Annihilation Operators: Prove af(a^\dagger)|n>=df(a^\dagger)/da|0>

[Shadowz]

So we all know about a and a^\dagger.

My problem says that if f(a^\dagger) is an arbitrary polynomial in a^\dagger then af(a^\dagger)|n> = \frac{df(a^\dagger)}{da}|0> where |0> is the ground state energy. How can I go about proving this?

A hint would be highly appreciated.

Thanks,

 

[ismaili]

So we all know about a and a^\dagger.

My problem says that if f(a^\dagger) is an arbitrary polynomial in a^\dagger then af(a^\dagger)|n> = \frac{df(a^\dagger)}{da}|0> where |0> is the ground state energy. How can I go about proving this?

A hint would be highly appreciated.

Thanks,

Since f(a^\dagger) is a polynomial.
It suffices to show that
a (a^\dagger)^n |0\rangle = n(a^\dagger)^{n-1}
where n = 0,1,2,\cdots

And you know the commutator of [a,a^\dagger] = ...

 

[Shadowz]

Since f(a^\dagger) is a polynomial.
It suffices to show that
a (a^\dagger)^n |0\rangle = n(a^\dagger)^{n-1}
where n = 0,1,2,\cdots

And you know the commutator of [a,a^\dagger] = ...

Maybe I don't quite get it but you said we can assume f(a^\dagger) be (a^\dagger)^n

But then if I consider the LHS, will that operator raises |n> to |2n-1>? (each a^\dagger would raise |n> to |n+1>) but the RHS has |0>.

Thank for your help,

 

[arkajad]

The formula

<br /> af(a^\dagger)|n&gt; = \frac{df(a^\dagger)}{da}|0&gt;<br />

Is written very sloppily. It should be written as

<br /> af(a^\dagger)|n&gt; = f&#039;(a^\dagger)|0&gt;<br />

Why? Because f(a^\dagger) does not depend on a.

 

[Shadowz]

Thank you for pointing that out.

So is it correct to consider

a(a^{\dagger})^{n-1}(a^{\dagger})|n&gt; = \sqrt{n}a(a^\dagger)^{n-1}|n+1&gt;

and then I have to find a way to do a|n+1&gt; = \sqrt{n}|n&gt; in order to get to the form n(a^\dagger)^{n-1} but the RHS has |0> so I don't know how to get it.

Thanks,

 

[arkajad]

Why not start with something like [A,BC]=A[B,C]+[A,B]C, thus

[a,(a^\dagger)^n]=a^\dagger [a,(a^\dagger)^{n-1}]+[a,a^\dagger](a^\dagger )^{n-1}=\ldots ...

in order to compute [a,(a^\dagger)^n]

 

[Shadowz]

Why not start with something like [A,BC]=A[B,C]+[A,B]C, thus

[a,(a^\dagger)^n]=a^\dagger [a,(a^\dagger)^{n-1}]+[a,a^\dagger](a^\dagger )^{n-1}=\ldots ...

in order to compute [a,(a^\dagger)^n]

If my algebra is correct then [a,(a^\dagger)^n] = (a^\dagger)^{n-1} +(n-1)(a^\dagger)^{n-1} = n(a^\dagger)^{n-1}Should I try to operate both sides with |0>?

 

[arkajad]

So, your algebra can be generalized to

[a,f(a^\dagger)]=f&#039;(a^\dagger)

Don't you think there is a mistake on the LHS of the equation you want to prove? Don't you think there should be |0> there rather than |n>? The RHS does not depend on n, how the LHS can?

 

[Shadowz]

Yes. So the RHS is close, but then we still have to break the commutator and somehow put in n and 0.

By the way, thank you!

and so I rearrange it to be

af(a^\dagger) = f(a^\dagger)a + f&#039;(a^\dagger)

but then how can we prove that (f(a^\dagger)a + f&#039;(a^\dagger))|n&gt; = f&#039;(a^\dagger)|0&gt;?

 

[Shadowz]

I am 95% sure that the expression I try to show is correct because it also says that " |0> is the ground state energy and |n> is any harmonic oscillator state."

PS: Although I think it would make a lot of sense to have |0> on both side since f(a^\dagger)a|0&gt; =0

 

[arkajad]

Just think of n=1000 and f(x)\equiv 1

Then on LHS you will have 999th excited state a|n>, on the LHS you will have just 0 (because f'=0 in this case).

 

[Shadowz]

Yes that is exactly what my 3rd post said (or not exactly, but similar idea)

Now I think that \frac{df(a^\dagger) }{da} can make a difference.

 

[naima]

Isn't there a relation between n and the degree of f?

 

[naima]

It seems to be true for any polynomial (of degree = n)
if you take |0> on each side of the equality.

 

[Shadowz]

It seems to be true for any polynomial (of degree = n)
if you take |0> on each side of the equality.

You're right. I got it. Thank all of you for help!

 

[Shadowz]

Hi,

How can I write \Delta x and \Delta p as operators? I want to show that \Delta x|\alpha&gt; = c\Delta p|\alpha&gt; where |\alpha&gt; is coherent state.

I feel like I have to write x and p in terms of annihilation operators, but I always think that \Delta x and \Delta p are numbers, not operators.

Thanks,

 

[arkajad]

It is not clear by itself what \Delta x,\Delta p can mean in this context. One can try to guess their meaning, but it is not evident.

 

[Shadowz]

So all I try to do was to show that the coherent state has minimum uncertainty equally distributed between x and p. And the hint given was to show that \Delta x |\alpha&gt; = x\Delta p |\alpha&gt;, and thus it makes me think that I should treat \Delta x and \Delta p as operators rather than numbers.

 

[arkajad]

For minimum uncertainty you need to calculate \langle \alpha|(\Delta x)^2|\alpha\rangle=\langle\Delta x\alpha|\Delta x\alpha\rangle, where \Delta x=x-\langle\alpha| x|\alpha\rangle . Now you can plug in your anihilation and creation operators. The same for \Delta p.