How does ladder operation in an anharmonic oscillator lead to a value of 3?

[rubertoda]

I wonder how <0|a_(a+a_ + a_a+)a+|0> = <0|(a_a+ + 2a_a+)|0> = 3??. Here <0| is the (unperturbed) ground state level of an anharmonic oscillator and a+ is the creation operator and a_ is the annihilation operator.

I would get from <0|a_(a+a_ + a_a+)a+|0> that this becomes:
<0|a_a+a_a+ + (a_)^2(a+)^2|0> or
<0|a_a+a_a+ + (a_)(a_)(a+)(a+)|0>.

How in the world could this equal to <0|(a_a+ + 2a_a+)|0> = 3

could anyone give a qualitative reason for this?

kind regards

 

[TSny]

Hi rubertoda.

Consider a_a_a+a+|0>. Just go step by step letting the operators operate consecutively starting with the operator on the far right. So, the first thing to do is to see what a+|0> yields.

 

[rubertoda]

ok, no, i wrote wrong. I meant <0|(a_a+a_a+ + (a_)(a_)(a+)(a+))|0> with paranthesis around everything...thx

 

[TSny]

ok, no, i wrote wrong. I meant <0|(a_a+a_a+ + (a_)(a_)(a+)(a+))|0> with paranthesis around everything...thx

OK. I was just considering simplifying the second term. Note that you could write <0|(a_a+a_a+ + (a_)(a_)(a+)(a+))|0> = <0|a_a+a_a+|0> + <0|(a_)(a_)(a+)(a+)|0>. So, I was trying to have you think about how the second term <0|(a_)(a_)(a+)(a+)|0> simplifies. But, if you prefer, start with the first term <0|a_a+a_a+|0>.

 

[rubertoda]

ok, thx. i will do. and one last question. is it true that every term where the number of a_'s and a+'s aren't equal, i. e for example <0|x_x_x+|0> become 0, because it changes the state?

 

[TSny]

ok, thx. i will do. and one last question. is it true that every term where the number of a_'s and a+'s aren't equal, i. e for example <0|x_x_x+|0> become 0, because it changes the state?

Yes, that's right.

 

[rubertoda]

ok thanks a lot