How do Bose operators transform in K-space homework equations?

[Petar Mali]

Homework Statement

How this would look in K space

-\sum_{\vec{n},\vec{m}}I_{\vec{n},\vec{m}}\hat{a}^+_{\vec{n}}\hat{a}_{\vec{n}}\langle \hat{a}^+_{\vec{n}}\hat{b}^+_{\vec{m}}\rangle

I need to get

-\sum_{\vec{k}}\hat{a}^+_{\vec{n}}\hat{a}_{\vec{n}}\frac{1}{N}\sum_{\vec{q}}J(\vec{q})\langle \hat{a}^+_{\vec{q}}\hat{b}^+_{-\vec{q}}\rangle

Homework Equations

\hat{a}, \hat{a}^+,\hat{b},\hat{b}^+ are Bose operators

I_{\vec{n},\vec{m}}=\frac{1}{N}\sum_{\vec{k}}J(\vec{k})e^{i\vec{k}(\vec{n}-\vec{m})

\hat{a}^+_{\vec{n}}=\frac{1}{\sqrt{N}}\sum_{\vec{k}}\hat{a}^+_{\vec{k}}e^{-i\vec{k}\cdot\vec{n}}

\hat{a}_{\vec{n}}=\frac{1}{\sqrt{N}}\sum_{\vec{k}}\hat{a}_{\vec{k}}e^{i\vec{k}\cdot\vec{n}}

\hat{b}^+_{\vec{n}}=\frac{1}{\sqrt{N}}\sum_{\vec{k}}\hat{b}^+_{\vec{k}}e^{-i\vec{k}\cdot\vec{n}}

\hat{b}_{\vec{n}}=\frac{1}{\sqrt{N}}\sum_{\vec{k}}\hat{b}_{\vec{k}}e^{i\vec{k}\cdot\vec{n}}

The Attempt at a Solution

Homework Statement

-\sum_{\vec{n},\vec{m}}I_{\vec{n},\vec{m}}\hat{a}^+_{\vec{n}}\hat{a}_{\vec{n}}\langle \hat{a}^+_{\vec{n}}\hat{b}^+_{\vec{m}}\rangle=-\sum_{\vec{n},\vec{m}}\frac{1}{N^2}\sum_{\vec{k}}J(\vec{k})e^{i\vec{k}(\vec{n}-\vec{m})}\sum_{\vec{k}_1}\hat{a}^+_{\vec{k}_1}e^{-i\vec{k}_1\cdot\vec{n}}\sum_{\vec{k}_2}\hat{a}_{\vec{k}_2}e^{i\vec{k}_2\cdot\vec{n}}\frac{1}{N}\langle\sum_{\vec{k}_3}\hat{a}^+_{\vec{k}_3}e^{-i\vec{k}_3\cdot\vec{n}}\sum_{\vec{k}_4}\hat{b}^+_{\vec{k}_4}e^{-i\vec{k}_4\cdot\vec{n}}\rangle=

=-\frac{1}{N^3}\sum_{\vec{k}}J(\vec{k})\sum_{\vec{k}_1,\vec{k}_2,\vec{k}_3,\vec{k}_4}\hat{a}^+_{\vec{k}_1}\hat{a}_{\vec{k}_2}\langle \hat{a}^+_{\vec{k}_3}\hat{b}^+_{\vec{k}_4}\rangle\sum_{\vec{n}}e^{i\vec{k}\cdot\vec{n}}e^{-i\vec{k}_1\cdot\vec{n}}e^{i\vec{k}_2\cdot\vec{n}}e^{-i\vec{k}_3\cdot\vec{n}}\sum_{\vec{m}}e^{-i\vec{k}\cdot\vec{m}}e^{-i\vec{k}_4\cdot\vec{m}

So I get

=-\frac{1}{N^3}\sum_{\vec{k}}J(\vec{k})\sum_{\vec{k}_1,\vec{k}_2,\vec{k}_3,\vec{k}_4}\hat{a}^+_{\vec{k}_1}\hat{a}_{\vec{k}_2}\langle \hat{a}^+_{\vec{k}_3}\hat{b}^+_{\vec{k}_4}\rangle N\delta_{\vec{k}+\vec{k}_2,\vec{k}_1+\vec{k}_3}N\delta_{\vec{k}_3,-\vec{k}_4}

And what now? I didn't get what I need to get!

=-\frac{1}{N}\sum_{\vec{k}}J(\vec{k})\sum_{\vec{k}_1,\vec{k}_2}\hat{a}^+_{\vec{k}_1}\hat{a}_{\vec{k}_2}\langle \hat{a}_{\vec{k}-\vec{k}_1+\vec{k}_2}\hat{a}_{-\vec{k}}\rangle

 

[Petar Mali]

Do someone no this? Thanks! Anyone?

 

[DrDu]

I need to get

-\sum_{\vec{k}}\hat{a}^+_{\vec{n}}\hat{a}_{\vec{n}}\frac{1}{N}\sum_{\vec{q}}J(\vec{q})\langle \hat{a}^+_{\vec{q}}\hat{b}^+_{-\vec{q}}\rangle

Why is there still an n as an index?

 

[Petar Mali]

Mistake. So I need to get

<br /> -\sum_{\vec{k}}\hat{a}^+_{\vec{k}}\hat{a}_{\vec{k}} \frac{1}{N}\sum_{\vec{q}}J(\vec{q})\langle \hat{a}^+_{\vec{q}}\hat{b}^+_{-\vec{q}}\rangle<br />

 

[shawl]

I did less work in this field.
But here I give my opinion.

The average of operators is equal to a number, that means the term <a+b+> should be viewed as a number.

Therefore, there are only two operators a+(k1) and a(k2) in the last equation you given: {a+(k1) a(k2) <a+(k-k1+k2) a+(-K)>}. Note that the condition for the conservation of momentum. It is delta(k1,k2) so that k1=-k2.

