Tuesday, May 13, 2014

The wave function 03

— 1.4. Normalization —

The Scroedinger equation is a linear partial differential equation. As such, if $ {\Psi(x,t)}$ is a solution to it, then $ {A\Psi(x,t)}$ (where $ {A}$ is a complex constant) also is a solution.

Does this mean that a physical problem has an infinite number of solutions in Quantum Mechanics? It doesn't! The thing is that besides the The Scroedinger equation one also has condition 11 to take into account. Stating 11 for the wave function:

$ \displaystyle \int_{-\infty}^{+\infty}|\Psi (x,t)|^2dx=1 \ \ \ \ \ (15)$

The previous equations states the quite obvious fact that the particle under study has to be in some place at a given instant.

Since $ {A}$ was a complex constant the normalization condition fixes $ {A}$ in absolute value but can't tell us nothing regarding its phase. Apparently once again one is haunted with the perspective of having an infinite number of solutions to any given physical problem. The things is that this time the phase doesn't carry any physical significance (a fact that will be demonstrated later) and thus we actually have just one physical solution.

In the previous discussion one is obviously assuming that the wave function is normalizable. That is to say that the function doesn't blow up and vanishes quickly enough at infinity so that the integral being computed makes sense.

At this level it is customary to say that these wave functions don't represent physical states but that isn't exactly true. A wave function that isn't normalizable because integral is infinite might represent a beam of particles in a scattering experiment. The fact that the integral diverges to infinity can then be said to represent the fact that beam is composed by an infinite amount of particles.

While the identically null wave function represents the absence of particles.

A question that now arises has to do with the consistency of our normalization and this is a very sensible question. The point is that we normalize the Schroedinger equation for a given time instant, so how does one know that the normalization holds for other times?

Let us look into the time evolution of our normalization condition 15.

$ \displaystyle \frac{d}{dt}\int_{-\infty}^{+\infty}|\Psi (x,t)|^2dx=\int_{-\infty}^{+\infty}\frac{\partial}{\partial t}|\Psi (x,t)|^2dx \ \ \ \ \ (16)$

Calculating the derivative under the integral for the right hand side of the previous equation

$ {\begin{aligned} \frac{\partial}{\partial t}|\Psi (x,t)|^2&=\frac{\partial}{\partial t}(\Psi^* (x,t)\Psi (x,t))\\ &=\Psi^* (x,t)\frac{\partial\Psi (x,t)}{\partial t}+\frac{\partial \Psi^* (x,t)}{\partial t}\Psi (x,t) \end{aligned}}$

The complex conjugate of the Schroedinger equation is

$ \displaystyle \frac{\partial \Psi^*(x,t)}{\partial t}=-\frac{i\hbar}{2m}\frac{\partial^2\Psi^*(x,t)}{\partial x^2}+\frac{i}{\hbar}V\Psi^*(x,t) \ \ \ \ \ (17)$

Hence for the derivative under the integral

$ {\begin{aligned} \frac{\partial}{\partial t}|\Psi (x,t)|^2&=\frac{\partial}{\partial t}(\Psi^* (x,t)\Psi (x,t))\\ &=\Psi^* (x,t)\frac{\partial\Psi (x,t)}{\partial t}+\frac{\partial \Psi^* (x,t)}{\partial t}\Psi (x,t)\\ &=\frac{i\hbar}{2m}\left( \Psi^*(x,t)\frac{\partial^2\Psi(x,t)}{\partial x^2}-\frac{\partial^2\Psi^*(x,t)}{\partial x^2}\Psi (x,t)\right)\\ &=\frac{\partial}{\partial x}\left[ \frac{i\hbar}{2m}\left( \Psi^*(x,t)\frac{\partial\Psi(x,t)}{\partial x}-\frac{\partial\Psi^*(x,t)}{\partial x}\Psi(x,t) \right) \right] \end{aligned}}$

Getting back to 16

$ \displaystyle \frac{d}{dt}\int_{-\infty}^{+\infty}|\Psi (x,t)|^2dx=\frac{i\hbar}{2m}\left[ \Psi^*(x,t)\frac{\partial\Psi(x,t)}{\partial x}-\frac{\partial\Psi^*(x,t)}{\partial x}\Psi(x,t) \right]_{-\infty}^{+\infty} \ \ \ \ \ (18)$

Since we're assuming that our wave function is normalizable the wave function (and its complex conjugate) must vanish for $ {+\infty}$ and $ {-\infty}$.

In conclusion

$ \displaystyle \frac{d}{dt}\int_{-\infty}^{+\infty}|\Psi (x,t)|^2dx=0$

Since the derivative vanishes one can conclude that the integral is constant.

In conclusion one can say that if one normalizes the wave equation for a given time interval it stays normalized for all time intervals.

Exercise 1 At time $ {t=0}$ a particle is represented by the wave function

$ \displaystyle \Psi(x,0)=\begin{cases} Ax/a & \text{if } 0\leq x\leq a\\ A(b-x)/(b-a) & \text{if } a\leq x\leq b \\ 0 & \text{otherwise}\end{cases} \ \ \ \ \ (19)$

where $ {A}$, $ {a}$ and $ {b}$ are constants.

