you can do the integration using mathematica type the
TRANSCRIPT
![Page 1: You can do the integration using Mathematica Type the](https://reader034.vdocuments.us/reader034/viewer/2022052620/628e543c8d037530fe4d5768/html5/thumbnails/1.jpg)
Problem 1 Griffiths 1.5
Ξ¨ π₯, π‘ = π΄πβπ|π₯|πβπππ‘
(a) 1 = |Ξ¨|2ππ₯
β
ββ
= ΨΨβππ₯β
ββ
= π΄2 πβ2π|π₯|ππ₯β
ββ
= 2π΄2 πβ2ππ₯ππ₯β
0
= 2π΄2πβ2ππ₯
β2π β0
=π΄2
π β π΄ = π
(b) π₯ = π₯|Ξ¨|2ππ₯
β
ββ
= π΄2 π₯πβ2π π₯ ππ₯ = 0 β
ββ
(πππ πππ‘ππππππ)
π₯2 = π₯2|Ξ¨|2ππ₯β
ββ
= π΄2 π₯2πβ2π π₯ ππ₯ = β
ββ
2π΄2 π₯2πβ2ππ₯ππ₯β
0
You can do the integration using Mathematica if you donβt want to do it by hand. Type the following line in Mathematica then βShift + Enterβ:
Integrate[π₯^2 β Exp[β2 β π β π₯ , π₯, 0, Infinity
You will get 1/4π3
So π₯2 =
2π΄2
4π3 =2π
4π3 =1
2π2
![Page 2: You can do the integration using Mathematica Type the](https://reader034.vdocuments.us/reader034/viewer/2022052620/628e543c8d037530fe4d5768/html5/thumbnails/2.jpg)
(c) π2 = π₯2 β π₯ 2 =
1
2π2 β π π‘ππππππ πππ£πππ‘πππ π =
1
2π
3 2 1 1 2 3
0.2
0.4
0.6
0.8
1.0
|Ξ¨|2
π₯ π βπ
This figure plots |Ξ¨|2 as a function of x for
Ξ» = 1. The two vertical lines shows the
positions of Ο =1
2and βΟ.
Probability outside the range (βΟ, π):
πππ’π‘ = |Ξ¨|2ππ₯βπ
ββ
+ |Ξ¨|2ππ₯β
π
= 2π΄2 πβ2ππ₯ππ₯β
π
=2π΄2
β2ππβ2ππ₯
βπ
= πβ 2 = 0.2431
![Page 3: You can do the integration using Mathematica Type the](https://reader034.vdocuments.us/reader034/viewer/2022052620/628e543c8d037530fe4d5768/html5/thumbnails/3.jpg)
Problem 2
(a) π» =π2
2π+
1
2ππ2π₯2
(b) π0 π₯ = π΄ expβπππ₯2
2β
1 = |π0|2ππ₯β
ββ
= π΄2 expβπππ₯2
βππ₯
β
ββ
= π΄2πβ
ππ β π΄ =
ππ
πβ
1/4
(c) π1 π₯ = π΅ + πΆπ₯ expβπππ₯2
2β
Orthogonality condition:
0 = π0 π1 = π0βπ1ππ₯
β
ββ
= π΄(π΅ + πΆπ₯)expβπππ₯2
βππ₯
β
ββ
= π΄π΅ expβπππ₯2
βππ₯
β
ββ
+ π΄πΆ π₯expβπππ₯2
βππ₯
β
ββ
= 0
= π΄π΅πβ
ππ β π΅ = 0
![Page 4: You can do the integration using Mathematica Type the](https://reader034.vdocuments.us/reader034/viewer/2022052620/628e543c8d037530fe4d5768/html5/thumbnails/4.jpg)
Normalization condition:
1 = π1 π1 = |π1|2ππ₯β
ββ
= πΆ2 π₯2expβπππ₯2
βππ₯
β
ββ
= πΆ2π
2
β
ππ
3/2
β πΆ =ππ
