What does this equation for a free particle mean?

[Rob Turrentine]

So there's a free particle with mass m.

\begin{equation}

\psi(x,0) = e^{ip_ox/\hbar}\cdot\begin{cases}

x^2 & 0 \leq x < 1,\\

-x^2 + 4x -2 & 1 \leq x < 3,\\

x^2 -8x +16 & 3 \leq x \leq 4, \\

0 & \text{otherwise}.

\end{cases}

\end{equation}

What does each part of the piecewise represent? And what are the boundary conditions representative of? Energy levels?

I'm used to working with non-piecewise functions, like \begin{equation} \psi = Ae^{ikx} + Be^{-ikx} \end{equation}
so I'm just not sure what to do.
The goal is to normalize it and then find <X> and <P> as functions of time.

Any help is greatly appreciated!

 

[Orodruin]

You are thinking of eigenfunctions of a Hamiltonian. A wave function can generally be any normalisable function satisfying the appropriate boundary conditions. It need not be an eigenfunction of a Hamiltonian.

 

[stevendaryl]

You should graph the part that multiplies $e^{i \frac{p_0 x}{\hbar}}$. I don't think there's any particular meaning to its shape. It's just a nice curve that is continuous, and which has a continuous first derivative (though not second derivative).

 

[Vanadium 50]

What does each part of the piecewise represent?

Why does it have to represent anything? This appears to be a problem intended to give you some facility with the mechanics of the sorts of calculations one does in QM.

 

[Rob Turrentine]

Okay, I think I'm getting it more, thank you! So when normalizing, do I follow this equation

$$\int_{a}^{b} \psi \psi^* dx$$

and set my piecewise accordingly? It appears that the complex conjugate causes $$\bigg(e^{\frac{ip_0x}{\hbar}}\bigg)^2 = \bigg(e^{\frac{ip_0x}{\hbar}}\bigg)\bigg(e^{\frac{ip_0x}{\hbar}}\bigg)^* = e^0 = 1$$
So I would end up at

\begin{equation}
\int_{a}^{b} \psi(x,0)^2dx =\begin{cases}
\int_{0}^{1} (x^4)dx & 0 \leq x < 1,\\
\int_{1}^{3} (x^4 - 8x^3d + 20x^2 +4)dx & 1 \leq x < 3,\\
\int_{3}^{4} (x^4 - 16x^3 +96x^2 - 256x + 256)dx & 3 \leq x \leq 4, \\
0 & \text{otherwise}.
\end{cases}
\end{equation}

Or am I going down the wrong road?

Once again, thank you for any help!

 

[MichPod]

You probably need to realize a fact that any function (some continuity limitations applied) may be a solution of time-dependent shrodinger equation at t=0. I.e. shredinger equation defines how psi function evolves in time, but at particular time t=0 all the values of pis in space may be just set arbitrarily.

 

[mfb]

So I would end up at

\begin{equation}
\int_{a}^{b} \psi(x,0)^2dx =\begin{cases}
\int_{0}^{1} (x^4)dx & 0 \leq x < 1,\\
\int_{1}^{3} (x^4 - 8x^3d + 20x^2 +4)dx & 1 \leq x < 3,\\
\int_{3}^{4} (x^4 - 16x^3 +96x^2 - 256x + 256)dx & 3 \leq x \leq 4, \\
0 & \text{otherwise}.
\end{cases}
\end{equation}

You mean the right thing, but you cannot write it like this.
The left side depends on a and b only, the right side depends on x (case bracket) but then has constant values within the x ranges.
If you directly take the integral from $-\infty$ to $+\infty$ you can just sum the integrals.

 

[Rob Turrentine]

By summing them, I got 70.1 plus a constant from the 0.

Is the equation now set up to find <X> and <P> as functions of time?

 

[Orodruin]

plus a constant from the 0.

The primitive function of 0 is a constant. This does not mean that a definite integral of 0 is non-zero.

 

[Rob Turrentine]

So I just end up at 70.1 without a constant? So by multiplying this times the equation is it normalized?

