• Support PF! Buy your school textbooks, materials and every day products Here!

SHO ladder operators & some hamiltonian commutator relations

  • Thread starter Hakkinen
  • Start date
  • #1
42
0

Homework Statement


For the SHO, find these commutators to their simplest form:
[itex] [a_{-}, a_{-}a_{+}] [/itex]
[itex]
[a_{+},a_{-}a_{+}]
[/itex]
[itex]
[x,H]
[/itex]
[itex]
[p,H]
[/itex]


Homework Equations





The Attempt at a Solution


I though this would be an easy problem but I am stuck on the first two parts. Here's what I did at first:
[itex]
[a_{-}, a_{-}a_{+}]\psi = a_{-}(n+1)\psi_{n} - a_{-}n\psi_{n} = a_{-}\psi_{n}
[/itex]
[itex]
= \sqrt{n}\psi_{n-1}
[/itex]

[itex]
[a_{+}, a_{-}a_{+}]\psi = a_{+}(n+1)\psi_{n} - a_{-}a_{+}\sqrt{n+1}\psi_{n+1}
[/itex]
[itex]
= (n+1)^{3/2}\psi_{n+1} - (n+1)^{3/2}\psi_{n+1} = 0
[/itex]


Now what I am confused about is the [itex]\psi_{n-1}[/itex] term in the first commutator. Surely there is a general form of the commutator without the test wavefunction? And I can't just drop this term and have root of n as the result. So did I do something wrong?

I tried the first part again using the explicit form of the ladder operators, in terms of H, p, x, with all of the constants. What I have gotten so far looks quite messy and involves [itex] [p,H] [/itex] and [itex] [x,H] [/itex], which I've yet to compute and are the last two parts of the problem... So it seems this route is not the easiest?

Any assistance is appreciated!
 
Last edited:

Answers and Replies

  • #2
42
0
Duh! The first answer is just [itex] a_{-}[/itex]

And I didnt realize you could just use the "product rule" for commutators to simplify the algebra a bit!

All set now
 

Related Threads on SHO ladder operators & some hamiltonian commutator relations

  • Last Post
Replies
2
Views
1K
Replies
17
Views
1K
Replies
1
Views
485
Replies
3
Views
416
Replies
1
Views
13K
Replies
1
Views
681
Replies
8
Views
819
Replies
2
Views
2K
Top