πβ
1/4 2ππ
β
π2 π₯ = π· + πΈπ₯ + πΉπ₯2 expβπππ₯2
2β
Orthogonality conditions: π0 π2 = 0 & π1 π2 = 0
0 = π0 π2 = π0βπ2ππ₯
β
ββ
= π΄(π· + πΈπ₯ + πΉπ₯2)expβπππ₯2
βππ₯
β
ββ
= π΄π· expβπππ₯2
βππ₯
β
ββ
+ π΄πΈ π₯expβπππ₯2
βππ₯
β
ββ
+ π΄πΉ π₯2expβπππ₯2
βππ₯
β
ββ
= 0
= π΄π·πβ
ππ+ π΄πΉ
π
2
β
ππ
3/2
β πΉ = β2π·ππ
β
![Page 5: You can do the integration using Mathematica Type the](https://reader034.vdocuments.us/reader034/viewer/2022052620/628e543c8d037530fe4d5768/html5/thumbnails/5.jpg)
0 = π1 π2 = π1βπ2ππ₯
β
ββ
= πΆπ₯(π· + πΈπ₯ + πΉπ₯2)expβπππ₯2
βππ₯
β
ββ
= πΆπ· π₯expβπππ₯2
βππ₯
β
ββ
+ πΆπΈ π₯2expβπππ₯2
βππ₯
β
ββ
+ πΆπΉ π₯3expβπππ₯2
βππ₯
β
ββ
= 0 = 0
= πΆπΈπ
2
β
ππ
3/2
β πΈ = 0
Normalization condition:
1 = π2 π2 = |π2|2ππ₯β
ββ
= (π· + πΉπ₯2)2expβπππ₯2
βππ₯
β
ββ
= π·2 expβπππ₯2
βππ₯
β
ββ
+ 2π·πΉ π₯2expβπππ₯2
βππ₯
β
ββ
+ πΉ2 π₯4expβπππ₯2
βππ₯
β
ββ
= π·2πβ
ππ+ 2π·πΉ
π
2
β
ππ
3/2
+ πΉ23 π
4
β
ππ
5/2
π πππππ πΉ = β2π·ππ
β
= 2π·2πβ
ππ βΉ π· =
1
2
ππ
πβ
14 and πΉ = β
2
π1/4
ππ
β
5/4
![Page 6: You can do the integration using Mathematica Type the](https://reader034.vdocuments.us/reader034/viewer/2022052620/628e543c8d037530fe4d5768/html5/thumbnails/6.jpg)
(d) π1 π₯ = πΆπ₯expβπππ₯2
2β with πΆ =
ππ
πβ
1/4 2ππ
β
π» =π2
2π+
1
2ππ2π₯2 = β
β2
2π
π2
ππ₯2 +1
2ππ2π₯2
π»π1 π₯ = ββ2
2π
π2
ππ₯2 +1
2ππ2π₯2 πΆπ₯exp
βπππ₯2
2β
= ββ2
2π
π2
ππ₯2 πΆπ₯expβπππ₯2
2β+
1
2ππ2π₯2πΆπ₯exp
βπππ₯2
2β
= ββ2
2πβ
3ππ
βπ₯ +
π2π2
β2 π₯3 πΆexpβπππ₯2
2β
+1
2ππ2π₯3πΆexp
βπππ₯2
2β
=3βπ
2πΆπ₯exp
βπππ₯2
2β
=3βπ
2π1 π₯ βΉ πΈ1 =
3βπ
2
![Page 7: You can do the integration using Mathematica Type the](https://reader034.vdocuments.us/reader034/viewer/2022052620/628e543c8d037530fe4d5768/html5/thumbnails/7.jpg)
Problem 3
π+ =1
2βππ(βππ + πππ₯) πβ =
1
2βππ(+ππ + πππ₯)
πβ, π+ =1
2βππ[ππ + πππ₯, βππ + πππ₯ =
1
2βππ{πππ π, π₯ β πππ[π₯, π }
=1
2βππ{πππ(βπβ) β πππ(πβ)} = 1
βπ π+πβ +1
2= βπ
1
2βππβππ + πππ₯ ππ + πππ₯ +
1
2
=1
2ππ2 + ππππ₯π β πππππ₯ + π2π2π₯2 +
1
2βπ
=1
2ππ2 + πππ[π₯, π + π2π2π₯2 +
1
2βπ
=1
2ππ2 β βππ + π2π2π₯2 +
1
2βπ
=π2
2π+
1
2ππ2π₯2 = π»
![Page 8: You can do the integration using Mathematica Type the](https://reader034.vdocuments.us/reader034/viewer/2022052620/628e543c8d037530fe4d5768/html5/thumbnails/8.jpg)
