Invariance of energy under change of origin

[cdog1350]

Suppose you have a particle in one dimension in an energy eigenstate, i.e. Hψ(x)=Eψ(x) for some E. For an observer B in a coordinate frame with the origin translated some distance K to the right, the wavefunction of the particle looks like ψ'(x) = ψ(x+K).

Surely, we expect the energy that B measures to be the same as you measure, so Hψ'(x) = Eψ'(x) or in other words Hψ(x+K) = Eψ(x+K). But how can we prove this?

 

[Bill_K]

<E> = ∫ψ*(x) H(x) ψ(x) dx = ∫ψ*(x') H(x') ψ(x') dx' where x' = x + k.

 

[HomogenousCow]

If you have a potential, it will produce some non physical factor in psi, which gets killed off by it's conjugate when we turn psi into a probability density function

 

[cdog1350]

<E> = ∫ψ*(x) H(x) ψ(x) dx = ∫ψ*(x') H(x') ψ(x') dx' where x' = x + k.

I'm not sure I understand this notation. The hamiltonian H is an operator, i.e. it's a functional that takes a function f(x) to some other function f'(x). I can see how it can be dependent on time like H(t) - that just means that the way it takes f(x) to f'(x) changes with time. But what does H(x) mean?

Also, even if you've proved the above result for <E>, that just says that the energy expectation values of ψ(x) and ψ'(x) are equal. Does that necessarily prove that ψ'(x) is an energy eigenvector with energy E?

 

[George Jones]

I am too busy to do the calculation, but I have a few thoughts.

Doesn't the Hamiltonian also have to be "shifted"? For example, consider a harmonic oscillator.

The chain rule might help.

Your example is a particular case of something much more general.

 

[tom.stoer]

usually

$$H(x) = -\partial_x^2 + V(x)$$

If you translate x to x+k in the wave function, then you have two options

1.)
$$x \to x+k$$
$$dx \to dx$$
$$\partial_x \to \partial_x$$
$$\psi(x) \to \psi(x+k)$$
$$V(x) \to V(x+k)$$
Then obviously this is a symmetry of the system and nothing changes

2.)
$$\psi(x) \to \psi(x+k)$$
$$V(x) \to V(x)$$
Then usually this is not a symmetry of the system; it corresponds to
$$\psi(x) \to \psi(x)$$
$$V(x) \to V(x-k)$$
and that means that you translate the oparticle w/o translating the potential, so the energy will change (only in the trivial case V(x) = V = const this is a symmetry)

 

[andrien]

<E> = ∫ψ*(x) H(x) ψ(x) dx = ∫ψ*(x') H(x') ψ(x') dx' where x' = x + k.

that is the nice way of doing it.