# Chapter 5. , r = r 1 r 2 (1) µ = m 1 m 2. r, r 2 = R µ m 2. R(m 1 + m 2 ) + m 2 r = r 1. m 2. r = r 1. R + µ m 1

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1 Tor Kjellsson Stockholm University Chpter 5 5. Strting with the following informtion: R = m r + m r m + m, r = r r we wnt to derive: µ = m m m + m r = R + µ m r, r = R µ m r 3 = µ m R + r, = µ m R r. 4 Finding R s function of r : R = m r + m r m + m R = m r + m r r m + m Rm + m = m + m r m r Rm + m + m r m + m = r R + m m + m r = r R + µ m r = r. Finding R s function of r : As n exercise, redo the steps but now mke the exchnge r = r + r to obtin: R µ m r = r.

2 Now we focus on showing tht = µ m R + r = µ m R r. To do this we first note the following: = x, y,. 5 z Let us define: R = X, Y, Z nd r = x, y, z 6 r = x, y, z nd r = x, y, z. 7 Let s proceed by finding : Note tht ce r nd r re symmetric in their coordintes we cn just consider one component of, sy the x-component. We will therefore focus on x = x now. Also note tht r depends on both r, R, i.e r r, R. It then follows tht we must mke use of the chin rule: x = X x X + x x x. 8 By combining eq. nd eq. we cn write R on the form X, Y, Z: R = µ r + µ µ r = x + µ µ x, y + µ µ y, z + µ z m m m m m m m m which gives: X x = x 9 [ µ x + µ x = µ. m m m In the sme mnner we now use eq. nd eq. to write r on the form x, y, z: r = r r = x x, y y, z z = x, y, z = x = [x x =. x x Now insert the results from eq. nd eq. into eq.8 to obtin: = µ x m X + x 3 If you don t see it now, just be ptient, you will see it t the end.

3 Becuse R nd r re completely symmetric with respect to the three components, it immeditely follows tht: = µ y m Y + y nd finlly inserting these into eq.5 gives our result: = µ z m Z + z 4 = µ m X, Y, Z + x, y, z = µ m R + r 5 If you wnt to test the steps by yourself you cn now redo them for : = µ m R r 6 bewre of the minus sign tht ppers from eq.. b The Hmiltonin is given by the sum of ech prticle s kinetic energy nd the totl potentil V r: Ĥ = m + m + V r The time independent Schrödinger eqution, Ĥψ = Eψ, then becomes: + ψ + V rψ = Eψ. 7 m m The gol is now to show tht eq.7 is equivlent with: m + m Rψ µ rψ + V rψ = Eψ nd we will show this by first evluting ψ: µ ψ = ψ = R + r m µ = R + r m µ = m µ m R + r µ m R ψ + r ψ Rψ + µ m R r ψ + µ m r R ψ + rψ ψ. nd ce we lwys consider functions with commuting prtil derivtives we get: µ ψ = m R ψ + µ m R r ψ + r ψ. 8 3

4 Similrly one gets do this s n exercise!: µ ψ = m R ψ µ m R r ψ + r ψ. 9 Multiplying these two with the respective fctor m ψ = m [ µ m m i : R ψ + µ m R r ψ + r ψ [ m ψ = µ R ψ µ R r ψ + r ψ m m m we see tht the middle terms will cncel if we dd the two lines. Doing this we obtin: [ m ψ m ψ = µ m m + µ m m R ψ + + r ψ m m [ m ψ m ψ = µ m + m m m + m m R ψ + r ψ m m [ m ψ m ψ = R ψ + r ψ m + m µ Inserting this into eq.7 now gives: m + m R ψ µ r ψ + V rψ = Eψ. c Assuming now tht the wve function cn be written s product of two wve functions: ψr, r = ψ R ψ r. We get: m + m R ψ R ψ r µ r ψ R ψ r + V rψ R ψ r = Eψ R ψ r. m + m ψ r R ψ R µ ψ R r ψ r + V rψ R ψ r = Eψ R ψ r division by ψ R ψ r gives: R ψ R r ψ r + V r = E. m + m ψ R µ ψ r 4