Also, after this am I ready to find <X> and <P> as functions of time?

 

[mfb]

So I just end up at 70.1 without a constant?

Yes.

So by multiplying this times the equation is it normalized?

Multiplication is not the right operation here.

The integral is now 70.1. You want the integral to become 1. How do you have to change ψ?

Also, after this am I ready to find <X> and <P> as functions of time?

I would start with the expectation values at t=0.

 

[Rob Turrentine]

I realized I miscalculated on my integral, and the actual answer is $\frac{92}{15} - 6.1333.$ So my normalized equation would now be
$$ \psi(x,0) = \sqrt{\frac{1}{6.1333}} \cdot e^{ip_ox/\hbar} \cdot
\begin{cases}
x^2 & 0 \leq x < 1,\\
-x^2 + 4x -2 & 1 \leq x < 3,\\
x^2 -8x +16 & 3 \leq x \leq 4, \\
0 & \text{otherwise}.
\end{cases}$$

Now, to turn $\psi$ into a function of time, my book says to multiply it by $e^{\frac{-iEt}{\hbar}}$. So we would get
$$ \psi(x,t) = \sqrt{\frac{1}{6.1333}} e^{-iEt/\hbar} \cdot e^{ip_ox/\hbar} \cdot
\begin{cases}
x^2 & 0 \leq x < 1,\\
-x^2 + 4x -2 & 1 \leq x < 3,\\
x^2 -8x +16 & 3 \leq x \leq 4, \\
0 & \text{otherwise}.
\end{cases}$$

Finding the expectation values of X and P requires the formula
$$<X> = \int_{-\infty}^{\infty} \psi^*(x,t) X \psi(x,t)dx$$

However, when I use this with $\psi$ I get an unbelievably long integral.

 

[Rob Turrentine]

Whoops forgot to add my question...

For the equation $e^{-iEt/\hbar}$ is it best to change E to another variable or keep it as is?

 

[Orodruin]

That time evolution is not correct. Each energy eigenstate evolves separately with its own phase contribution. You need to write your initial state as a linear combination of energy eigenstates and evolve them separately. However, you should not really need the exact time evolution of the state to find the time evolution of the expectation values.

 

[Rob Turrentine]

I've seen the equation $<X> = <\psi|X|\psi>$, but wouldn't I need to have $\psi$ as a function of time to find $<X>$ as a function of time? $\psi(x,t)$ instead of $\psi(x,0)$?

Going off of that, I have to find both $<X>$ and $\dot{<X>}$.
Ehrenfest's theorem is $\frac{d}{dt}<X> = (-i\hbar)[X,H]$

Would this equation work? I'm not sure if the expectation value of $X$ in that is just $\dot{X}$, so I would need a new way to find $<X>$ instead of $\dot{<X>}$

 

[Orodruin]

No. Starting from the general time evolution you should be able to relate the time evolution of the expectation values to their values at t=0.

Also, in LaTeX < and > are relations and are typeset as such. The correct braket delimiters are \langle ($\langle$) and \rangle ($\rangle$).

 

[Rob Turrentine]

Ah okay, thank you for the tip!

Okay, I'm still a little bit confused. So you're saying that I can use the equation $\langle{X}\rangle = \langle\psi|X|\psi\rangle$ to find X at t=0, but somehow relate the time evolution expectation values to those at t=0?

 

[mfb]

It is a free particle - what do you know about the evolution of its momentum?
Start there, then work on <X(t)> later.

 

[Rob Turrentine]

Since there's no force, it should just be $p_o$, right?

 

[mfb]

You should have a formula to calculate it.

 

[Orodruin]

Let me put it this way: For any operator $A$, the expectation value at a time $t$ is given by
$$
\langle A\rangle(t) = \langle \psi(t)| A | \psi(t)\rangle,
$$
where $|\psi(t)\rangle$ is the state at time $t$, which follows the Schrödinger equation (all of this in the Schrödinger picture). What is the time derivative of this expectation value?