Expectation Value of Position for Even Wavefunction

[hellsteiger]

Homework Statement

Hello, I need to calculate the expectation value for position and momentum for a wavefunction that fulfills the following relation:

ψ₀(-x)=ψ₀(x)=ψ^*₀(x)

The wave function is normalised.

Homework Equations

There is also a second wave equation that is orthogonal to the first:

ψ₁=Nd₀/dx

I also need to calculate <x> and <p> for this wavefunction, also normalised.

The Attempt at a Solution

I am tempted to go ahead and say that the expectation value for position for both wavefunctions is 0, as they are both even. As for momentum I am not certain as to how i should proceed, inserting the relevant operator just seems to indicate to me momentum will be zero.

 

[hellsteiger]

Correction:

ψ1=Ndψ₀/dx

 

[hellsteiger]

I should point out that ψ₁ is actually an odd function, being the derivative of an even function, it is not even as i stated earlier.

 

[vanhees71]

Just write down the expectation values of position and momentum in terms of the wave function (the position representation of the quantum state). Then discuss the mathematics given the symmetry properties, i.e., either even or odd, of the wave functions.

 

[paranormal]

Hello! It appears that you are working with a system that has two wavefunctions, ψ0 and ψ1, which are related by the given equation. In this case, the expectation value for position, <x>, can be calculated as follows:

<x> = ∫ ψ*0(x) x ψ0(x) dx

Since ψ0 is an even function, we can rewrite this as:

<x> = 2∫ ψ*0(x) x ψ0(x) dx

Using the given relation between ψ0 and ψ1, we can rewrite this as:

<x> = 2∫ ψ*0(x) x ψ1(x) dx

Now, since ψ1 is the derivative of ψ0 with respect to x, we can rewrite this as:

<x> = 2∫ ψ*0(x) x (dψ0/dx) dx

Using integration by parts, we can simplify this to:

<x> = 2ψ*0(x) ψ0(x) | - ∫ ψ0(x) (dψ*0/dx) dx

Since the wave function is normalized, we know that the integral of ψ0(x) (dψ*0/dx) dx must be equal to 0. Therefore, the final result for <x> is:

<x> = 2ψ*0(x) ψ0(x)

Similarly, the expectation value for momentum, <p>, can be calculated as:

<p> = ∫ ψ*0(x) (-iħ d/dx) ψ0(x) dx

Using the same steps as above, we can simplify this to:

<p> = -iħ 2ψ*0(x) (dψ0/dx) | - ∫ (dψ*0/dx) (dψ0/dx) dx

Again, since the wave function is normalized, we know that the integral of (dψ*0/dx) (dψ0/dx) dx must be equal to 0. Therefore, the final result for <p> is:

<p> = -2iħ ψ*0(x) (dψ0/dx)

I hope this helps with your calculations. Keep in mind that these results are specific to the given wavefunctions