Time to find particle left of the origin maximal

84 Views Asked by At

We study a quantum particle conned to a one-dimensional box with walls at the positions ±1. The Hilbert-space of this system in the Schrödinger representation is again given by $L_2(-1,1)$. In this Hilbert space, we consider the two functions $g_0(x):=(1+i)exp(i \pi x)$ and $g_1(x):=exp(i 2 \pi x)$. I shown that $g_0$ and $g_1$ are orthogonal in $L_2([-1,1])$ and that the orthonormal states $\phi_0=\frac{g_0(x)}{2}$ and $\phi_1=\frac{g_1(x)}{\sqrt{2}}$.

Now we now consider a general superposition $\Psi=\alpha \phi_0+\beta \phi_1$ of the two states $\phi_0=\frac{g_0(x)}{2}$ and $\phi_1=\frac{g_1(x)}{\sqrt{2}}.$

And now I have to find out for which time i'st the probability to find the particle left of the origin maximal? And find out what the expected value of a position measurement in that case.

I'm not sure how to deal with it. I think that the probability to find the particle left of the origin can be found by $\int_{-1}^{0}|\Psi(x)|^2\,\mathrm{d}x$. But how can I use this to find the time i'st the probability to find the particle left of the origin maximal? And find out what the expected value of a position measurement in that case? Can anyone help me with some hints to find that?

1

There are 1 best solutions below

2
On

As you said, the probability of the position of the system on a state $\psi$ to be on the left the origin is $$\int_{(-1,0)}|\psi(x)|^2 dx.$$ Now, finding the solution of the Shrodinger equation gives as the time evolution of $\psi$, namely for each $t \in \mathbb{R}$, we have a state $\psi_t=\psi(x,t)$. Thus, the probability of finding the particle on the left of the origin at a time $t$ is $$p_t := \int_{(-1,0)}|\psi_t(x)|^2 dx.$$ This defines a function $$p: \mathbb{R} \rightarrow [0,1]$$$$t \mapsto p_t$$ which under some restrictions is differentiable. So, what you want now is to find the maximum points of $p$. If $p$ is nice enough, as often is the case in exercises, you can find a maximum by looking for the points in which $\frac{d}{dt}p(t)=0.$

The expected value of the position operator $X$ at a state $\psi$, as with any observable, is $\langle X\rangle_\psi := \langle \psi, X\psi \rangle $, since we know how the state $\psi$ changes in time, we also know how this value changes with time, namely $\langle \psi_t, X\psi_t \rangle. $