# Vanishing Wavefunction: Show Expectation Values of x and p Vanish

• misterpickle
In summary: The integral you are trying to do is \int_{-h}^{h} x \psi^{*}(x-\langle x \rangle) \psi(x-\langle x \rangle)\,dx.
misterpickle

## Homework Statement

If <x> and <p> are the expectation values of x and p formed with the wave-function of a one-dimensional system, show that the expectation value of x and p formed with the wave-function vanishes. The wavefunction is:

$$\phi(x)=exp(-\frac{i}{h}\langle p\rangle x)\psi(x+\langle x\rangle)$$Basically I don't know how to start this problem. Do I plug in the integral forms of the expectation values into the exponential and distribute the psi over x and $$\langle x\rangle$$?

You left out some important text, but I found the problem set with a google search. $$\langle x\rangle, \langle p\rangle$$ are computed with the wavefunction $$\psi(x)$$, i.e.

$$\langle x\rangle = \int x \psi^*(x) \psi(x) \, dx, \ldots$$

You are asked to compute expectation values with the wavefunction $$\phi(x)$$:

$$\langle x\rangle_\phi = \int x \phi^*(x) \phi(x) \, dx, \ldots$$

These expressions can be reduced to expectation values for certain quantities in the wavefunction $$\psi(x)$$. You don't have to do any integrals explicity, just some algebra and derivatives.

I've set it up the way you suggested, and the exponential terms computing $$\langle x \rangle$$ cancel. However there is a $$\langle x \rangle$$ term in $$\psi(x+\langle x \rangle)$$. Is this not a discrepancy to have the value you are trying to calculate inside the equation for calculating its value?

Specifically,

$$\int{ x \psi^{*}(x+\langle x \rangle) \psi(x+\langle x \rangle)\,dx}$$

misterpickle said:
I've set it up the way you suggested, and the exponential terms computing $$\langle x \rangle$$ cancel. However there is a $$\langle x \rangle$$ term in $$\psi(x+\langle x \rangle)$$. Is this not a discrepancy to have the value you are trying to calculate inside the equation for calculating its value?

Specifically,

$$\int{ x \psi^{*}(x+\langle x \rangle) \psi(x+\langle x \rangle)\,dx}$$

Well $$\langle x \rangle$$ is just a number (it came out of another integral over x, so it doesn't depend on x). There's still something you have to do to this integral though.

If $$\psi(x+\langle x \rangle)$$ is distributable (meaning $$f(x+y)=f(x)+f(y)$$) then I come up with the following:

$$\int{ x \psi^{*}(x+\langle x \rangle) \psi(x+\langle x \rangle)\,dx}$$
$$\int{ x[ (\psi^{*}x+\psi^{*}\langle x \rangle)( \psi x+\psi\langle x \rangle)\,dx}$$
$$\int{ (\psi^{*}\psi x^{3}+2\psi^{*}\psi x^{2}\langle x \rangle+\psi^{*}\psi x\langle x \rangle^{2})\,dx}$$
$$\langle x^{3}\rangle +2\langle x^{2}\langle x \rangle\rangle + \langle x\langle x\rangle^{2}\rangle$$

...which doesn't make any sense to me. Also the "distributable" assumption I made does not work for exponentials, since $$f(x+y)=f(x)f(y)$$ for exponential functions.

Yeah wavefunctions will almost never have that property since they're always related to exponential functions. Think about how you can convert this to an expectation value in $$\psi(x)$$ by making a substitution in the integration variable.

## 1. What is the vanishing wavefunction and why is it important?

The vanishing wavefunction refers to the probability amplitude of a particle being at a specific location in space, represented by the function Ψ(x). It is important because it allows us to calculate the probability of a particle being found at a certain position or having a certain momentum. It also plays a crucial role in understanding the behavior of quantum systems.

## 2. How do you calculate the expectation values of x and p from the vanishing wavefunction?

The expectation value of x (position) is calculated by taking the integral of x multiplied by the squared absolute value of the wavefunction Ψ(x) over all possible positions. The expectation value of p (momentum) is calculated by taking the integral of the momentum operator (represented by the variable p) multiplied by the squared absolute value of the wavefunction Ψ(x) over all possible positions.

## 3. Why do the expectation values of x and p vanish for the vanishing wavefunction?

The expectation values of x and p vanish for the vanishing wavefunction because the wavefunction itself is symmetrically distributed around the origin, meaning that there is an equal probability of finding the particle at any position. This results in a cancelation of the positive and negative values, resulting in a net value of zero.

## 4. Can the expectation values of x and p ever be non-zero for the vanishing wavefunction?

No, the expectation values of x and p will always be zero for the vanishing wavefunction. This is due to the fundamental principles of quantum mechanics, where the position and momentum of a particle cannot be precisely known at the same time.

## 5. How does the vanishing wavefunction relate to the Heisenberg uncertainty principle?

The vanishing wavefunction is closely related to the Heisenberg uncertainty principle, which states that the more precisely we know the position of a particle, the less precisely we can know its momentum, and vice versa. The vanishing wavefunction represents a state where both the position and momentum are completely uncertain, resulting in zero expectation values for both quantities.

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