# Homework Help: Quantum mechanics, adding a phase factor

1. Apr 8, 2013

### Cogswell

Here's the question:

1. In classical mechanics, we know that the evolution of a system, that is the trajectories of the particles and objects, does not depend on where we chose the xero-point of the potential energy. Here we analyse what happens in quantum mechanics. Assume that you add a constant C0 to the potential energy of an arbitrary potential V(x)

Show that the wave function now differs from the one you would get without the C0 added by having a phase factor $$e^{-i \frac{C_0 t}{\hbar}}$$ multiplied to it.
How will this phase-factor affect the expectation value of a dynamical variable?

So I was thinking of using: $$\Psi (x,t) = \psi (x) \varphi (x)$$
to solve it.

Can someone guide me through this (like with a hint or 2)? I'm quite confused about quantum mechanics.

2. Apr 8, 2013

### Staff: Mentor

Start with an eigenstate $\psi$ of a Hamiltonian $\hat{H}$, with eigenenergy $E$. What is the time evolution of $\psi$? (This should be in your textbook.)

Then, arbitrarily shift the zero of energy by $C_0$. How is the time evolution of $\psi$ affected?

Then, calculate the expectation value of an operator $\hat{A}$,
$$\langle \hat{A} \rangle = \int_{-\infty}^{\infty} \psi(x,t)^* \hat{A} \psi(x,t) dx$$
for both wave functions (energy shifted and not shifted), and compare.

3. Apr 10, 2013

### Cogswell

Sorry, I'm not getting what you mean. Maybe I should have posted this in the Introductory Physics section of this forum.

In my textbook, it says I can use a separation of variables to solve the Schrodinger equation.
$$\Psi (x,t) = \psi (x) + \varphi (t)$$

$$i \hbar \dfrac{\partial \Psi (x,t)}{\partial t} = -\dfrac{\hbar ^2}{2m} \dfrac{\partial ^2 \Psi (x,t)}{\partial x^2} + V \Psi (x,t)$$

$$i \hbar \psi (x) \dfrac{\partial \varphi (t)}{\partial t} = -\dfrac{\hbar ^2}{2m} \dfrac{\partial ^2 \psi (x)}{\partial x^2} \varphi (t) + V \psi (x) \varphi (t)$$

And then dividing through by $\psi (x) + \varphi (t)$ I will get:

$$i \hbar \dfrac{1}{\varphi (t)} \dfrac{\partial \varphi (t)}{\partial t} = -\dfrac{\hbar ^2}{2m} \dfrac{1}{\psi (x)} \dfrac{\partial ^2 \psi (x)}{\partial x^2} + V$$

It then uses a separation constant E to separate out the left and right hand sides (since one depends on time and the other depends on position).

$$E = -\dfrac{\hbar ^2}{2m} \dfrac{1}{\psi (x)} \dfrac{\partial ^2 \psi (x)}{\partial x^2} + V$$

So I guess what you're saying is to multiply everything by $\psi (x)$ whch will give me:

$$E \psi (x) = -\dfrac{\hbar ^2}{2m} \dfrac{\partial ^2 \psi (x)}{\partial x^2} + V \psi (x)$$

And the right hand side can be defined as the Hamiltonian operator $\hat{H}$ with an eigenstate $\psi (x)$

$$E \psi (x) = \hat{H} \psi (x)$$

And the time evolution of $\psi (x)$ is constant because it is only dependent on position and not time?

"Then, arbitrarily shift the zero of energy by C0. How is the time evolution of ψ affected?"
What do you mean? I'm confused as to where to go from here.
Do I add C0 to V which will give me:
$$(E + C_0) \psi (x) = -\dfrac{\hbar ^2}{2m} \dfrac{\partial ^2 \psi (x)}{\partial x^2} + (V+C_0) \psi (x)$$
and then solve this second order differential equation?

4. Apr 10, 2013

### Staff: Mentor

You're getting along fine. You just need to go back to the time-dependent Schrödinger equation (TDSE) to get the full wave function:

Indeed, for $\psi (x)$ an eigenfunction of $\hat{H}$, the separation constant $E$ is the energy. Use this to solve for $\varphi(t)$ in the equation above.

Yes, it corresponds to that equation, but you don't need to actually solve it. Again go back to solving for $\varphi(t)$, and see how the full, time-dependent wave function chnages with this $C_0$ added to the energy.

5. Apr 10, 2013

### Cogswell

Right so with the time dependent part,

$$E = i \hbar \dfrac{1}{\varphi (t)} \dfrac{\partial \varphi (t)}{\partial t}$$

If I re-arrange:

$$\dfrac{E}{i \hbar} = \dfrac{1}{\varphi (t)} \dfrac{\partial \varphi (t)}{\partial t}$$

And the integrate with respect to (t) I'll get:

$$\dfrac{E}{i \hbar} t = \ln(\varphi (t))$$

And then re-arranging I get:

$$\varphi (t) = e^{\dfrac{-i E t}{ \hbar}}$$

So if the separation constant E is the energy, I'll get my other differential equation with C0 added to it:

$$E + C_0 = i \hbar \dfrac{1}{\varphi (t)} \dfrac{\partial \varphi (t)}{\partial t}$$

$$\dfrac{E + C_0}{i \hbar} = \dfrac{1}{\varphi (t)} \dfrac{\partial \varphi (t)}{\partial t}$$

$$\dfrac{E + C_0}{i \hbar} t = \ln (\varphi (t))$$

And then re-arranging:

$$e^\dfrac{-iEt -iC_0 t}{\hbar} = \varphi (t)$$

Which becomes $e^\dfrac{-iEt}{\hbar} e^\dfrac{-iC_0 t}{\hbar} = \varphi (t)$

This is just like the original one, but with the phase factor of $e^\dfrac{-iC_0 t}{\hbar}$ added to it. So I guess that's the answer?

With the second part of the question: "How will this phase-factor affect the expectation value of a dynamical variable?"

Does it mean to calculate $<x> = \int \varphi (t)^* (x) \varphi (t) dx$ and compare the one with and without the C0?