A particle moves in a one-dimensional box with a small potential dip E(0) ²² 2m/2 Quortion 4 V= ∞o for x<0 and x>1. V = -b for 0 <2< (1/2)l₂ V = 0 for (1/2) { 1, V = 0 for 0 < x < 1). Find the first order energy of the ground state. The ground state energy and wavefunction is given by . "(x) = 25 (0) 0 -b √√sin H. 0 [10]
A particle moves in a one-dimensional box with a small potential dip E(0) ²² 2m/2 Quortion 4 V= ∞o for x<0 and x>1. V = -b for 0 <2< (1/2)l₂ V = 0 for (1/2) { 1, V = 0 for 0 < x < 1). Find the first order energy of the ground state. The ground state energy and wavefunction is given by . "(x) = 25 (0) 0 -b √√sin H. 0 [10]
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Question
![A particle moves in a one-dimensional box with a small potential dip
E(0) -
77²1²
2m/2
Question 4
Treat the potential dip as a perturbation to a regular rigid box (V=∞0 for x<0 and x > 1, V = 0
for 0 < x < 1). Find the first order energy of the ground state. The ground state energy and
wavefunction is given by
V
z
V= ∞ for x<0 and x>1,
V = -b for 0 < 2 < (1/2) ł,
V = 0 for (1/2) { <x< 1.
(0) (1) = 1
√√sin
0
1
[10]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F0f71bc4f-b443-409b-900a-c5e680e620b6%2F0c47527f-6dc8-4bd7-85d0-052d3dc86f31%2F080o69vc_processed.png&w=3840&q=75)
Transcribed Image Text:A particle moves in a one-dimensional box with a small potential dip
E(0) -
77²1²
2m/2
Question 4
Treat the potential dip as a perturbation to a regular rigid box (V=∞0 for x<0 and x > 1, V = 0
for 0 < x < 1). Find the first order energy of the ground state. The ground state energy and
wavefunction is given by
V
z
V= ∞ for x<0 and x>1,
V = -b for 0 < 2 < (1/2) ł,
V = 0 for (1/2) { <x< 1.
(0) (1) = 1
√√sin
0
1
[10]
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