PH12b 2010 Solutions HW#3

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1 PH 00 Solutions HW#3. The Hmiltonin of this two level system is where E g < E e The experimentlist sis is H E g jgi hgj + E e jei hej j+i p (jgi + jei) j i p (jgi jei) ) At t 0 the stte is j (0)i j+i, then, the stte will evolve s j (t)i p jgi + e i!t jei where! (E e E g )h nd we used the fct j (t)i is de ned up to overll phse fctor, i.e. j (t)i e iegth j (t)i Then the proility of + outcome nd - outcome if j (0)i j+i re P t + (+) jh+ j (t)ij + e i!t!t ( +!t) P t + ( ) jh j (t)ij e i!t!t (!t) P t + (+) + P t + ( ) s expected. ) First, it is esy to see tht if t t 0 the stte is j (0)i j - outcome if j (0)i j i re i then the proility of + outcome nd P t (+) P + t ( ) P t ( ) P + t (+) Now, we wnt to nd P t + (+ ) nd P t + ( ) the proility of + outcome nd - outcome fter mesurements ech done t intervl t if j (0)i j+i We cn get + outcome in two wys, the rst mesurement is + nd the second + or the rst mesurement is - nd the second is +. Rememer tht if the rst mesurement is + or - the stte will collpse to + or - respectively (i.e. j (t)i j+i or j i) nd will strt evolving gin. Then, similrly P + t (+ ) P + t (+) P + t (+) + P + t ( ) P t (+) P + t (+) P + t (+) + P + t ( ) P + t ( ) ( +!t) + (!t) +!t P + t ( ) P + t (+) P + t ( ) + P + t ( ) P t ( ) P + t (+) P + t ( )!t ) We re going to perform N mesurements t equl t intervls. Since we wnt the proility tht in ll the mesurements we get + outcome this is equl to the proility of doing Nth times the experiment in ) nd getting + outcome in ech one, ecuse if we get + outcome we re collpg the stte to j+i so this is like if we strt the experiment gin. Then P + N P + t (+) N N!t

2 d) Ug T Nt we get P + N N!t!T N N!T N + O N N "! T!T + O N N 3 N + O N # N Then f(wt )! T We cn conclude tht wtched quntum stte never evolve ecuse P + N! s N! with!t xed.. ) The proility of n inconclusive result is P (exit). Then, the proility of otining the inconclusive result N times in succession is just P () N ) If the om is good nd we repite the test s mny times s necessry until tht eventully either the om explodes or the test yields conclusive result, then the proility P Explode tht the om eventully explodes is P Explode P (explode) + P (inconclusive)p Explode + P Explode where P (explode) nd P (inconclusive) P (exit). Then we get, P Explode 3 Similrly, the proility P Conclusive tht the test eventully yields to conclusive result is P Conclusive P (conclusive) + P (inconclusive)p Conclusive + P Conclusive ) P Conclusive 3 Note tht P Conclusive + P Explode s expected. c) First notice tht ( + ) ( + ) ( + ) ( + ) then it is esy to see tht V N N N N N N 0 0 d) If the input stte is V in ( 0) T then fter the rst em splitter V out N T N This mens tht if the om is good then it will explode with the proility of nding the stte jouti tht is P (explode) N If it does not explode then, P (not explode) N nd the postmesurement stte will collpse to jouti This mens tht the input stte for the next em splitter is gin V in ( 0) T, so we re just repeting the experiment gin. Therefore, the proility tht it does not

3 explode fter N em splitters is the sme thn the proility tht it does not explode fter mking the gle em splitter experiment N times. Then then P (not explode) N N N N + O N N + O N c e) If the om does not explode the stte tht enters the lst em splitter will e jouti nd it will e trnsform y the lst em splitter to jouti! N jexiti+ N jexiti then P (exit). N And P (exitjnot explode) P (exit) N N + O N Then, 3. Alice sis is where 0 Bo sis is j i je i d j i je i where 0 ) The experiment hs four possile outcomes with proilities prepremesure he j he j j i P jhe j ij P jhe j ij j i P jhe j ij P jhe j ij where P P is the proility tht the stte ws prepre y Alice in j i or j i The o digonl outcomes in the tle represent the wrong guesses of Bo, therefore the proility tht Bo guess is wrong is P error jhe j ij + jhe j ij ) The overlps etween the sis re he j i ( + ) he j i ( ) then P error ( + ) + ( ) + c) Becuse 0 then 0 This implies tht P error is minimized when 0, then min 0 P min error () d) For 0 Perror min (0) 0, this mke sense ce for 0 j i i je i i nd h j i 0 so the sis re orthogonl. For Perror min (), this mke sense ce for the two sttes re the sme j i j i p nd there is equl chnce of Alice hving prepred j i or j i. 3

