Showing the expectation values of a system are real quantities

[FisiksIdiot]

Homework Statement

A one-dimension system is in a state described by the normalisable wave function Ψ(x,t) i.e. Ψ → 0 for x → ±∞.

(a) Show that the expectation value of the position ⟨x⟩ is a real quantity. [1]

(b) Show that the expectation value of the momentum in the x-direction ⟨p⟩ is a real quantity, too. Hint: using integration by parts and normalisability show that ⟨pˆx⟩ = ⟨p⟩∗. [4]

Homework Equations

1=N∫ψ*ψ.dx

<x>=∫ψ* xψ.dx over all space

<p>=∫ψ* -ih dψ/dx.dx over all space

The Attempt at a Solution

The difficult aspect of this for me is determining what the correct wave function is. Using the information given I assumed that the correct wave function was e^-ax2/2 e^iEt/h (where a/2 is an arbitrary constant) as it fits the above requirements (I could be wrong.)

However, upon normalising and calculating <x> and <p>, the values obtained will of course will be 0 and therefore real as my assumption was a symmetric wave function. This is all well and good, however the question explicitly states to use integration by parts to solve for <p>.

<p>=N∫(e^-ax2/2 e^iEt/h) -ih d/dx(e^-ax2/2 e^-iEt/h).dx over all space

which gave:

<p>=N iha∫xe^-ax2.dx over all space

which cannot be integrated by parts as far as I understand-perhaps it can?. Have I got the wrong end of the stick somewhere in my thinking?

Thanks in advance.

 

[Bryson]

You wave function is incorrect. Do not plug it in until you have done the math. You do not even need a "test" wavefunction for part a.

 

[FisiksIdiot]

Thanks for the prompt reply, but I don't quite follow. Could you clarify what you mean by 'not needing a wave function'? I feel I am approaching the problem from the wrong angle.

 

[Bryson]

Hmm, perhaps forget what I mentioned previously. You simply need to reconsider your wave function. Perhaps try something of the form $\psi = e^{\pm i(kx-E t/ \hbar)}$. . .

 

[Nickie]

Thank you for your question. It seems like you have the right idea, but there are a few things that need to be clarified. First, the wave function given in the question is not necessarily the correct one, it is just an example of a normalizable wave function that satisfies the given conditions. So, you can use any other normalizable wave function as well.

Now, for part (a), to show that the expectation value of position is a real quantity, you can simply use the definition of the expectation value and the fact that the wave function is normalized. So, you should get:

<x>=∫Ψ* x Ψ dx = ∫Ψ* Ψ x dx = ∫|Ψ|2 x dx

Since the wave function is normalized, the integral of |Ψ|2 over all space is equal to 1. Therefore, <x> is a real quantity.

For part (b), you are on the right track. To show that the expectation value of momentum is a real quantity, you can use integration by parts as follows:

<p>=∫Ψ* (-ih dΨ/dx) dx = ih∫(dΨ*/dx) Ψ dx = ih[Ψ* Ψ] - ih∫Ψ* (dΨ/dx) dx

Since the wave function goes to 0 at infinity, the first term in the above equation is equal to 0. And by the same logic as in part (a), the integral of |Ψ|2 over all space is equal to 1. Therefore, <p> is a real quantity.

I hope this helps clarify your understanding. Keep up the good work!