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Calculate the expectation value of $L_{z}=-ih \left(x\frac{d}{dy}-y\frac{d}{dx} \right)$ for the state : $\phi(x,y,z) = a^{5/2} e^{-a(|x|+|y|+|z|) } (x+iy),$ where $a>0$. So, I need to calculate the integral: $ \langle L_{z} \rangle = \int_{-\infty}^{\infty}\overline{\phi}L_{z} \phi \hspace{1mm} dx \hspace{1mm} dy \hspace{1mm} dz = \int a^{5/2}e^{-a(|x|+|y|+|z|)}(x\frac{d}{dy}-y\frac{d}{dx})a^{5/2}e^{-a(|x|+|y|+|z|)}(x+iy) \hspace{1mm} dx \hspace{1mm}dy \hspace{1mm} dz .$

The possible cases are:

  1. $x>0, y>0, z>0$;
  2. $x<0, y<0, z<0$;
  3. $x>0, y<0, z<0$;
  4. $x>0, y>0, z<0$;
  5. $x<0, y>0, z<0$;
  6. $x<0, y<0, z>0$;
  7. $x>0, y<0, z>0$

which gives the six integrals:

  1. $\displaystyle \int a^{5} e^{-2a(x+y+z)}(x^{2}+y^{2}\;dx\;dy\;dz$

  2. $\displaystyle \int a^5 e^{-2a(-x-y-z)} (x^2+y^2)\;dx\;dy\;dz$

  3. $\displaystyle \int a^5 e^{-2a(x-y-z)} (x^2+y^2)\;dx\;dy\;dz$

  4. $\displaystyle \int a^5 e^{-2a(x+y-z)} (x^2+y^2)\;dx\;dy\;dz$

  5. $\displaystyle \int a^5 e^{-2a(-x+y-z)} (x^2+y^2)\;dx\;dy\;dz$

  6. $\displaystyle \int a^5 e^{-2a(-x-y+z)} (x^2+y^2)\;dx\;dy\;dz$

  7. $\displaystyle \int a^5 e^{-2a(x-y+z)} (x^2+y^2)\;dx\;dy\;dz$

    Does anybody see how to reduce this to a integral of one variable? Please do tell.

Edit: there is a horrible mistake, I forgot to multiply within the integral by $L_{z}$, Srivatsans solutions for the "wrong" integral stays true, though.

I come to:

$\displaystyle{ \int \int \int (x-iy)a^{5/2}e^{-a(|x|+|y|+|z|)}(-i\hbar(x\frac{d}{dy}-y\frac{d}{x})e^{-a(|x|+|y|+|z|)}a^{5/2}(x+iy)}dxdydz$

$z\frac{d}{dx}(e^{-a(|x|+|y|+|z|}(x+iy))$, according to wolframalpha this gives an a' in the derivative… how can this be?

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    There should be 8 cases: three variables and depending on whether they are positive or not. However, The integral are symmetric with respect to $x$ and $y$, i.e. the integral for x>0, y<0 is the same as the integral for x<0, y>0. This help us to reduces to the following cases: (i) x>0, y>0, z>0 (ii) x>0, y>0, z>0 (iii) x<0, y<0, z>0 (iv) x<0, y<0, z<0 (v) x>0, y<0, z>0 (vi) x>0, y<0, z<0.2011-11-21

1 Answers 1

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You can do away with case analysis completely. Notice that the integral is completely symmetric w.r.t. the eight orthants; i.e., if you flip the sign of any variable, the integrand remains unchanged. So, immediately you can rewrite the integral as $ \iiint_{\mathbb R^3} a^5 e^{-2a(|x|+|y|+|z|)}(x^{2}+y^{2}) \ dx \ dy \ dz = 8 \iiint_{D} a^5 e^{-2a(|x|+|y|+|z|)}(x^{2}+y^{2}) \ dx \ dy \ dz $ where $D$ is the positive orthant $\{ (x,y,z) \mid x, y, z \geqslant 0 \}$. You can further simplify this to: $ 8 \iiint_D a^5 e^{-2a(x+y+z)}(x^2+y^2) \ dx \ dy \ dz. $

Further manipulations: In what follows, I manipulate the integral systematically to evaluate it. Before reading on, make sure you have attempted the integral and then compare our methods.

Split the integral into the sum $ 8 \iiint_D a^5 e^{-2a(x+y+z)} x^2 \ dx \ dy \ dz + 8 \iiint_D a^5 e^{-2a(x+y+z)} y^2 \ dx \ dy \ dz . $ The above two integrals are equal (by interchanging the variables $x \leftrightarrow y$). So the integral can be written simply as $ 16 \iiint_D a^5 e^{-2a(x+y+z)} x^2 \ dx \ dy \ dz = 16a^5 \int_0^\infty \int_0^\infty \int_0^\infty (e^{-2ax} x^5) \cdot e^{-2ay} \cdot e^{-2az} \ dx \ dy \ dz. $ Now we can split the integral as a product of three integrals: $ 16a^5 \left( \int_0^\infty e^{-2ay} \ dy \right) \left( \int_0^\infty e^{-2az} \ dz \right) \left( \int_0^\infty e^{-2ax} x^2 \ dx \right). $ The first two integrals are easy and standard: they evaluate to $\frac{1}{2a}$. Plugging in this value, we can simplify the above expression to $ 16a^5 \cdot \frac{1}{2a} \cdot \frac{1}{2a} \cdot \int_0^\infty e^{-2ax} x^2 \ dx = 4a^3 \int_0^\infty e^{-2ax} x^2 \ dx . $ Make the substitution $u = 2a x$ (typo corrected) to get $ 4a^3 \int_0^\infty e^{-u} \frac{u^2}{4a^2} \ \frac{1}{2a} du = \int_0^\infty e^{-u} \cdot \frac{u^2}{2} du. $ Integrating the above by parts, we get $ -\left. e^{-u} \left( \frac{u^2}{2} + u + 1 \right) \right|_{u=0}^{u=\infty} = 1. $

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    @VVV Full solutio$n$ with lots and lots of manipulation; enjoy! =)2011-11-21