π» π+ππ = βπ π+πβ +1
2π+ππ = βπ(π+πβπ+ + 1/2π+) ππ
= βππ+ πβπ+ +1
2ππ = βππ+ π+πβ + 1 +
1
2ππ
= π+ π» + βπ ππ = π+ πΈ + βπ ππ = πΈ + βπ π+ππ
Therefore π+ππ is an eigenstate of H with energy πΈ + βπ . But it may not be
normalized. It is related to the normalized eigenstate ππ+1 up to a coefficient:
π+ππ = πΆππ+1
1 = ππ+1 ππ+1 =1
|πΆ|2π+ππ π+ππ =
1
|πΆ|2ππ πβπ+ππ
=1
|πΆ|2ππ (π+πβ + 1)ππ =
1
|πΆ|2ππ (π»/βπ + 1/2)ππ =
1
πΆ 2 (π + 1)
βΉ πΆ = π + 1 and π+ππ = π + 1 ππ+1 Similarly πβππ = π ππβ1
π»ππ = βπ π+πβ +1
2ππ = βπ π+πβππ +
1
2ππ = βπ π+ π ππβ1 +
1
2ππ
= βπ π πππ +1
2ππ = βπ π +
1
2ππ
![Page 9: You can do the integration using Mathematica Type the](https://reader034.vdocuments.us/reader034/viewer/2022052620/628e543c8d037530fe4d5768/html5/thumbnails/9.jpg)
Problem 4
π0 π₯ = π΄ expβπππ₯2
2β Ground state: π΄ =
ππ
πβ
1/4
with
π₯ = π0β π₯ π₯π0 π₯ ππ₯
β
ββ
= π΄2 π₯expβπππ₯2
βππ₯
β
ββ
= 0
π₯2 = π0β π₯ π₯2π0 π₯ ππ₯ππ₯
β
ββ
= π΄2 π₯2expβπππ₯2
βππ₯
β
ββ
= π΄2π
2
β
ππ
3/2
=β
2ππ
ππ₯ = π₯2 β π₯ 2 =β
2ππ
π = π0β π₯ βπβ
π
ππ₯π0 π₯ ππ₯
β
ββ
= π΄2 expβπππ₯2
2βπππ π₯exp
βπππ₯2
2βππ₯
β
ββ
= ππππ΄2 π₯expβπππ₯2
βππ₯
β
ββ
= 0
π2 = π0β π₯ ββ2
π2
ππ₯2 π0 π₯ ππ₯β
ββ
![Page 10: You can do the integration using Mathematica Type the](https://reader034.vdocuments.us/reader034/viewer/2022052620/628e543c8d037530fe4d5768/html5/thumbnails/10.jpg)
= βπ΄2 β2 expβπππ₯2
2β
π2π2
β2 π₯2 βππ
βexp
βπππ₯2
2βππ₯
β
ββ
= βπ΄2 π2π2 π₯2expβπππ₯2
βππ₯
β
ββ
+ π΄2βππ expβπππ₯2
βππ₯
β
ββ
= βπ΄2 π2π2π
2
β
ππ
3/2
+ π΄2βπππβ
ππ=
1
2βππ
ππ = π2 β π 2 =1
2βππ
ππ₯ππ =β
2ππ
1
2βππ =
β
2 Minimum uncertainty
![Page 11: You can do the integration using Mathematica Type the](https://reader034.vdocuments.us/reader034/viewer/2022052620/628e543c8d037530fe4d5768/html5/thumbnails/11.jpg)
1st excited state:
π1 π₯ = πΆπ₯expβπππ₯2
2β with πΆ =
2
π1/4
ππ
β
3/4
π₯ = π1β π₯ π₯π1 π₯ ππ₯
β
ββ
= πΆ2 π₯3expβπππ₯2
βππ₯
β
ββ
= 0
π₯2 = π1β π₯ π₯2π1 π₯ ππ₯
β
ββ
= πΆ2 π₯4expβπππ₯2
βππ₯
β
ββ
= πΆ23 π
4
β
ππ
5/2
=3β
2ππ
ππ₯ = π₯2 β π₯ 2 =3β
2ππ
π = π1β π₯ βπβ
π
ππ₯π1 π₯ ππ₯
