# Invariancy of H(x,p) to translation and conservation of momentum

1. Aug 19, 2007

### Ene Dene

I have two similar questions.
1. Show the relationship between invariancy of hamiltonian to translation and conservation of momentum. (In QM)
2. Show the relationship between invariancy of hamiltonian to rotation and conservation of angular momentum. (In QM)
I have no idea how to prove this. I was wondering if I could use Heisenberg operator equation on momentum operator so I could prove that operator doesn't change with time, but I don't how to input hamiltonian invariancy to translation in equation and even then is that the real way of doing it.

2. Aug 19, 2007

### Dick

p is the infinitesimal generator of translations. So basically H(x+a)=exp(iap)H(x)exp(-iap). I'm being sloppy about the signs and factors of hbar here. Fix them please. If H(x)=H(x+a) (translation invariant) you can expand that relation at first order to conclude [p,H]=0. So translation invariance means p commutes with the Hamiltonian.

3. Aug 20, 2007

### Ene Dene

I don't understand your answer. First, I don't understand why I need this:
H(x+a)=exp(iap)H(x)exp(-iap)
Second, how do I expand H in Taylor series (and why?) and how to connect that with [p,H]=0, and how does that prove the connection between hamiltonian invariantcy to translation and conservation of momentum?

4. Aug 20, 2007

### dextercioby

In the Schroedinger picture an observable is conserved iff it commutes with the time indep hamiltonian. So p is conserved iff $[p,H]=0$.

exp(iap) is the operator for spatial translations. H(x) is the time indep hamiltonian.

5. Aug 20, 2007

### Dick

You need to show that translation invariance of H means H commutes with p. To do this you notice that exp(a*d/dx)f(x)=f(x+a), since the left hand side is the power series expression of f(x+a). Now p is essentially d/dx. You can now use this to express the translated hamiltonian in terms of the original hamiltonian and the translation operator exp(iap) (like I said you'll need to throw some i's and hbar's around here). H(x+a)=exp(iap)H(x)exp(-iap) to lowest order is H(x+a)=(1+iap)H(x)(1-iap). If H(x+a)=H(x), what can you conclude about the commutator of H and p?

6. Aug 20, 2007

### ChaoticOrder

Here is a nice derivation of how an we get the exponentiated operator to generate translations:

Consider the differentiable function f at x. Translating f to the right means changing the argument to get f(x-a). Now take the displacement a and cut it up in N pieces, each piece having length Dx. Hence N Dx = a. The translated function now reads f(x-Dx-Dx-...-Dx), where there are N Dx's. Take the definition of a derivative and rearrange it to read:

f(x-Dx) = f(x) - f'(x)*Dx
= [1 - Dx (d/dx)]f(x) in the limit as Dx ---> 0

Doing this procedure to the translated function gives us

f(x-a) = [1 - Dx (d/dx)]f(x-a+Dx)
= [1 - Dx (d/dx)][1 - Dx (d/dx)]f(x-a+2Dx)
= [1 - Dx (d/dx)][1 - Dx (d/dx)][1 - Dx (d/dx)]f(x-a+3Dx)
...
= [1 - Dx (d/dx)]^N f(x)
= [1 - a/N (d/dx)]^N f(x)

In quantum mechanics we learned that -i*hbar (d/dx) = p, the momentum operator. Hence we have

f(x-a) = [1 - iap/N ]^N f(x)

We now allow the number N to approach infinity, which basically means that we are subdividing the original interval a into an infinitum of infinitesimals. We then utilize the definition of the number e to see that

f(x-a) = exp(-iap)f(x)

Read from right to left, we have just discovered that the exponentiated momentum operator, when applied to a function, generates a translation. A similiar thing can be done for the anguilar momentum operator, but you will need to consider a function of two variables, f(x,y), and examine how both x and y change when the function is rotated.

Now back to the Hamiltonian, H(x). How do we derive its transformation properties under translations? Consider the state vector |psi(x)>. To find the energy of this state we take the Dirac bracket of the Hamiltonian. The enregy of this state is independent of the origin of the coordinate system that we use, so we have

E = <psi(x)|H(x)psi(x)>
= <psi(x-a)|H(x-a)psi(x-a)>
= <exp(-iap)psi(x)|H(x-a)exp(-iap)psi(x)>
= <psi(x)|exp(iap)H(x-a)exp(-iap)psi(x)>

By examining the first and fourth lines we are lead to the transformation properties of the Hamiltonian

H(x) = exp(iap)H(x-a)exp(-iap) or H(x-a) = exp(-iap)H(x)exp(iap)

Now we do what the last post says. We invoke the translational invariance of the Hamiltonian, H(x)=H(x-a), expand the transformation law derived above to first order, and voila! We have the commutation ralation between the Energy and Momentum operators.

7. Aug 28, 2007

### Ene Dene

Thanks I got it now!