Harmonic oscillator in 2D - applying operators

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| ax+ay++ax+ay+axay++axay|0,1> = <1,0|1,0>

Homework Equations

3. The Attempt at a Solution
We have that |0,1> = ay+ |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
Mentor
Can you figure out what
(ax+ay++ax+ay+axay++axay)|0,1>
is?

(Hint: try it term by term.)

ax+ay+|1,0 > =ax+ay+ay+|0,0 >
ax+ay|1,0 >=ax+ayay+|0,0 >
axay+|1,0 >=axay+ay+|0,0 >
axay|1,0 >=axayay+|0,0 >

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

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

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

DrClaude
Mentor
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.

ay|1,0> = sqrt(0)|1, 0-1> = 0

DrClaude
Mentor
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 ay|1,0> = 0.

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

ax+ay++ax+ay+axay++axay|0,1> = ax+ay++axay+|1,0>
ax+ay+|1,0>=|2,1>
axay+|1,0>=|0,1>

There is a mistake somewhere it is
ax+ay+|0,1>=|1,2>
axay+|0,1>=0

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

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

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

DrClaude
Mentor
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.

Thank you very much !