Translation Operators: Raising & Lowering Equations

[dyn]

α I have been studying translation operators of the type T = exp ( -ipx₀/ hbar) where p is the momentum operator which leads to T⁺xT = x+x₀. I am ok with that but then I came across the following equation concerning raising and lowering operators exp(-alpha a⁺) a exp(alpha a⁺) = a + alpha. Is this equation related to the translation operator or is it something different ?

 

[strangerep]

α I have been studying translation operators of the type T = exp ( -ipx₀/ hbar) where p is the momentum operator which leads to T⁺xT = x+x₀. I am ok with that but then I came across the following equation concerning raising and lowering operators exp(-alpha a⁺) a exp(alpha a⁺) = a + alpha. Is this equation related to the translation operator or is it something different ?

It involves similar math. Typically ##a, a^\dagger## are linear combinations of the x,p operators.

 

[dyn]

Is it a standard result like with the translation operator or would it need to be calculated ?

 

[vanhees71]

I guess you mean the annihilation and creation operators for phonons of a simple harmonic oscillator. Then you have
$$\hat{a}=\sqrt{\frac{m \omega}{2 \hbar}} \hat{x} + \frac{\mathrm{i}}{\sqrt{2m \omega \hbar}}\hat{p}.$$
Now just evaluate ##\hat{T}^{\dagger} \hat{a} \hat{T}## using the known commutation relations of this translation operator with ##\hat{x}## (given in your posting) and ##\hat{p}## (which is straight forward).

 

[dyn]

ħ Could someone please explain how the following result is obtained step by step. I have seen 2 explanations so far and I don't understand the steps involved.
<x-x₀| exp( i p₀ x / hbar) = exp( ip₀(x-x₀)/hbar) <x-x₀| Thanks

 

[strangerep]

Is it a standard result like with the translation operator or would it need to be calculated ?

All these sorts of formulas are (or can be derived) from "Baker-Campbell-Hausdorff" formula(s). Most textbooks on QM and symmetries usually discuss it.

 

[strangerep]

ħ Could someone please explain how the following result is obtained step by step. I have seen 2 explanations so far and I don't understand the steps involved.
<x-x₀| exp( i p₀ x / hbar) = exp( ip₀(x-x₀)/hbar) <x-x₀|

I have a policy of not putting much more effort into my answers than the questioner puts into his/her questions.

Try either posting a link to a derivation and saying which step(s) you don't follow. Better still, reproduce (in Latex) the key steps of the derivation you're having trouble with. You'll tend to get more more and better answers if you don't create unnecessary work for potential helpers.

 

[dyn]

hi strangerep. If you don't want to answer my question that's up to you. I have a policy of only accepting help from people who wish to help me.

 

[berkeman]

Thread closed temporarily for Moderation...

 

[Dale]

The thread is reopened. We remind all participants about the desire to create a helpful and friendly atmosphere and also the forum rules which require posters to demonstrate their own effort and require respondents to avoid giving complete solutions.

 

[dyn]

This wasn't a homework question. It only involves 2 or 3 steps which I don't understand. I can't demonstrate my effort because I don't understand how to do it !

 

[Avodyne]

<x-x₀| exp( i p₀ x / hbar) = exp( ip₀(x-x₀)/hbar) <x-x₀| Thanks

My interpretation: on the left-hand side, x is an operator. This operator is acting to the left on a position eigenstate (in bra form). On the right-hand side, the operator x has been replaced by its eigenvalue x-x₀, where x is now a number.

In general, a function of an operator, acting on an eigenstate of that operator, can be replaced with the same function of the corresponding eigenvalue.

It would be better notation to distinguish between an operator and an eigenvalue of that operator.

 

[dyn]

Thank you Avodyne

 

[dyn]

when the exponential as a number is taken outside the bracket would it not have to complex conjugated as it is on the bra side ?

 

[dyn]

I have got it now.