Triangular wave function: probabilities

Reference: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 1.4.

The square modulus of the wave function which is the solution to the Schrödinger equation is interpreted as a probability density. As an example consider the wave function given by

\displaystyle  \Psi\left(x,0\right)=\begin{cases} A\frac{x}{a} & 0\le x\le a\\ A\frac{b-x}{b-a} & a\le x\le b\\ 0 & \mbox{otherwise} \end{cases} \ \ \ \ \ (1)

We can normalize {\Psi} by requiring

\displaystyle  \int_{0}^{b}\left|\Psi\right|^{2}dx=1 \ \ \ \ \ (2)

Plugging in the formula and doing the integral gives

\displaystyle   \int_{0}^{b}\left|\Psi\right|^{2}dx \displaystyle  = \displaystyle  \left|A\right|^{2}\left[\int_{0}^{a}\frac{x^{2}}{a^{2}}dx+\int_{a}^{b}\left(\frac{b-x}{b-a}\right)^{2}dx\right]\ \ \ \ \ (3)
\displaystyle  \displaystyle  = \displaystyle  \left|A\right|^{2}\frac{b}{3}\ \ \ \ \ (4)
\displaystyle  A \displaystyle  = \displaystyle  \sqrt{\frac{3}{b}} \ \ \ \ \ (5)

where we’ve taken the positive real root for {A}. Note that {A} could also be multiplied by a phase factor {e^{i\delta}} for any real {\delta} without affecting normalization. This can be important in some applications where we need to add together wave functions.

Given this value for {A}, we can plot 1. Here, we’ve taken {a=1} and {b=3}:

Since {\Psi} has its maximum at {x=a}, that is where the particle is most likely to be found. The probability of the particle being found to the left of {x=a} is

\displaystyle  P_{x<a}=\frac{3}{b}\int_{0}^{a}\frac{x^{2}}{a^{2}}dx=\frac{a}{b} \ \ \ \ \ (6)

If {b=a}, then {\Psi} drops to zero at {x=a} so {P_{x<a}=1}. If {b=2a}, then {\Psi} is an isosceles triangle symmetric about {x=a} so {P_{x<a}=\frac{1}{2}}.

The expectation value of {x} is

\displaystyle  \left\langle x\right\rangle =\int x\left|\Psi\right|^{2}dx=\frac{a}{2}+\frac{b}{4} \ \ \ \ \ (7)

where we used Maple to simplify the integration. If {b=2a}, then {\left\langle x\right\rangle =a} as expected.

8 thoughts on “Triangular wave function: probabilities

  1. Deniska

    Oh the wave function given in the question is the solution of schrodinger equation ( which may or may not be the actual wave function of the particle) but after normalizing we get the wave function(actual/correct), is my reasoning correct?

    Reply
    1. gwrowe Post author

      Essentially, yes. The normalization is required to make the wave function consistent with the probabilistic interpretation, that is, equation 2 says that the probability of finding the particle somewhere in space must add up to 1.

      Reply
    1. gwrowe Post author

      If you plug equation 1 into the integral in 7, the gory details are:

      \displaystyle   \int x\left|\Psi\right|^{2}dx \displaystyle  = \displaystyle  \int_{0}^{a}x\left(\sqrt{\frac{3}{b}}\frac{x}{a}\right)^{2}dx+\int_{a}^{b}x\left(\sqrt{\frac{3}{b}}\frac{b-x}{b-a}\right)^{2}dx
      \displaystyle  \displaystyle  = \displaystyle  \frac{3}{4}\,{\frac{{a}^{2}}{b}}+\frac{3}{4}\,{\frac{-{a}^{4}+{b}^{4}}{b\left(-b+a\right)^{2}}}-2\,{\frac{-{a}^{3}+{b}^{3}}{\left(-b+a\right)^{2}}}+\frac{3}{2}\,{\frac{b\left(-{a}^{2}+{b}^{2}\right)}{\left(-b+a\right)^{2}}}
      \displaystyle  \displaystyle  = \displaystyle  \frac{3}{4}\,{\frac{{a}^{2}}{b}}+\frac{-\frac{3}{4}\,{a}^{4}+\frac{3}{4}\,{b}^{4}-2\,b\left(-{a}^{3}+{b}^{3}\right)+\frac{3}{2}\,{b}^{2}\left(-{a}^{2}+{b}^{2}\right)}{b\left(-b+a\right)^{2}}
      \displaystyle  \displaystyle  = \displaystyle  \frac{3}{4}\,{\frac{{a}^{2}}{b}}+\frac{-\frac{3}{4}\,{a}^{4}+\frac{1}{4}\,{b}^{4}+2\,{a}^{3}b-\frac{3}{2}\,{a}^{2}{b}^{2}}{b\left(-b+a\right)^{2}}
      \displaystyle  \displaystyle  = \displaystyle  \frac{3}{4}\,{\frac{{a}^{2}}{b}}-\frac{1}{4}\frac{\left(3\,a+b\right)\left(-b+a\right)^{3}}{b\left(-b+a\right)^{2}}
      \displaystyle  \displaystyle  = \displaystyle  \frac{3}{4}\,{\frac{{a}^{2}}{b}}-\frac{1}{4}\,{\frac{3\,{a}^{2}-2\,ab-{b}^{2}}{b}}
      \displaystyle  \displaystyle  = \displaystyle  \frac{a}{2}+\frac{b}{4}

      Now you can see why I use Maple to do the integrals and algebra(!)

      Reply
        1. gwrowe Post author

          You just have to multiply the integrands out and then do the integrals. It’s a lot of work, but there’s nothing fancy about it.

          Reply

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