β
ββ
= βπβπΆ2 π₯ 1 βππ
βπ₯2 exp
βπππ₯2
βππ₯
β
ββ
= 0
![Page 12: You can do the integration using Mathematica Type the](https://reader034.vdocuments.us/reader034/viewer/2022052620/628e543c8d037530fe4d5768/html5/thumbnails/12.jpg)
π2 = π1β π₯ ββ2
π2
ππ₯2 π1 π₯ ππ₯β
ββ
= βπΆ2 β2 π2π2
β2 π₯4 β3ππ
βπ₯2 exp
βπππ₯2
βππ₯
β
ββ
= βπΆ2 π2π2 π₯4expβπππ₯2
βππ₯
β
ββ
+ 3πΆ2βππ π₯2expβπππ₯2
βππ₯
β
ββ
= β3
2βππ + 3βππ =
3
2βππ
ππ = π2 β π 2 =3
2βππ
ππ₯ππ =3β
2ππ
3
2βππ =
3β
2>
β
2
![Page 13: You can do the integration using Mathematica Type the](https://reader034.vdocuments.us/reader034/viewer/2022052620/628e543c8d037530fe4d5768/html5/thumbnails/13.jpg)
Problem 4: operator approach: MUCH simpler!
π₯ =β
2ππ(π+ + πβ) π = π
βππ
2(π+ β πβ)
π π₯ π =β
2πππ π+ + πβ π = 0
π π₯2 π =β
2πππ π+π+ + π+πβ + πβπ+ + πβπβ π =
β
πππ + 1/2
π π π = πβππ
2π π+ β πβ π = 0
π π2 π = ββππ
2π π+π+ β π+πβ β πβπ+ + πβπβ π = βππ π + 1/2
Ground state: ππ₯ππ =β
2ππ
1
2βππ =
β
2
ππ₯ππ =3β
2ππ
3
2βππ =
3β
2 1st excited state:
![Page 14: You can do the integration using Mathematica Type the](https://reader034.vdocuments.us/reader034/viewer/2022052620/628e543c8d037530fe4d5768/html5/thumbnails/14.jpg)
Problem 5
(a) Probability to be in the ground state =
| π1(π₯) ππ΄(π₯, π‘ = 0) |2 = π1(π₯)1
2π1 π₯ +
1
2π2 π₯
2
=1
2
(b) ππ΄ π₯, π‘ =
1
2π1 π₯ πβππΈ1π‘/β +
1
2π2 π₯ πβππΈ2π‘/β
For a 1D infinite square well πΈπ =π2π2β2
2ππ2 , πΈ1 =π2β2
2ππ2 , πΈ2 =2π2β2
ππ2 ,
ππ΄ π₯, π‘ =1
2π1 π₯ exp β
ππ2βπ‘
2ππ2 +1
2π2 π₯ exp β
π2π2βπ‘
ππ2
(c) Density distribution at π‘ = 0 is:
|ππ΄ π₯, π‘ = 0 |2 = ππ΄β π₯, π‘ = 0 ππ΄ π₯, π‘ = 0
=1
2π1
β π₯ +1
2π2
β π₯1
2π1 π₯ +
1
2π2 π₯
![Page 15: You can do the integration using Mathematica Type the](https://reader034.vdocuments.us/reader034/viewer/2022052620/628e543c8d037530fe4d5768/html5/thumbnails/15.jpg)
=1
2|π1|2 + |π2|2 + 2π1π2 since π1 & π2 are real
|ππ΄ π₯, π‘ |2 = ππ΄β π₯, π‘ ππ΄ π₯, π‘
=1
2π1
β π₯ πππΈ1π‘/β +1
2π2
β π₯ πππΈ2π‘/β1
2π1 π₯ πβππΈ1π‘/β +
1
2π2 π₯ πβππΈ2π‘/β
=1
2|π1|2 + |π2|2 + π1
βπ2πβπ(πΈ2βπΈ1)π‘/β+π1π2βππ(πΈ2βπΈ1)π‘/β
=1
2|π1|2 + |π2|2 + 2π1π2cos ((πΈ2βπΈ1)π‘/β)
