2 2 2 2 0 0 0 0 2 2 2 2 0 0 0 0 3 3 3 3 0 0 0 0 a a a ar r r r e e e e a a a a r r r r a a a a r r r r a a a a + + + + = = = = 0 0 0 0 0 0 0 0 2 2 2 2 10101010 2 2 2 2 11111111 = = = =r r r r drdrdrdr d d d d 0 0 0 0 11111111= = = = r r r r为几率最小处。 0 0 0 0 0 0 0 011111111 2 2 2 2 10101010 2 2 2 2 drdrdrdr d d d d 0 0 0 0= = = =r r r r为几率最小位置。 0 0 0 0 3 3 3 3 2 2 2 2 6 6 6 6 7 7 7 7 0 0 0 0 2 2 2 2 3232323232323232 98415984159841598415 8 8 8 8 ) ) ) )( ( ( ( : : : :3 3 3 3 a a a a r r r r e e e er r r r a a a a R R R Rr r r rd d d d = = = = = = = 0 0 0 0 3 3 3 3 2 2 2 2 5 5 5 5 0 0 0 0 7 7 7 7 0 0 0 0 32323232 ) ) ) ) 3 3 3 3 2 2 2 2 5 5 5 5( ( ( ( 98415984159841598415 8 8 8 8 a a a a r r r r e e e er r r r a a a a r r r r a a a adrdrdrdr d d d d = = = = 令0 0 0 0 32323232 = = = = drdrdrdr d d d d ，得 , , , , 0 0 0 0 31313131 = = = =r r r r 0 0 0 032323232 9 9 9 9a a a ar r r r= = = = 同理可知0 0 0 0 31313131 = = = =r r r r为几率最小处。 0 0 0 032323232 9 9 9 9a a a ar r r r= = = =为几率最大处。 11、求一维谐振子处在第一激发态时几率最大的位置。 解： 2 2 2 22 2 2 2 2 2 2 2 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 ) ) ) )( ( ( ( x x x x xexexexex x x x = = = = 2 2 2 22 2 2 2 2 2 2 2 3 3 3 3 2 2 2 2 1 1 1 11 1 1 1 2 2 2 2 ) ) ) )( ( ( () ) ) )( ( ( ( x x x x e e e ex x x xx x x xx x x x = = = = = = = 2 2 2 22 2 2 2 ) ) ) )( ( ( ( 4 4 4 4 3 3 3 32 2 2 2 3 3 3 3 1 1 1 1x x x x e e e ex x x xx x x x dxdxdxdx d d d d = = = = 2 2 2 22 2 2 2 ) ) ) )1 1 1 1( ( ( ( 4 4 4 4 2 2 2 22 2 2 2 3 3 3 3 x x x x xexexexex x x x = = = = 2 2 2 22 2 2 2 ) ) ) )2 2 2 25 5 5 51 1 1 1( ( ( ( 4 4 4 4 4 4 4 44 4 4 42 2 2 22 2 2 2 3 3 3 3 2 2 2 2 1 1 1 1 2 2 2 2 x x x x e e e ex x x xx x x x dxdxdxdx d d d d + + + + = = = = 令0 0 0 0 1 1 1 1 = = = = dxdxdxdx d d d d ，得 52 0 0 0 0 1 1 1 1= = = = x x x x， 0 0 0 0 0 0 0 0 2 2 2 2 2 2 2 2 1 1 1 1 x x x xx x x x = = = = = = = = = = = = 0 0 0 0 0 0 0 0