[a, at]=[b, b]= 1, [a, b] = [a, b]=[a, b] = [a, b]=0 ata = (a+ax +ața, +iața, — iata,), - b+b= (a+ax +atay - iața, +iața,) H = (ata+b+b+1)hw = (na + n + 1)hw = (n+1)hw, n = 0, 1, 2,... L₂ = (ata — b¹b)ħ = (na - nɩ)ħ = mh - For each n = n+n, there are n + 1 Exercise 6: Derive all of these equations. possible values for m = n₁-n: m = n m=n-2 na = n nan-1 пь = 0 n₂ = 1 na n 2 n₂ = 2 m=n-4 na = 1 n₁ = n−1 m = −(n-2) na = 0 nb = n m=-n m = n, n-2, n-4, ..., -n +2, -n
[a, at]=[b, b]= 1, [a, b] = [a, b]=[a, b] = [a, b]=0 ata = (a+ax +ața, +iața, — iata,), - b+b= (a+ax +atay - iața, +iața,) H = (ata+b+b+1)hw = (na + n + 1)hw = (n+1)hw, n = 0, 1, 2,... L₂ = (ata — b¹b)ħ = (na - nɩ)ħ = mh - For each n = n+n, there are n + 1 Exercise 6: Derive all of these equations. possible values for m = n₁-n: m = n m=n-2 na = n nan-1 пь = 0 n₂ = 1 na n 2 n₂ = 2 m=n-4 na = 1 n₁ = n−1 m = −(n-2) na = 0 nb = n m=-n m = n, n-2, n-4, ..., -n +2, -n
Algebra: Structure And Method, Book 1
(REV)00th Edition
ISBN:9780395977224
Author:Richard G. Brown, Mary P. Dolciani, Robert H. Sorgenfrey, William L. Cole
Publisher:Richard G. Brown, Mary P. Dolciani, Robert H. Sorgenfrey, William L. Cole
Chapter4: Polynomials
Section4.1: Exponents
Problem 24OE
Related questions
Question
![[a, at]=[b, b]= 1,
[a, b] = [a, b]=[a, b] = [a, b]=0
ata = (a+ax +ața, +iața, — iata,),
-
b+b= (a+ax +atay - iața, +iața,)
H = (ata+b+b+1)hw = (na + n + 1)hw = (n+1)hw, n = 0, 1, 2,...
L₂ = (ata — b¹b)ħ = (na - nɩ)ħ = mh
-
For each n = n+n, there are n + 1
Exercise 6: Derive all of
these equations.
possible values for m = n₁-n:
m = n
m=n-2
na = n
nan-1
пь = 0
n₂ = 1
na n 2
n₂ = 2
m=n-4
na = 1
n₁ = n−1 m = −(n-2)
na =
0
nb = n
m=-n
m = n, n-2, n-4, ..., -n +2, -n](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F542c5f9b-31fa-4e41-923c-eed948cd7206%2Fcfa1b385-ae05-47cd-a5c5-affbeb2cf1e9%2F7z0w4y4_processed.jpeg&w=3840&q=75)
Transcribed Image Text:[a, at]=[b, b]= 1,
[a, b] = [a, b]=[a, b] = [a, b]=0
ata = (a+ax +ața, +iața, — iata,),
-
b+b= (a+ax +atay - iața, +iața,)
H = (ata+b+b+1)hw = (na + n + 1)hw = (n+1)hw, n = 0, 1, 2,...
L₂ = (ata — b¹b)ħ = (na - nɩ)ħ = mh
-
For each n = n+n, there are n + 1
Exercise 6: Derive all of
these equations.
possible values for m = n₁-n:
m = n
m=n-2
na = n
nan-1
пь = 0
n₂ = 1
na n 2
n₂ = 2
m=n-4
na = 1
n₁ = n−1 m = −(n-2)
na =
0
nb = n
m=-n
m = n, n-2, n-4, ..., -n +2, -n
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