Therefore |ππ΄ π₯, π‘ |2 = |ππ΄ π₯, π‘ = 0 |2 when cos ((πΈ2βπΈ1)π‘/β) = 1
Or (πΈ2βπΈ1)π‘/β = 2π, 4π, 6π, β¦
π‘ = 2ππβ/(πΈ2βπΈ1) = 2ππβ/2π2β2
ππ2 βπ2β2
2ππ2 =4πππ2
3πβ where π = 1,2,3, β¦
The smallest π‘1 =4ππ2
3πβ
![Page 16: You can do the integration using Mathematica Type the](https://reader034.vdocuments.us/reader034/viewer/2022052620/628e543c8d037530fe4d5768/html5/thumbnails/16.jpg)
(d)
0.2 0.4 0.6 0.8 1.0
0.5
1.0
1.5
2.0
2.5
3.0
0.2 0.4 0.6 0.8 1.0
0.5
1.0
1.5
2.0
2.5
0.2 0.4 0.6 0.8 1.0
0.5
1.0
1.5
2.0
0.2 0.4 0.6 0.8 1.0
0.5
1.0
1.5
2.0
0.2 0.4 0.6 0.8 1.0
0.5
1.0
1.5
2.0
2.5
0.2 0.4 0.6 0.8 1.0
0.5
1.0
1.5
2.0
2.5
3.0
0.2 0.4 0.6 0.8 1.0
0.5
1.0
1.5
2.0
2.5
0.2 0.4 0.6 0.8 1.0
0.5
1.0
1.5
2.0
0.2 0.4 0.6 0.8 1.0
0.5
1.0
1.5
2.0
0.2 0.4 0.6 0.8 1.0
0.5
1.0
1.5
2.0
2.5
0.2 0.4 0.6 0.8 1.0
0.5
1.0
1.5
2.0
2.5
3.0
t=0 t=0.1*t1 t=0.2*t1 t=0.3*t1
t=0.4*t1 t=0.5*t1 t=0.6*t1 t=0.7*t1
t=0.8*t1 t=0.9*t1 t=t1
![Page 17: You can do the integration using Mathematica Type the](https://reader034.vdocuments.us/reader034/viewer/2022052620/628e543c8d037530fe4d5768/html5/thumbnails/17.jpg)
(e)
ππ΅ π₯, π‘ =1
2π1 π₯ πβππΈ1π‘/β β
1
2π2 π₯ πβππΈ2π‘/β
Density distribution at π‘ = 0 is:
|ππ΅ π₯, π‘ = 0 |2 = ππ΅β π₯, π‘ = 0 ππ΅ π₯, π‘ = 0
=1
2π1
β π₯ β1
2π2
β π₯1
2π1 π₯ β
1
2π2 π₯
=1
2|π1|2 + |π2|2 β 2π1π2 since π1 & π2 are real
|ππ΅ π₯, π‘ |2 = ππ΅β π₯, π‘ ππ΅ π₯, π‘
=1
2π1
β π₯ πππΈ1π‘/β β1
2π2
β π₯ πππΈ2π‘/β1
2π1 π₯ πβππΈ1π‘/β β
1
2π2 π₯ πβππΈ2π‘/β
=1
2|π1|2 + |π2|2 β π1
βπ2πβπ(πΈ2βπΈ1)π‘/ββπ1π2βππ(πΈ2βπΈ1)π‘/β
=1
2|π1|2 + |π2|2 β 2π1π2cos ((πΈ2βπΈ1)π‘/β)
π‘1 ππ π‘βπ π πππ ππ π‘βπ ππππ£πππ’π πππ π
![Page 18: You can do the integration using Mathematica Type the](https://reader034.vdocuments.us/reader034/viewer/2022052620/628e543c8d037530fe4d5768/html5/thumbnails/18.jpg)
(f)
0.2 0.4 0.6 0.8 1.0
0.5
1.0
1.5
2.0
2.5
3.0
ππ΄
0.2 0.4 0.6 0.8 1.0
0.5
1.0
1.5
2.0
2.5
3.0
ππ΅
0 a a 0
0.3a 0.7a
The two initial states can be determined by measuring the position of the particle:
If the measured value of x is around 0.3a then the initial state is ππ΄;
If the measured value of x is around 0.7a then the initial state is ππ΅.