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Commutation in coupled Harmonic Oscillators

Starting with a coupled Harmonic Oscillator problem $$ H = \frac{p_1^2 + p_2^2}{2m} + \frac{K}{2}\left[x_1^2 + x_2^2 + \left(x_1 - x_2\right)^2\right] = \left(\frac{p_1^2}{2m} + \frac{2K}{2}x_1^2\...
0
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1answer
72 views

If $\psi$ acting on $a_+$ and $a_-$ operator just moves up and down the ladder, why is $[a_-, a_+] = 1$ and not 0? [duplicate]

If $\psi$ is acted upon by both the operators one by one, it should return the same wave function. Thus order in which you increase or decrease the energy shouldn't matter. Then why is it so that the ...
0
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1answer
116 views

The creation and annihilation operators in quantum mechanics

What is the result of the commutation relation between the creation operator and a power of the annihilation operators in simple harmonic oscillator problem?
2
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2answers
184 views

Physical meaning of commutation

I was reading the solution to quantum harmonic oscillator by J.J. Sakurai. He uses the annihilation and creation operators and there's a key step (I think) which is $$[a,a^{\dagger}]=1$$ I know we can ...
-1
votes
1answer
50 views

What is the Significance of $\hat{N}$ and the Raising/Lowering operators not Commuting?

I just calculated the commutators for the number operator and the lowering/raising operators of the harmonic oscillator and they are non zero. Is there any significance to that? I got $[\hat{N},\hat{a}...
0
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0answers
70 views

How can $\hat p = - i \hbar \partial_q$ be derived starting from the definitions of $\hat q$ and $\hat p$ in terms of creation/destruction operators? [duplicate]

Consider the position and momentum operators $\hat q$ and $\hat p$, defined respectively in terms of creation and destruction operators in the usual way: $$ \hat q = c (\hat a + \hat a^\dagger), \...
3
votes
2answers
853 views

Why do the ladder operators in harmonic oscillators work?

The Hamiltonian can be diagonalized by transforming $x$ and $p$ to $a$ and $a^\dagger$. I understand how one proceeds from there to find the spectrum of $a^\dagger a$, the ground state $|0\rangle$ and ...
-1
votes
1answer
51 views

Deriving eigen values of $\hat{N}$

So let's say we have an operator $\hat{a}$ (ladder operator), where $\left[\hat{a},\hat{a}^\dagger\right] = 1$, and $\hat{a}^2 |\phi\rangle = 0$. How do I show that the eigenvalues of $\hat{N}=\hat{...
2
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2answers
2k views

Symmetry and degeneracy in quantum mechanics

If an operator commutes with the Hamiltonian of a problem, must it always admit degeneracy? It appears that not necessarily. For example, the parity operator commutes with the Hamiltonian of a free ...
3
votes
1answer
167 views

Proving $[a_k^\dagger, a_q^\dagger]=0$

I am trying to prove the commutation relations between the creation and annihilation operators in field theory. I was already able to show that $[a_k, a_q^\dagger]=i\delta(k-q)$. I want to show that $[...
8
votes
2answers
1k views

Dilation operator in CFT viewed as 'hamiltonian'?

From the commutation relations for the conformal Lie algebra, we may infer that the dilation operator plays the same role as the Hamiltonian in CFTs. The appropriate commutation relations are $[D,P_{...
9
votes
3answers
5k views

Why Don't the Ladder Operators Commute?

I have two problems with ladder operators. The first is that I feel they should somehow result in measurable things. The asymmetry of applying the plus operator versus the minus operator is very ...
1
vote
1answer
212 views

Energy eigenvalues of a Q.H.Oscillator with $[\hat{H},\hat{a}] = -\hbar \omega \hat{a}$ and $[\hat{H},\hat{a}^\dagger] = \hbar \omega \hat{a}^\dagger$

I just finished deriving the commutators: \begin{align} [\hat{H}, \hat{a}] &= -\hbar \omega \hat{a}\\ [\hat{H}, \hat{a}^\dagger] &= \hbar \omega \hat{a}^\dagger\\ \end{align} On the ...
6
votes
2answers
2k views

Proof for commutator relation $[\hat{H},\hat{a}] = - \hbar \omega \hat{a}$

I know how to derive below equations found on wikipedia and have done it myselt too: \begin{align} \hat{H} &= \hbar \omega \left(\hat{a}^\dagger\hat{a} + \frac{1}{2}\right)\\ \hat{H} &= \...