Q28MP - Mathematical Methods in Physical Sciences

Question

Evaluate the following absolute square of a complex number (which arises in a problem in quantum mechanics). Assume a and b are real. Express your answer in terms of a hyperbolic function.

${\left|\frac{{\left(a+\mathrm{bi}\right)}^2{e}^b-{\left(a-\mathrm{bi}\right)}^2{e}^{-b}}{4{\mathrm{abie}}^{-\mathrm{ia}}}\right|}^{\mathbf{2}}$

Step-by-Step Solution

Answer

The absolute square of the complex number is 

$\mathrm{w}=1+{\sinh }^2\left(\mathrm{b}\right){\left(\frac{{\mathrm{a}}^2+{\mathrm{b}}^2}{2\mathrm{ab}}\right)}^2$

Step 1: Given Information.

The given expression is ${\left|\frac{{\left(\mathrm{a}+\mathrm{bi}\right)}^2{\mathrm{e}}^{\mathrm{b}}-{\left(\mathrm{a}-\mathrm{bi}\right)}^2{\mathrm{e}}^{-\mathrm{b}}}{4{\mathrm{abie}}^{-\mathrm{ia}}}\right|}^2$ .

Step 2: Meaning of rectangular form.

Represent the complex number in rectangular form means writing the given complex number in the form of x+iy in which x is the real part and y is the imaginary part.

Step 3: Simplify.

$$\begin{gathered} w={\left|\frac{{\left(\mathrm{a}+ib\right)}^2{\mathrm{e}}^{\mathrm{b}}-{\left(\mathrm{a}-ib\right)}^2{\mathrm{e}}^{\left(-\mathrm{b}\right)}}{4{\mathrm{abie}}^{\left(-\mathrm{ia}\right)}}\right|}^2 \\ w={\left|\frac{{\left(\mathrm{a}+ib\right)}^2{\mathrm{e}}^{\mathrm{b}}-{\left(\mathrm{a}-ib\right)}^2{\mathrm{e}}^{\left(-\mathrm{b}\right)}}{4\mathrm{ab}}\right|}^2\times \frac{1}{{\left|i{e}^{\left(-ia\right)}\right|}^2} \end{gathered}$$

 

Find the value of the second term

$$\begin{gathered} {\left|i{e}^{\left(\theta \right)}\right|}^2={\left|i\right|}^2\times \left[{\cos }^2\theta +{\sin }^2\theta \right] \\ {\left|i{e}^{\left(\theta \right)}\right|}^2=1 \end{gathered}$$     

 

Substitute this value in the above equation.

$w={\left|\frac{{\left(\mathrm{a}+ib\right)}^2{\mathrm{e}}^{\mathrm{b}}-{\left(\mathrm{a}-ib\right)}^2{\mathrm{e}}^{\left(-\mathrm{b}\right)}}{4\mathrm{ab}}\right|}^2$

 

Let us consider.

$u=\frac{{\left(\mathrm{a}+ib\right)}^2{\mathrm{e}}^{\mathrm{b}}-{\left(\mathrm{a}-ib\right)}^2{\mathrm{e}}^{\left(-\mathrm{b}\right)}}{4\mathrm{ab}}$

Step 4: Simplify.

$$\begin{gathered} u=\frac{{\left(\mathrm{a}+ib\right)}^2{\mathrm{e}}^{\left(\mathrm{b}\right)}-{\left(\mathrm{a}-bi\right)}^2{\mathrm{e}}^{\left(-\mathrm{b}\right)}}{4\mathrm{ab}} \\ u=\frac{\left[a^2+2abi-b^2\right]e^{\left(b\right)}-\left[a^2-2abi-b^2\right]e^{\left(-b\right)}}{4ab} \\ u=\frac{a^2\left[e^{\left(b\right)}-e^{\left(-b\right)}\right]+2abi\left[e^{\left(b\right)}+e^{\left(-b\right)}\right]-b^2\left[e^{\left(b\right)}-e^{\left(-b\right)}\right]}{4ab} \end{gathered}$$

 

Divide 2 in the numerator and the denominator.

$$\begin{gathered} u=\frac{a^2\left[e^{\left(b\right)}-e^{\left(-b\right)}\right]/2+2abi\left[e^{\left(b\right)}+e^{\left(-b\right)}\right]-b^2\left[e^{\left(b\right)}-e^{\left(-b\right)}\right]/2}{4ab/2} \\ U=\frac{a^2\sin h\left(b\right)+2abi\cos h\left(b\right)-b^2\sin \left(b\right)}{2ab} \\ u=\left(\frac{a^2-b^2}{2ab}\right)\sin h\left(b\right)+i \cos h\left(b\right) \\ u=x \sin h\left(b\right)+i \cos h\left(b\right) \end{gathered}$$

 

In the above equation$x=\left(\frac{a^2-b^2}{2ab}\right)$ for simplification.

Step 5: Substitute the value.

$$\begin{gathered} w={\left|x \sin h\left(b\right)+i \cos h\left(b\right)\right|}^2 \\ w=x^2\sin h^2\left(b\right)+\cos h^2\left(b\right) \\ w=x^2\sin h^2\left(b\right)+\left(1+\sin h^2\left(b\right)\right) \\ w=1+\sin h^2\left(1+x^2\right) \end{gathered}$$

 

Put the value of x.

$$\begin{gathered} w=1+\sin h^2(b=1+\sin h^2\left(b\right)\left[\frac{a^2+2a^2{b}^2+b^4}{4a^2{b}^2}\right])\left[1+\frac{a^4-2a^2{b}^2+b^4}{4a^2{b}^2}\right] \\ w=1+\sin h^2\left(b\right)\left[\frac{a^2+2a^2{b}^2+b^4}{4a^2{b}^2}\right] \\ w=1+\sin h^2\left(b\right){\left(\frac{a^2+b^2}{2ab}\right)}^2 \end{gathered}$$

 

Hence the absolute square of the complex number is 

$w=1+\sin h^2\left(b\right){\left(\frac{a^2+b^2}{2ab}\right)}^2.$