Question

Psi(x) = (alpha/pi) ^ (1/4) * e ^ (- (alpha * x ^ 2)/2)

Find wave function
1- p*2 =
2- x =
3- X*2 =

Answer 1

The wave function can be obtained by taking the square root of the probability density function. In this case, the probability density function is given by:

Psi(x) = (alpha/pi)^(1/4) * e^(-(alpha * x^2)/2)

So the wave function is:

phi(x) = sqrt(Psi(x)) = sqrt((alpha/pi)^(1/4) * e^(-(alpha * x^2)/2))

We can simplify this expression by using the fact that the square root of a product is equal to the product of the square roots:

phi(x) = sqrt(alpha/pi)^(1/4) * sqrt(e^(-(alpha * x^2)/2))

phi(x) = (alpha/pi)^(1/8) * e^(-(alpha * x^2)/4)

Therefore, the wave function is:

phi(x) = (alpha/pi)^(1/8) * e^(-(alpha * x^2)/4)

Now, to answer the other questions:

To find p^2, we need to square the wave function:

phi(x)^2 = ((alpha/pi)^(1/8) * e^(-(alpha * x^2)/4))^2

phi(x)^2 = (alpha/pi)^(1/4) * e^(-(alpha * x^2)/2)

So p^2 is (alpha/pi)^(1/4) * e^(-(alpha * x^2)/2).

The variable x is already given in the function Psi(x). It represents the position of a particle in one dimension.

To find X^2, we need to use the operator for the position squared:

X^2 = x^2

So X^2 is simply x squared, which in this case would be:

X^2 = x^2

Answer 2

1.
`\( p^2 \Psi(x) \)`=
`\( p^2 \Psi(x) = -\Psi''(x) \)`.

2.
`\( x \Psi(x) \)`=
`\( p^2 \Psi(x) = -\Psi''(x) \).`

3.
`\( x^2 \Psi(x) \)`=
`\[x^2 \Psi(x) = x^2 \left((\alpha)/(\pi)\right)^(1/4) e^{-(\alpha x^2)/(2)}\]`

To find the expressions for
`\( p^2 \Psi(x) \)`,
`\( x \Psi(x) \)`, and
`\( x^2 \Psi(x) \)` given the wave function
`\(\Psi(x)` =
`\left((\alpha)/(\pi)\right)^(1/4) e^{-(\alpha x^2)/(2)}\)`, we need to perform the appropriate operations. Let's break down each step.

1.
`\( p^2 \Psi(x) \)`

The momentum operator in quantum mechanics, represented in position space, is given by
`\( p = -i \hbar (d)/(dx) \)` (where
`\( \hbar \)` is the reduced Planck's constant and i is the imaginary unit). Thus,
`\( p^2 = -\hbar^2 (d^2)/(dx^2) \)`. However, if we ignore constants like
`\( \hbar \)` and
`\( i \)`,
`\( p^2 \)` simply becomes
`\( -(d^2)/(dx^2) \)`. Therefore,
`\( p^2 \Psi(x) \)`is the second derivative of
`\(\Psi(x)\)` with a negative sign.

Let's calculate
`\( p^2 \Psi(x) \)`:

- First, we find the first derivative of
`\(\Psi(x)\)` with respect to x:

`\[ \Psi'(x) = (d)/(dx) \left[ \left((\alpha)/(\pi)\right)^(1/4) e^{-(\alpha x^2)/(2)} \right] = -\alpha x \left((\alpha)/(\pi)\right)^(1/4) e^{-(\alpha x^2)/(2)} \]`

- Next, we find the second derivative:

`\[ \Psi''(x) = (d)/(dx) \left[ -\alpha x \left((\alpha)/(\pi)\right)^(1/4) e^{-(\alpha x^2)/(2)} \right] = \left((\alpha)/(\pi)\right)^(1/4) \alpha^(1.25) (\alpha x^2 - 1) e^{-(\alpha x^2)/(2)} \]`

- Finally,
`\( p^2 \Psi(x) = -\Psi''(x) \).`

2.
`\( x \Psi(x) \)`

This is simply the product of x and
`\(\Psi(x)\)`:
`\(\Psi(x)\):`

`\[x \Psi(x) = x \left((\alpha)/(\pi)\right)^(1/4) e^{-(\alpha x^2)/(2)}\]`

3.
`\( x^2 \Psi(x) \)`

Similarly, this is the product of
`\(x^2\)` and
`\(\Psi(x)\):`

`\[x^2 \Psi(x) = x^2 \left((\alpha)/(\pi)\right)^(1/4) e^{-(\alpha x^2)/(2)}\]`

These expressions capture the desired operations on the wave function
`\(\Psi(x)\)`.