By direct substitution, show that the wavefunction in the figure satisfies the timedependent Schrodinger equation (provided that En takes the value derived in figure).
By direct substitution, show that the wavefunction in the figure satisfies the timedependent Schrodinger equation (provided that En takes the value derived in figure).
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The odd parity eigenstates of the infinte square well , with potential V = 0 in the range −L/2 ≤ x ≤ L/2, are given by :
(see figure)
and have Ψn(x, t) = 0 elsewhere , for n=2 , 4 , 6 etc
I have got the expectation value of momentum for ⟨p⟩ and ⟨p 2⟩ for n = 2 (see figures)
By direct substitution, show that the wavefunction in the figure satisfies the timedependent Schrodinger equation (provided that En takes the value derived in figure).
Transcribed Image Text:√sin (2)
and have ₂(x, t) = 0 elsewhere, for n = 2, 4, 6, etc.
Yn(x, t)
=
-iEnt/ħ
for - sest
![(b) Momentum operator is,
=and
(x, t)=√√sin (1x)e-/ h
**(x,1)=√sin (1-x)+1/
So, the expectation value of momentum is defined as,
p> = √(x. ). p. (x, t)dx
sin(x)/x sin(x) dx
sin(x) cos(x). e-/+/hdx
LT/sin(x) cos(x)-1-dx
=
=
LT-0
(c) Momentum operator is,
-18
<p> = Tex
So,
and
*(x ) = √sin (1x)e-
**(x,1)=√sin(x)+¹/
So, the expectation value of momentum is defined as,
p²> = S(x. t)-p²(x. t)dx
-E/M
=
=
=
=
=
=
( odd function)
=
√√sin(x) -¹² √sin(x)¹² dx
hax
1/2-sin(x) (7) sin(x) e-Et/h+iEt/hdx
(L/sin(x) sin(x) - 1 - dx
(sin(x) dx
(7) S/21 - cos (21x)]-
-dx
(7)
mm(4)
(7) sin(²-)---sin (²²-(-))]]
()
sin(²77)+ -sin (21²)
(17) (4-0)
Lh
-
(7) L-2sin (²1)
(L-2sin (17)]
(n)²
It is also true for every value of 'n' (i.e. n=1, 2, 3, 4)
(n=1, 2, 3, ...)
c
2
En = n²π1²³ h ²2²
и п
2mL²
n=1,2,3,](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3b765a29-fcb0-4895-9cef-5ef60c256f5b%2F7976826a-e979-47a6-838a-a18e9c60159c%2F7itfdjc_processed.jpeg&w=3840&q=75)
Transcribed Image Text:(b) Momentum operator is,
=and
(x, t)=√√sin (1x)e-/ h
**(x,1)=√sin (1-x)+1/
So, the expectation value of momentum is defined as,
p> = √(x. ). p. (x, t)dx
sin(x)/x sin(x) dx
sin(x) cos(x). e-/+/hdx
LT/sin(x) cos(x)-1-dx
=
=
LT-0
(c) Momentum operator is,
-18
<p> = Tex
So,
and
*(x ) = √sin (1x)e-
**(x,1)=√sin(x)+¹/
So, the expectation value of momentum is defined as,
p²> = S(x. t)-p²(x. t)dx
-E/M
=
=
=
=
=
=
( odd function)
=
√√sin(x) -¹² √sin(x)¹² dx
hax
1/2-sin(x) (7) sin(x) e-Et/h+iEt/hdx
(L/sin(x) sin(x) - 1 - dx
(sin(x) dx
(7) S/21 - cos (21x)]-
-dx
(7)
mm(4)
(7) sin(²-)---sin (²²-(-))]]
()
sin(²77)+ -sin (21²)
(17) (4-0)
Lh
-
(7) L-2sin (²1)
(L-2sin (17)]
(n)²
It is also true for every value of 'n' (i.e. n=1, 2, 3, 4)
(n=1, 2, 3, ...)
c
2
En = n²π1²³ h ²2²
и п
2mL²
n=1,2,3,
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