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

$$af(a^\dagger)|n> = \frac{df(a^\dagger)}{da}|0>$$

Is written very sloppily. It should be written as

$$af(a^\dagger)|n> = f'(a^\dagger)|0>$$

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

 

[Shadowz]

Thank you for pointing that out.

So is it correct to consider

$$a(a^{\dagger})^{n-1}(a^{\dagger})|n> = \sqrt{n}a(a^\dagger)^{n-1}|n+1>$$

and then I have to find a way to do $$a|n+1> = \sqrt{n}|n>$$ 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

$$[a,f(a^\dagger)]=f'(a^\dagger)$$

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

$$af(a^\dagger) = f(a^\dagger)a + f'(a^\dagger)$$

but then how can we prove that $$(f(a^\dagger)a + f'(a^\dagger))|n> = f'(a^\dagger)|0>$$?

 

[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 $$f(a^\dagger)a|0> =0$$

 

[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 $$\Delta x|\alpha> = c\Delta p|\alpha>$$ where $$|\alpha>$$ 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 $$\Delta x |\alpha> = x\Delta p |\alpha>$$, 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 $\Delta p$.