Solving Quantum Mechanics Integral Equation: How to Get from (1) to (2)?

[physics bob]

The book on quantum mechanics that I was reading says:
d<x>/dt = d/dt ∫^∞_-∞ |ψ(x,t)|² dx
=iħ/2m ∫^∞_-∞ x∂/∂x [ψ∂ψ^*/∂x+ψ^*∂ψ/∂x]dx (1)
=-∫^∞_-∞ [ψ∂ψ^*/∂x+ψ^*∂ψ/∂x]dx (2)
I want to know how to get from (1) to (2)

The book says you use integration by part:
∫^a_bfdg/dx dx = [fg]^a_b - ∫^a_bdf/df dg dx
I chose f = x and g = [ψ∂ψ^*/∂x+ψ^*∂ψ/∂x]
This gave me:
iħ/2m ∫^∞_-∞ x∂/∂x [ψ∂ψ^*/∂x+ψ^*∂ψ/∂x]dx = x[ψ∂ψ^*/∂x+ψ^*∂ψ/∂x]|^∞_-∞ - ∫^∞_-∞ [ψ∂ψ^*/∂x+ψ^*∂ψ/∂x]dx

The book says that the since lim_x->∞ ψ∂ψ^*/∂x = 0, the first term is 0. But I don't think you can make such assumption because firs term is a product of x.
Therefore you should need to show that |dx/dx|=1 < [ψ∂ψ^*/∂x+ψ^*∂ψ/∂x]/dx as x tend towards infinity.

I've been stuck on this problem for days please help. Thanks

 

[PeroK]

The book says that the since lim_x->∞ ψ∂ψ^*/∂x = 0, the first term is 0. But I don't think you can make such assumption because firs term is a product of x.
Therefore you should need to show that |dx/dx|=1 < [ψ∂ψ^*/∂x+ψ^*∂ψ/∂x]/dx as x tend towards infinity.

It's difficult to follow what you've typed, but in general these proofs in QM require the wavefunction to go to 0 faster than $1/x$ or $1/x^2$ etc.

You're right that the first term in the integration by parts won't be 0 for some functions. So, you need the assumption that $\psi$ tends to 0 quickly enough to kill off that term.

 

[physics bob]

It's difficult to follow what you've typed, but in general these proofs in QM require the wavefunction to go to 0 faster than $1/x$ or $1/x^2$ etc.

You're right that the first term in the integration by parts won't be 0 for some functions. So, you need the assumption that $\psi$ tends to 0 quickly enough to kill off that term.

Thanks for the reply and I'm sorry for the messy question. Is there a reason why we can assume that wave function go to 0 faster than 1/x ? or do we just assume for convenience?

 

[PeroK]

Thanks for the reply and I'm sorry for the messy question. Is there a reason why we can assume that wave function go to 0 faster than 1/x ? or do we just assume for convenience?

I would go with the idea that, since it describes the probability of finding a particle in a certain region, then in most or all physical cases the wave function must eventually go to 0 like $e^{-x^2}$.

There's a discussion here:

https://physics.stackexchange.com/q...wave-functions-required-to-vanish-at-infinity

 

[vanhees71]

It's in the very foundations of quantum theory that a pure state in the position representation is described by a square-integrable function in the sense of the Hilbert space of square integrable functions to begin with.

Now, this is only part of the mathematical subtlety, because you also have to deal with the description of the observables by operators that must be self-adjoint, which means that they have to be definable on a dense linear subspace of the Hilbert space, and they also must map the vectors in this dense subspace again to vektors in this space.

Take, e.g., the most simple case of a particle moving in one dimension with the observables position and momentum. These are described by the operators
$$\hat{x} \psi(x)=x \psi(x), \quad \hat{p} \psi(x)=-\mathrm{i} \hbar \partial_x \psi(x).$$
No it's clear that there are a lot of square-integrable functions, for which $\hat{x} \psi(x)$ is not square integrable. E.g., obviously $\psi(x)=\sin x/x$ is square integrable since $\int_{\mathbb{R}} \mathrm{d} x \sin^2/x^2=\pi$ exists, but $\hat{x} \psi(x)=\sin x$ is not square-integrable.

So for the formal rules to evaluate expectation values, the domain of validity for the corresponding operators is restricted to a dense subspace of the Hilbert space, where the operator doesn't lead out of this dense subspace. For position and momentum you can choose the Schwartz space of quickly falling functions, i.e., those square-integrable functions that go to 0 for $x \rightarrow \pm \infty$ faster than any power of $x$. Then all the expectation values necessary to define the moments of the position distribution are well defined.

For your calculation you must assume that the same ideas hold also for the Hamiltonian, i.e., for the free particle $\hat{H}=\hat{p}^2/(2m)=-\hbar^2 \partial_x^2/(2m)$. Then the manipulations of your integrals are well defined, leading to the conclusion that the operator describing the velocity, i.e., the time derivative of position, is given by
$$\hat{v}=\mathring{x}=\frac{1}{\mathrm{i} \hbar} [\hat{x},\hat{H}]=\frac{1}{m} \hat{p}.$$