5 Since the right hnd side is constnt nd the two terms on the left hnd side re solely functions of seprte vribles, ssumed to be non-dependent of ech other, ech term must be constnt. Thus: R ψ R = E R m + m ψ R with E R + E r = E. r ψ r + V r = E r 3 µ ψ r Note tht eq. is the Schrödinger eqution for free prticle no potentil while eq.3 is the Schrödinger eqution for prticle of mss µ moving in potentil V r. 5.4 Q. If ψ nd ψ b re orthogonl nd both normlized, wht is the constnt A in: ψ ± r, r = A [ψ r ψ b r ± ψ r ψ b r 4 Requiring tht ψ ± itself is normlized we obtin: = A [ψ r ψ b r ± ψ r ψ b r [ψ r ψ b r ± ψ r ψ b r dr dr [ A = ψ r dr } {{ } ± ψ b r dr } {{ } ± ψ r ψ b r ψ r ψ b r dr dr + ψr ψb r ψ r ψ b r dr dr + ψ r dr } {{ } ψ b r dr } {{ } [ A = ± ψr ψb r ψ r ψ b r dr dr + ψ r ψ b r ψ r ψ b r dr dr We will now show tht ech double integrl in the squre brcket is by focug on the left one: 5

6 ψr ψb r ψ r ψ b r dr dr = ψb r ψ r dr ψr ψ b r dr } {{ }} {{ } becuse ψ r i nd ψ b r i re orthogonl. Do the sme for the other double integrl s n exercise. Our condition is now: A = ± A = 5 b Q. Wht is A if ψ = ψ b nd ψr, r is normlized? In this cse we hve: ψr, r = A [ψ r ψ r + ψ r ψ r = Aψ r ψ r 6 which gives: 4 A = ψ r ψ r dr dr = ψ r dr ψ r dr = A =. 7 6

7 5.5 Q. Write down the Hmiltonin for two nonintercting identicl prticles in the infinite squre well. Verify tht the fermion ground stte: ψ x, x = [πx / πx / πx / πx / 8 is n eigenfunction of H with the pproprite eigenvlue. why is ψ not the ground stte? The reson why ψ cnnot be the ground stte for fermions is tht the two prticles would then occupy the sme stte. We see tht eq.8 would be in tht cse. Now, from the Hmiltonin for generl two prticle system moving in n unknown potentil: H = + V r, r, t m m we see tht for the cse of identicl prticles we must hve the sme mss: H = [ m + + V r, r, t. 9 Furthermore, in our cse we know the potentil. The infinite squre well potentil is inside the well nd ce the prticles re non-intercting the totl potentil is lso. Therefore: H = [ m + [ d = m dx + d dx 3 where the lst step comes from the fct tht we re only considering the D infinite squre well. Now we check if eq. 8 is n eigenfunction of our Hmiltonin opertor: [ Hψ = d m dx + d dx ψ = Eψ? by letting the Hmiltonin in eq.3 work on the wve function eq.8: If we neglect spin. 7

8 [ [ Hψ = d m dx + d πx dx Hψ = m [ d dx [ πx πx πx d dx [ πx πx πx [ d πx πx d + dx dx πx πx + [ πx Hψ = m [ [ π πx + Hψ = 5π m πx [ 4π πx πx [ 4π πx πx πx + [ π πx [ πx πx πx πx = 5 π m ψ which indeed shows tht ψ is n eigenfunction of H. The fctor in front, the eigenvlue, is lso wht we expect ce we expect it to be of the form: n }{{} + n π }{{} m = 5 π m 4 3 b Q. Give the wvefunctions nd energies of the two next excited sttes for ech of these three cses: distinguishble prticles, identicl bosons nd identicl fermions. To tidy up the nottion we define: K = π m. 3 The two next excited sttes would then correspond to sttes with eigenvlues bove 5K. 8

