How Does Momentum Translate in Quantum Mechanics?

[Karliski]

Homework Statement

Prove the following:

$$|\vec{p}\rangle = \exp\left[i\vec{p} \cdot \hat{\vec{x}} / \hbar\right] |\vec{p}=\vec{0}\rangle$$

Where $$|\vec{p}\rangle$$ is any eigenstate of the momentum operator $$\hat{\vec{p}}$$.

Homework Equations

See above.

The Attempt at a Solution

I guess one has to follow the same argument as when one shows that momentum is the generator of translation. Would this mean that position is the generator of momentum?

When following the argument, one ends up with:

$$|\vec{p} + \vec{p}'\rangle = e^{-i \hat{\vec{K}} \cdot \vec{p}}' |\vec{p}$$

I just chose the 'translation' operator as $$\hat{K}$$ again, not to be confused with the one when working with translation of position. I don't really know if this argument will carry on to prove the result or how to do it. Any tips would be appreciated.

 

[Jhero]

Thank you for your post. I am a scientist and I would be happy to help you with your proof.

First, let's clarify some notation. In quantum mechanics, we often use the bra-ket notation to represent states and operators. In this notation, |\vec{p}\rangle represents the eigenstate of the momentum operator \hat{\vec{p}} with momentum \vec{p}. The operator \hat{\vec{x}} is the position operator and \hbar is the reduced Planck constant.

Now, to prove the given statement, we need to show that |\vec{p}\rangle can be written as \exp\left[i\vec{p} \cdot \hat{\vec{x}} / \hbar\right] |\vec{p}=\vec{0}\rangle. To do this, we will use the fact that \hat{\vec{p}} is the generator of translations in position space. This means that applying the operator \exp\left[i\vec{p} \cdot \hat{\vec{x}} / \hbar\right] to the state |\vec{p}=\vec{0}\rangle will give us a state |\vec{p}\rangle with momentum \vec{p}.

To see this, let's consider the action of \hat{\vec{p}} on a position eigenstate |\vec{x}\rangle. We know that \hat{\vec{p}}|\vec{x}\rangle = -i\hbar \vec{\nabla}|\vec{x}\rangle = \vec{p}|\vec{x}\rangle, where \vec{\nabla} is the gradient operator. This means that the state \hat{\vec{p}}|\vec{x}\rangle is an eigenstate of the momentum operator with momentum \vec{p}.

Now, let's apply the operator \exp\left[i\vec{p} \cdot \hat{\vec{x}} / \hbar\right] to the state |\vec{x}\rangle. We get:

\exp\left[i\vec{p} \cdot \hat{\vec{x}} / \hbar\right]|\vec{x}\rangle = \sum_{n=0}^\infty\frac{(i\vec{p} \cdot \hat{\vec{x}} / \hbar)^n}{n!}|\vec{x}\rangle = \sum_{n