Harmonic oscillator in 2D - applying operators

[Dassinia]

Hello, I juste don't know how this was done it is on the solutionnary of a very long exercise and i am not getting this calculation

1. Homework Statement

<1,0| a_x⁺a_y⁺+a_x⁺a_y+a_xa_y⁺+a_xa_y|0,1> = <1,0|1,0>

Homework Equations

3. The Attempt at a Solution
We have that |0,1> = a_y⁺ |0,0>
I don't understand how they did the simplification to get to the <1,0|1,0>, i am missing something
Thanks [/B]

 

[DrClaude]

Can you figure out what
(a_x⁺a_y⁺+a_x⁺a_y+a_xa_y⁺+a_xa_y)|0,1>
is?

(Hint: try it term by term.)

 

[Dassinia]

a_x⁺a_y⁺|1,0 > =a_x⁺a_y⁺a_y⁺|0,0 >
a_x⁺a_y|1,0 >=a_x⁺a_ya_y⁺|0,0 >
a_xa_y⁺|1,0 >=a_xa_y⁺a_y⁺|0,0 >
a_xa_y|1,0 >=a_xa_ya_y⁺|0,0 >

But then I don't know how to simplify all that

 

[DrClaude]

You're going backwards: reduce the number of operators. The same way that a_y⁺|0,0 > = |0,1 >, what is a_y|1,0 > ?

 

[Dassinia]

I am so confused, are you saying that I should not replace |1,0> by ay+| 0 0 > ?
ay|1,0 > = 0 ?

 

[DrClaude]

I am so confused, are you saying that I should not replace |1,0> by ay+| 0 0 > ?

Indeed. The final expression, <1,0|1,0> has no more operators. What you need to do is to apply the operators successively to get the resulting state vector (ket).

ay|1,0 > = |1,0> ?

That's not correct. You need to revise creation and annihilation operators.

 

[Dassinia]

I edited my answer
ay|1,0> = sqrt(0)|1, 0-1> = 0

 

[DrClaude]

I edited my answer
ay|1,0> = sqrt(0)|1, 0-1> = 0

I don't like the intermediate step, as there is no such thing as |1, -1>. The result is simply a_y|1,0> = 0.

So now go back to my question in post #2.

 

[Dassinia]

a_x⁺a_y⁺+a_x⁺a_y+a_xa_y⁺+a_xa_y|0,1> = a_x⁺a_y⁺+a_xa_y⁺|1,0>
a_x⁺a_y⁺|1,0>=|2,1>
a_xa_y⁺|1,0>=|0,1>

 

[Dassinia]

There is a mistake somewhere it is
a_x⁺a_y⁺|0,1>=|1,2>
a_xa_y⁺|0,1>=0

So I get to |1,2> and not |1,0>

 

[Dassinia]

Oh my god I have to see this again I applied ay to |1,0> and not |0,1> so i canceled them
I am left with

<1,0|1,2>=0 but why ? and <1,0|1,0>=1

 

[DrClaude]

I am left with
<1,0|1,2>=0 but why ?

Because eigenstates of the harmonic oscillator are orthogonal. It is the y part, <0|2>, that is zero.

and <1,0|1,0>=1

For normalized states, yes.

 

[Dassinia]

Thank you very much !