100% satisfaction guarantee Immediately available after payment Both online and in PDF No strings attached
logo-home
Homework 6 Solutions Temple University PHYSICS 3701 $8.99   Add to cart

Exam (elaborations)

Homework 6 Solutions Temple University PHYSICS 3701

 6 views  0 purchase
  • Course
  • Institution

Department of Physics Temple University Introduction to Quantum Mechanics, Physics 3701 Instructor: Z.-E. Meziani Solution set for homework # 6 April 16, 2013 Exercise #2, Complement FVI, page 765 Consider an arbitrary physical system whose four-dimensional state space is spanned by a basis of...

[Show more]

Preview 2 out of 9  pages

  • April 16, 2023
  • 9
  • 2022/2023
  • Exam (elaborations)
  • Questions & answers
avatar-seller
Department of Physics Temple University
Introduction to Quantum Mechanics, Physics 3701 Instructor: Z.-E. Meziani




Solution set for homework # 6
April 16, 2013




Exercise #2, Complement FVI , page 765

Consider an arbitrary physical system whose four-dimensional state space is spanned by a basis of four
eigenvectors |j, mz i common to Jˆ2 and Jz (j = 0 or 1; −j ≤ mz ≤ +j), of eigenvalues j(j + 1)h̄2 and mz h̄,
such that: q
J± |j, mz >= h̄ j(j + 1) − mz (mz ± 1|j, mz ± 1 > (1)
J+ |j, j >= J− |j, −j >= 0 (2)
• a) Express in terms of the kets |j, mz >, the eigenstates common to Jˆ2 and Jˆx to be denoted by
|j, mx >.

We must first form the matrix of the operator Jˆx in the basis {|j, mz >}. If we recall the following
relation
1 ˆ
Jˆx = (J+ + Jˆ− ) (3)
2
then we may use Eqs. ?? and ?? to write the matrix of the Jˆx operator in the given basis. We first
calculate the individual matrix elements. First, the Jˆ+ terms

Jˆ+ |1, 1 > = 0 (4)

Jˆ+ |1, 0 > = 2h̄|1, 1 > (5)

Jˆ+ |1, −1 > = 2h̄|1, 0 > (6)
Jˆ+ |0, 0 > = 0 (7)

then the J− terms

Jˆ− |1, 1 > = 2h̄|1, 0 > (8)

ˆ
J− |1, 0 > = 2h̄|1, −1 > (9)
Jˆ− |1, −1 > = 0 (10)
Jˆ− |0, 0 > = 0 (11)

Now we may write the operator Jˆx in matrix form, using Eq. ??. in the basis {|1, 1 >, |1, 0 > and
|1, −1 >, |0, 0 >}
0 1 0 0
 
h̄ 1 0 1 0
Jˆx = √ 

 (12)
2 0 1 0
 0
0 0 0 0



This study source was downloaded by 100000850872992 from CourseHero.com on 04-15-2023 23:56:45 GMT -05:00


https://www.coursehero.com/file/8707035/Homework-6-Solution/

, To find the eigenvalues of this matrix, it is necessary to diagonalize it.
−λ √h̄ 0 0
2
√h̄ −λ √h̄ 0
2 2 =0 (13)
0 √h̄ −λ 0
2
0 0 0 −λ

The determinant is solved in the usual way, resulting in a characteristic equation given by
−λ √h̄ 0 √h̄ √h̄ 0
2 h̄ 2 2
−λ √h̄ −λ 0 0−√ −λ 0 = 0
2 2
0 0 −λ 0 0 −λ
−λ h̄ √h̄ 0 h̄ h̄ −λ 0
 
0
−λ −λ −√ 2 −√ √ =0
0 −λ 2 0 −λ 2 2 0 −λ
!
2 2 h̄2 h̄2
λ λ − − =0
2 2
 
λ2 λ2 − h̄2 = 0 (14)

This produces four roots, two of which are zero. This λ = 0 eigenvalue is thus two-fold degenerate.
The other two eigenvalues are +h̄ and −h̄. Substitution of these eigenvalues into Eq. ?? allows us to
solve for the eigenvectors.
0 1 0 0 a a
    
h̄ 1 0 1 0b b
Jˆx |φ >= λ|φ >⇒ √ 
   
= λ  (15)
2 0 1 0 0   c   c
0 0 0 0 d d

We are therefore able to write the eigenvectors in the new basis, |j, mx > in terms of the old basis,
|j, mz >. The following are the eigenvalues that correspond to the eigenvectors in both bases, in the
order: |j, mx >, |j, mz >
0 : |0, 0 >x = |0, 0 > (16)
1
0 : |1, 0 >x = √ (|1, −1 > −|1, −1 >) (17)
2
1
+h̄ : |1, 1 >x = √ (|1, 1 > +|1, −1 > +|1, 0 >) (18)
2
1
−h̄ : |1, −1 >x = √ (|1, 1 > +|1, −1 > −|1, 0 >) (19)
2
These are the eigenstates common to the operators Jˆ2 and Jˆx
• b) Consider a system in the normalized state:


|ψ >= α|j = 1, mz = 1 > +β|j = 1, mz = 0 > +γ|j = 1, mz = −1 > +δ|j = 0, mz = 0 > (20)
Note that this state is normalized. Therefore, we must have the following relation between the
coefficients:
|α|2 + |β|2 + |γ|2 + |δ|2 = 1 (21)




This study source was downloaded by 100000850872992 from CourseHero.com on 04-15-2023 23:56:45 GMT -05:00


https://www.coursehero.com/file/8707035/Homework-6-Solution/

The benefits of buying summaries with Stuvia:

Guaranteed quality through customer reviews

Guaranteed quality through customer reviews

Stuvia customers have reviewed more than 700,000 summaries. This how you know that you are buying the best documents.

Quick and easy check-out

Quick and easy check-out

You can quickly pay through credit card or Stuvia-credit for the summaries. There is no membership needed.

Focus on what matters

Focus on what matters

Your fellow students write the study notes themselves, which is why the documents are always reliable and up-to-date. This ensures you quickly get to the core!

Frequently asked questions

What do I get when I buy this document?

You get a PDF, available immediately after your purchase. The purchased document is accessible anytime, anywhere and indefinitely through your profile.

Satisfaction guarantee: how does it work?

Our satisfaction guarantee ensures that you always find a study document that suits you well. You fill out a form, and our customer service team takes care of the rest.

Who am I buying these notes from?

Stuvia is a marketplace, so you are not buying this document from us, but from seller ExamsConnoisseur. Stuvia facilitates payment to the seller.

Will I be stuck with a subscription?

No, you only buy these notes for $8.99. You're not tied to anything after your purchase.

Can Stuvia be trusted?

4.6 stars on Google & Trustpilot (+1000 reviews)

75323 documents were sold in the last 30 days

Founded in 2010, the go-to place to buy study notes for 14 years now

Start selling
$8.99
  • (0)
  Add to cart