4 e) Evluting we get Perror min (8) p 06 The sketch is just je i in the x-xis, je i in the y-xis, j i mking n ngle of 8 rdins with the x-xis, nd j i mking n ngle of 8 with the y-xis. All four vectors hve unit length.. The wlls of the one-dimensionl ox re t x nd The wvefunction is r x (x) hxj i x ) The proility density of the prticle to e t some x is The proility tht scttered light is detected is P (x) j (x)j (x) (x) x P scttered P (x)dx ( ) + ) [ ] [ 38], sustituting in ) we get P scttered 8 p 0078 c) (Elorte Answer) First we hve to de ne the opertor tht mke the mesurement. The opertor tht detects prticle t some point x 0 is given y jx 0 i hx 0 j dx. Then the Hermitin opertor tht detects prticle in the intervl [ ] is The wve function is O j i jx 0 i hx 0 j dx 0 (y) jyi dy The eigenvlue eqution for this opertor is O j i j i this is O j i (y) jx 0 i hx 0 jyi dydx 0 (y) jx 0 i (x 0 y) dydx 0 (y) jyi dy (y) jyi dy the lst line cn just e stisfy if 0 nd (y) is zero in the intervl [ ] ie just hve support in the complement of the intervl [ ] or if nd (y) just hve support in the intervl [ ] Then the eigenvlues nd the degenerte eigenvector of the opertor re 0 jx 0 i hx 0 j dx 0 x 0 [ ) [ ( ] jx 0 i hx 0 j dx 0 x 0 [ ]

5 From the lecture notes "Quntum postultes III" we know tht if the outcome 0 is otined, the normlized stte right fter the mesurement is Q0 j i fter q h j Q 0 j i where Q 0 is the orthogonl projector to the spce of eigenvectors with 0, i.e. Then we hve Q 0 j i Q 0 jx 0 i hx 0 j dx 0 + (y) hx 0 j yi jx 0 i dx 0 dy + (y)(x 0 y) jx 0 i dx 0 dy + (x 0 ) jx 0 i dx 0 + jx 0 i hx 0 j dx 0 (x 0 ) jx 0 i dx 0 (y) hx 0 j yi jx 0 i dx 0 dy (y) jx 0 i (x 0 y)dx 0 dy where we used hx 0 j yi (x 0 y) nd (x 0 y) is Dirc Delt functionthis just mens tht the wve function fter the mesurement is proportionl to hxj i hxj Q (x) x [ ) [ ( ] 0 j i 0 x [ ] Finlly we hve to normlized the wve function fter the mesurement, this is done y the fctor q h j Q 0Q0 j i p hj i where we used Q 0Q0 Q 0 Now, it is esy to see tht q h j Q 0 j i Therefore, hj i P scttered ( ) (x) x [ ) [ ( ] fter (x) hxj fter p Pscttered ( ) 0 x [ ] For [ ] [ ] P scttered ( ) 087 we hve ( q q x fter (x) ( ) x [ ) [ ( ] 0 x [ ] d) Becuse the wve function fter the mesurement in the intervl [ ] [ 38] is proportionl to the originl wve function we hve tht P fter scttered ( 38) P scttered( 38) where P scttered ( 38) comes from ). e) Chrlie s rgument is wrong. Although the light nd prticle do not interct this is mesurement ecuse we re getting the informtion tht the prticle is not in certin intervl. From the postultes of Quntum mechnics we know tht ny mesurement will collpse the wvefunction to new wvefunction s explin in c). In this process we need to normlize gin the wvefunction nd this will chnge the proility s we clculte in d). If Chrlie were right then the proility of nding the prticle in the entire ox fter the mesurement will not e one which is oviously wrong. 5