9 Distinguishble prticles: In this cse the wvefunction is just the product of two gle prticle wvefunctions: ψx, x = ψx ψx of the D infinite squre well: ψ n x = nπx/. The first excited stte bove ψ is ψ : ψ = ψ x ψ x = πx / πx / with E = + K = 8K The second excited stte bove ψ cn be either of these two sttes: 33 ψ 3 = ψ x ψ 3 x = πx / 3πx / 34 ψ 3 = ψ 3 x ψ x = 3πx / πx / 35 both with the corresponding eigenvlue E = K this stte is thus doubly degenerte. Identicl bosons: Recll tht for bosons the wve function hs the generl ppernce: ψx, x = A [ψ x ψ b x + ψ b x ψ x 36 where A = if b nd A = if = b. Since bosons cn occupy the sme stte, the first excited stte with energy bove E = 5K is tht for which = b = : ψ = πx / πx / with E = 8K The second excited stte with energy E > 5K is given when either or b is nd the other 3: ψ 3 = [πx / 3πx / + 3πx / πx /. Note tht for identicl prticles, ψ b nd ψ b re not distinct eigensttes. For bosons they re even mthemticlly identicl. Thus the stte ψ 3 is not degenerte with ψ 3 - they re the sme stte. The energy of this stte is E = K. 9

10 Identicl fermions: Identicl fermions cnnot occupy the sme stte so in this cse b. The first level is given by: ψ 3 = [πx / 3πx / 3πx / πx / nd the second level is given by: ψ 3 = [πx / 3πx / 3πx / πx / where the negtive signs come from the fct tht we re deling with fermions. The energies of the two sttes re E 3 = K nd E 3 = 3K. Note here, s in the discussion concluding the Identicl bosons, tht ψ 3 depends on ψ 3. Unlike for bosons where ψ 3 = ψ 3, fermionic wvefunctions re ntisymmetric with respect to the interchnging of two sttes: ψ 3 = ψ 3. This however does not men tht E 3 nd E 3 re degenerte eigenvlues. Degenercy only occurs when two linerly dependent eigensttes hve the sme eigenvlue. 5.6 Q. Imgine two nonintercting prticles, ech of mss m, in the infinite squre well where the potentil is for x nd infinite elsewhere.. If one is in the stte: nπ ψ n = x 37 nd the other in the stte ψ l l n, clculte x x ssuming tht: They re dinstinguishble prticles. b They re identicl bosons. c They re identicl fermions. In this cse the totl wve function is: ψx, x = ψ n x ψ l x = nπ x with n l. Note now tht: lπ x 38 x x = x + x x x x x = x + x x x 39

11 so we study ech of the three terms on the RHS seprtely. Strting with the two first terms we obtin: x = ψ x, x x ψx, x dx dx = x ψnx ψ n x dx ψl x ψ l x dx x x } {{} x = ψ x, x x ψx, x dx dx = ψnx ψ n x dx x ψl x ψ l x dx. x } x {{} More compctly this is: x = x ψnx ψ n x dx = x n. x 4 x = x ψl x ψ l x dx = x l. x 4 Note tht we dropped the indices i =, on x i. Even if they re seprte coordintes they both spn the sme spce the distnce within the squre well. Therefore, when you integrte over the whole spce the index is not importnt. Now we insert our ψ n into eq.4: x n = x ψnxψ n xdx = x nπ x dx 4 nd note to our horror tht there does not seem to be redy mde expression in either Physics Hndbook nor Mthemtics Hndbook for x Ax. However 3, the primitive function of x cosax is listed nd we cn rewrite Ax = cosax. Therefore: x n = x nπ x dx = nπ x x cos x dx = [ x 3 3 [ x nπ/ cos = [ 3 3 nπ x + [ 3 n π x nπ/ nπ/ 3 = 3 n π nπ x We conclude: x n = 3 n π Thnks to Dniel Blixt for pointing this out!