  1. Normalize $ {\Psi}$.

    $ {\begin{aligned} 1&=\int_{-\infty}^{+\infty} |\Psi|^2\,dx\\ &=\int_0^a|\Psi|^2\,dx+\int_a^b|\Psi|^2\,dx\\ &=\dfrac{|A|^2}{a^2}\int_0^a|x^2\,dx+\dfrac{|A|^2}{(b-a)^2}\int_a^b(b-x)^2\,dx\\ &=\dfrac{|A|^2}{a^2}\left[ \dfrac{x^3}{3} \right]_0^a+\dfrac{|A|^2}{(b-a)^2}\left[ \dfrac{(b-x)^3}{3} \right]_a^b\\ &=\dfrac{|A|^2a}{3}+\dfrac{|A|^2}{(b-a)^2}\dfrac{(b-a)^3}{3}\\ &=\dfrac{|A|^2a}{3}+|A|^2\dfrac{b-a}{3}\\ &=\dfrac{b|A|^2}{3} \end{aligned}}$

    Hence for $ {A}$ it is

    $ \displaystyle A=\sqrt{\dfrac{3}{b}} $

  2. Sketch $ {\Psi(x,0)}$

    In $ {0\leq x \leq a}$ $ {\Psi(x,0)}$ is a strictly increasing function that goes from $ {0}$ to $ {A}$.

    In $ {a \leq x \leq b}$ $ {\Psi(x,0)}$ is strictly decreasing function that goes from $ {A}$ to $ {0}$.

    Hence the plot of $ {\Psi(x,0)}$ is (choosing the following values $ {a=1}$, $ {b=2}$ and $ {A=\sqrt{b}=\sqrt{2}}$):

  3. Where is the particle most likely to be found at $ {t=0}$? Since $ {x=a}$ is maximum of the $ {\Psi}$ function the most likely value for the particle to be found is at $ {x=a}$.

  4. What is the probability of finding the particle to the left of $ {a}$? Check the answers for $ {b=a}$ and $ {b=2a}$.

    $ {\begin{aligned} P(x<a)&=\int_0^a|\Psi|^2\,dx\\ &=\dfrac{|A|^2}{a^2}\int_0^a x^2\,dx\\ &=\dfrac{|A|^2}{a^2}\left[ \dfrac{x^3}{3} \right]_0^a\\ &=\dfrac{|A|^2}{3}a\\ &=\dfrac{3}{3b}a\\ &=\dfrac{a}{b} \end{aligned}}$

    At first let us look into the $ {b=a}$ limiting case. We can imagine that this is the end result of $ {b}$ getting nearer and nearer to $ {a}$. That is to say that the domain of the strictly decreasing part of $ {\Psi(x,0)}$ is getting shorter and shorter and when finally $ {b=a}$ $ {\Psi(x,0)}$ doesn't have a domain where its is strictly decreasing and $ {\Psi(x,0)}$ is defined by its strictly increasing and vanishing features (in the appropriate domains). That is to say that to the right of $ {a}$ the function is $ {0}$. Hence the probability of the particle being found to the left of $ {a}$ is $ {1}$.

    From the previous calculation $ {P(x<a)_{b=a}=1}$ which is indeed the correct result.

    The $ {b=2a}$ case can be analyzed in a different way. In this case:

    • $ {x=a}$ is the half point of the domain of $ {\Psi(x,0)}$ where $ {\Psi(x,0)}$ is non vanishing (end points of the domain are excluded).
    • $ {\Psi(x,0)}$ is strictly increasing in the first half of the domain ($ {0\leq x\leq a}$).
    • $ {\Psi(x,0)}$ is strictly decreasing in the second half of the domain ($ {a\leq x\leq b}$).
    • $ {\Psi(x,0)}$ is continuous.

    Thus one can conclude that $ {\Psi(x,0)}$ is symmetric around $ {a}$ and consequently the probability of the particle being found to the left of $ {a}$ has to be $ {1/2}$.

    From the previous calculation $ {P(x<a)_{b=2a}=1/2}$ which is indeed the correct result.

  5. What is the expectation value of $ {x}$?

    $ {\begin{aligned} <x>&= \int_a^b x|\Psi|^2\,dx\\ &=\dfrac{|A|^2}{a^2}\int_0^a x^3\,dx+\dfrac{|A|^2}{(b-a)^2}\int_a^b x(b-x)^2\,dx\\ &=\dfrac{|A|^2}{a^2}\left[ \dfrac{x^4}{4} \right]_0^a+\dfrac{|A|^2}{(b-a)^2}\left[ 1/2x^2b^2-2/3x^3b+x^4/4 \right]_a^b\\ &=\dfrac{2a+b}{4} \end{aligned}}$

Exercise 2 Consider the wave function

$ \displaystyle \Psi(x,t)=Ae^{-\lambda |x|}e^{-i\omega t} \ \ \ \ \ (20)$

where $ {A}$, $ {\lambda}$ and $ {\omega}$ are positive real constants.