Finally, you got the desired equation.

PS: Sometimes, if the time dependent operator you are considering, please add the condition for the conservation of energy.

 

[Petar Mali]

I did less work in this field.
But here I give my opinion.

The average of operators is equal to a number, that means the term <a+b+> should be viewed as a number.

Therefore, there are only two operators a+(k1) and a(k2) in the last equation you given: {a+(k1) a(k2) <a+(k-k1+k2) a+(-K)>}. Note that the condition for the conservation of momentum. It is delta(k1,k2) so that k1=-k2.

Finally, you got the desired equation.

PS: Sometimes, if the time dependent operator you are considering, please add the condition for the conservation of energy.

I'm not sure how to do that?

Can you explain me bold part?

\langle \hat{a}_{\vec{k}-\vec{k}_1+\vec{k}_2}\hat{a}_{-\vec{k}}\rangle=?

 

[shawl]

When you considered the condition for the conservation of momentum, you will derive k1=-k2.

then the last equation is changed
from {a+(k1) a(k2) <a+(k-k1+k2) a+(-k)>}
to...{a+(k1) a(k1) <a+(k) a+(-k)>}.
That is the formalism you desired.

In generally, double creation and destruction operators should be taken into account for the condensed term in Hamiltonian.

If the condensed system is in its ground state, the total momentum is 0. (sum of all k is zero) Double creation and destruction process do not affect the total momentum which is 0.

Therefore, the double creation and destruction operators is required to have the anti-paralleled momentum, i.e. in your equation <a+(q)a+(-q)>.

 

[shawl]

The conservation of momentum and for the conservation of energy are both the essential physical condition for our world.

But they are not contained in Hamiltonian in our quantum mechanism which is the formalism of mathematics.

That's why we need add the conditions by hand.

 

[Petar Mali]

When you considered the condition for the conservation of momentum, you will derive k1=-k2.

then the last equation is changed
from {a+(k1) a(k2) <a+(k-k1+k2) a+(-k)>}
to...{a+(k1) a(k1) <a+(k) a+(-k)>}.
That is the formalism you desired.

In generally, double creation and destruction operators should be taken into account for the condensed term in Hamiltonian.

If the condensed system is in its ground state, the total momentum is 0. (sum of all k is zero) Double creation and destruction process do not affect the total momentum which is 0.

Therefore, the double creation and destruction operators is required to have the anti-paralleled momentum, i.e. in your equation <a+(q)a+(-q)>.

How to see the law conservation of momentum? This, what I wrote, is one of the term od the Hamiltonian. I'm not quite sure how to see this conservation? I how Hamiltonian which I wrote in https://www.physicsforums.com/showthread.php?t=445228

And term \hat{H}_4 I transform in last post in that subject. And this is Fourrier transformation of one of the terms. Can you explain me just how to see this law of conservation of momentum? Maybe in some easier case?

 

[shawl]

The Hamiltonian accompanied by the conditions of conservation of momentum or energy can be analogous to the differential equation accompanied by the boundary condition or the initial condition.

Yes, In the term H4, it is not easy to understand why we should consider the conservation of momentum. That's because H4 is NOT written in momentum space (k space). But the conservation is still behind the formalism.

For a toy example, Here is a Hamiltonian consists of a pair of operators a+(i) a(j), where i, j can be arbitrary numbers. Behind the Hamiltonian, there is a law of conservation of momentum and energy, which tells us that the created particle must have the same momentum and energy with the destructed one.

 

[shawl]

Briefly speaking, the law of the conservation of momentum and energy can not be naturally written in the Hamiltonian. We should add it to our equation because it really exists in our world.

It's hard to understand the law when Hamiltonian is not written in k space. Although i, j can be the arbitrary numbers, the energy and momentum of particle i and particle j are related. To derive a realistic result, we should add the law by hand.

 

[Petar Mali]

All of that is ok but I still don't understand how to calculate <br /> \langle \hat{a}^+_{\vec{k}-\vec{k}_1+\vec{k}_2}\hat{a}^+_{-\vec{k}}\rangle=?<br />

Can I say just (if I suppose this)

<br /> \langle \hat{a}^+_{\vec{k}-\vec{k}_1+\vec{k}_2}\hat{a}^+_{-\vec{k}}\rangle=\langle\hat{a}^+_{\vec{k}}\hat{a}^+_{-\vec{k}}\rangle\delta_{\vec{k}_1,\vec{k}_2}<br />?

Then I will have <br /> -\frac{1}{N}\sum_{\vec{k}}J(\vec{k})\sum_{\vec{k}_1 ,\vec{k}_2}\hat{a}^+_{\vec{k}_1}\hat{a}_{\vec{k}_2 }\langle \hat{a}^+_{\vec{k}-\vec{k}_1+\vec{k}_2}\hat{a}^+_{-\vec{k}}\rangle=-\frac{1}{N}\sum_{\vec{k}}J(\vec{k})\sum_{\vec{k}_1 ,\vec{k}_2}\hat{a}^+_{\vec{k}_1}\hat{a}_{\vec{k}_2 }\langle\hat{a}^+_{\vec{k}}\hat{a}^+_{-\vec{k}}\rangle\delta_{\vec{k}_1,\vec{k}_2}=-\frac{1}{N}\sum_{\vec{k}}J(\vec{k})\sum_{\vec{k}_1 }\hat{a}^+_{\vec{k}_1}\hat{a}_{\vec{k}_1 }\langle\hat{a}^+_{\vec{k}}\hat{a}^+_{-\vec{k}}\rangle<br />

 

[shawl]

Yes, you are right. That is what i mean

 

[Petar Mali]

Thanks for helping me!