12 It obviously follows tht: x l = 3 l π. 44 Now we evlute the mixed term x x : x x = ψ x, x x x ψx, x dx dx = x ψnx ψ n x dx x x ψl x ψ l x dx = x n x l. x Following the sme structure s bove, we now clculte x n : Use Ax = cosax : x n = x nπ x dx = nπ x x cos x dx Integrte the first term nd look up the second term: x n = [ x [ cos nπ x nπ/ = [ [ + x nπ x nπ/ = x n = + = 45 This gives us: x x = x + x x x x x = 3 n π + 3 l π = = 6 n π + l π. 46

13 b To sve some time we refer the reder to the derivtion in the text book section 5.. of: where: x x = x n + x l x n x l x nl 47 x nl = xψnxψ l xdx. 48 Compring eq.47 nd the one t the top of this pge we see tht the only difference is the lst term. Let us therefore focus on this one: x nl = xψnxψ l xdx = nπ lπ x x x dx. 49 Now we mke use of the following trigonometric identity: nπ lπ x x = [ πn l πn + l cos x cos x nd from Physics hndbook we find: 5 which gives: xψ nxψ l xdx = gx = x cosbx = Gx = cosbx b cos πn l πn l x + x πn l πn l + x bx b x πn+l πn+l cos x x 5 πn+l πn+l x The second nd fourth terms re while the rest evlutes to: cos πn l x cos πn+l { x = n+l n+l if l + n = even = else. πn l πn+l ce cosπn ± l = n+l. This mens tht the dditionl term now is: πn l x nl = 8 n l π 4 n l 4 πn+l x x = 6 n π + l π 8n l π 4 n l nl π n l 3

14 if n + l is even. Otherwise we get the sme nswer s in the previous problem. c The only difference for fermions lies in the lst sign: x x = x + x b x x b + x b 53 so we immeditely get the nswer: x x = 6 n π + l π + 8n l π 4 n l if n + l is even. Otherwise we get the sme nswer s in problem. Note tht: The verge distnce between two prticles being in fixed sttes might differ depending on if they re distinguishble or not. If they re indistinguishble they re either closer bosons or further fermions prt from distinguishble prticles in the sme stte. 5.7 Q. Construct the totl wvefunction for three prticles in the orthonorml sttes ψ, ψ b, ψ c for the cses of distinguishble, b bosonic nd c fermionic prticles. For distinguishble prticles the totl wvefunction is just the simple product: ψx, x, x 3 = ψ x ψ b x ψ c x 3 55 nd ce ψ, ψ b, ψ c re ll normlized the product hs norm. compre this result to eq.5.5 in the text book. 4

15 b For identicl bosons we hve to form completely symmetric combintion with respect to the interchnge of ny two of the three prticles. The number of terms will be the sme s the number of permuttions, so for n sttes this will give n! number of terms. In our cse we hve three prticles so the totl wvefunction is product of six terms: [ ψx, x, x 3 = A ψ x ψ b x ψ c x 3 +ψ x ψ c x ψ b x 3 +ψ b x ψ x ψ c x 3 +ψ b x ψ c x ψ x 3 +ψ c x ψ x ψ b x 3 +ψ c x ψ b x ψ x 3 The remining thing now is to find the normliztion constnt A. This is esy ce ech term hs norm compre with the nswer in so A = 6. Our normliztion constnt is then A =. 6 c For fermions we lso need n! terms but the totl wve function should be ntisymmetric. This mens tht it hs to chnge sign for ech interchnge of two fermions. You could do this either by ug the result from problem b nd just chnge sign on three terms to mke the totl wve function ntisymmetric, or use fster wy; the Slter determinnt ψx, x, x 3 = 3! ψ x ψ b x ψ c x ψ x ψ b x ψ c x ψ x 3 ψ b x 3 ψ c x 3 56 ψx, x, x 3 = [ ψ x ψ b x ψ c x 3 +ψ b x ψ c x ψ x 3 +ψ c x ψ x ψ b x 3 6 ψ c x ψ b x ψ x 3 ψ b x ψ x ψ c x 3 ψ x ψ c x ψ b x 3 57 which works for ny number of elements note tht the normliztion then becomes N!. 5

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