  1. Normalize $ {\Psi}$

    $ {\begin{aligned} 1&=\int_{-\infty}^{+\infty} |\Psi|^2\,dx\\ &=\int_{-\infty}^{+\infty} |A|^2e^{-2\lambda |x|}\,dx\\ &=2|A|^2\int_0^{+\infty}e^{-2\lambda |x|}\,dx \\ &=2|A|^2\int_0^{+\infty}e^{-2\lambda x}\,dx \\ &=-\dfrac{|A|^2}{\lambda}\left[ e^{-2\lambda x} \right]_0^{+\infty}\\ &=\dfrac{|A|^2}{\lambda} \end{aligned}}$

    Hence it is

    $ \displaystyle A=\sqrt{\lambda} $

  2. Determine $ {<x>}$ and $ {<x^2>}$

    $ {\begin{aligned} <x>&=\int_{-\infty}^{+\infty} x|\Psi|^2\,dx\\ &=|A|^2\int_{-\infty}^{+\infty} xe^{-2\lambda |x|}\,dx\\ &=0 \end{aligned}}$

    The integral is vanishing because we're calculating the integral of an odd function between symmetrical limits.

    $ {\begin{aligned} <x^2>&=\int_{-\infty}^{+\infty} x^2|\Psi|^2\,dx\\ &=2\lambda\int_0^{+\infty} x^2e^{-2\lambda x}\,dx\\ &=2\lambda\int_0^{+\infty} \dfrac{1}{4}\dfrac{\partial^2}{\partial \lambda ^2}\left( e^{-2\lambda x} \right)\,dx\\ &=\dfrac{\lambda}{2} \dfrac{\partial^2}{\partial \lambda ^2}\int_0^{+\infty}e^{-2\lambda\,dx} x \,dx\\ &= \dfrac{\lambda}{2} \dfrac{\partial^2}{\partial \lambda ^2} \left[ -\dfrac{e^{-2\lambda\,dx}}{2\lambda} \right]_0^{+\infty}\\ &= \dfrac{\lambda}{2}\dfrac{\partial^2}{\partial \lambda ^2}\left(\dfrac{1}{2\lambda} \right)\\ &=\dfrac{\lambda}{2}\dfrac{\partial}{\partial \lambda}\left(-\dfrac{1}{\lambda ^2} \right)\\ &=\dfrac{\lambda}{2}\dfrac{1}{\lambda^3}\\ &= \dfrac{1}{2\lambda^2} \end{aligned}}$

  3. Find the standard deviation of $ {x}$. Sketch the graph of $ {\Psi ^2}$. What is the probability that the particle will be found outside the range $ {[<x>-\sigma,<x>+\sigma]}$?

    $ \displaystyle \sigma ^2=<x^2>-<x>^2=\frac{1}{2\lambda ^2}-0=\frac{1}{2\lambda ^2}$

    Hence the standard deviation is

    $ \displaystyle \sigma=\dfrac{\sqrt{2}}{2\lambda}$

    The square of the wave function is proportional to $ {e^{-2\lambda |x|}}$. Dealing for piecewise definitions of the square of the wave function, its first derivative in order to $ {x}$ and its second derivative in order to $ {x}$

    $ \displaystyle |\Psi|^2=\begin{cases} e^{2\lambda x} & \text{if } x < 0\\ e^{-2\lambda x} & \text{if } x \geq 0 \end{cases} \ \ \ \ \ (21)$

    $ \displaystyle \dfrac{\partial}{\partial x}|\Psi|^2=\begin{cases} 2\lambda e^{2\lambda x} & \text{if } x < 0\\ -2\lambda e^{-2\lambda x} & \text{if } x \geq 0 \end{cases} \ \ \ \ \ (22)$

    $ \displaystyle \dfrac{\partial ^2}{\partial x ^2}|\Psi|^2=\begin{cases} 4\lambda ^2 e^{2\lambda x} & \text{if } x < 0\\ 4\lambda ^2 e^{-2\lambda x} & \text{if } x \geq 0 \end{cases} \ \ \ \ \ (23)$

    As we can see the first derivative of $ {|\Psi|^2}$ changes its sign on $ {0}$ from positive to negative. Hence it was strictly increasing before $ {0}$ and it is strictly decreasing after $ {0}$. Hence $ {0}$ is a maximum of $ {|\Psi|^2}$.

    The second derivative is always positive so $ {|\Psi|^2}$ is always concave up (convex).

    Hence its graphical representation is:

    The probability that the particle is to be found outside the range $ {[<x>-\sigma, <x>+\sigma ]}$ is

    $ {\begin{aligned} P(<x>-\sigma, <x>+\sigma)&= 2\int_\sigma^{+\infty}|\Psi|^2\,dx\\ &= 2\lambda\int_\sigma^{+\infty}e^{2\lambda x}\\ &= 2\lambda\left[ -\dfrac{e^{2\lambda x}}{2\lambda} \right]_\sigma^{+\infty}\\ &=\lambda \dfrac{e^{2\lambda x}}{2\lambda}\\ &=e^{-2\lambda\dfrac{\sqrt{2}}{2\lambda}}\\ &=e^{-\sqrt{2}} \end{aligned}}$


Related Posts