Expectation values of x and x^2

[cyberdeathreaper]

Given the wave function:
$$\psi (x,t) = Ae^ {-\lambda \mid x\mid}e^ {-(i ) \omega t}$$
where A, $\lambda$, and $\omega$ are positive real constants

I'm asked to find the expectation values of x and x^2.

I know that the values are given by
$$<x> = \int_{-\infty}^{+\infty} x(A^2)e^ {-2\lambda \mid x\mid} dx$$
and
$$<x^2> = \int_{-\infty}^{+\infty} (x^2)(A^2)e^ {-2\lambda \mid x\mid} dx$$
However, when calculated, I get <x> = <x^2> = 0. Since this would yield a standard deviation of zero, I'm thinking I've made a mistake (the reasoning being that the function does have some spread).

Does this seem correct, or should I be getting a non-zero value for one of the expectation values?

 

[xanthym]

Given the wave function:
$$\psi (x,t) = Ae^ {-\lambda \mid x\mid}e^ {-(i ) \omega t}$$
where A, $\lambda$, and $\omega$ are positive real constants

I'm asked to find the expectation values of x and x^2.

I know that the values are given by
$$<x> = \int_{-\infty}^{+\infty} x(A^2)e^ {-2\lambda \mid x\mid} dx$$
and
$$<x^2> = \int_{-\infty}^{+\infty} (x^2)(A^2)e^ {-2\lambda \mid x\mid} dx$$
However, when calculated, I get <x> = <x^2> = 0. Since this would yield a standard deviation of zero, I'm thinking I've made a mistake (the reasoning being that the function does have some spread).

Does this seem correct, or should I be getting a non-zero value for one of the expectation values?

SOLUTION HINTS:

For <x>:

$$1: \ \ \ \ <x> \ \ = \ \ \int_{-\infty}^{+\infty} x(A^2)e^ {-2\lambda \mid x\mid} \, dx \ \ = \ \$$

$$2: \ \ \ \ \ \ \ = \ \ \int_{-\infty}^{0} x(A^2)e^ {-2\lambda \mid x\mid} \, dx \ \ + \ \ \int_{0}^{+\infty} x(A^2)e^ {-2\lambda \mid x\mid} \, dx \ \ = \ \$$

$$3: \ \ \ \ \ \ \ = \ \ \color{red}\int_{0}^{+\infty} \color{black} \color{red}(-x)\color{black}(A^2)e^ {-2\lambda \mid x\mid} \, dx \ \ + \ \ \int_{0}^{+\infty} x(A^2)e^ {-2\lambda \mid x\mid} \, dx \ \ = \ \ \color{red}\mathbf{(0)}$$


For <x²>:

$$4: \ \ \ \ <x^{2}> \ \ = \ \ \int_{-\infty}^{+\infty} (x^{2})(A^2)e^ {-2\lambda \mid x\mid} \, dx \ \ = \ \$$

$$5: \ \ \ \ \ \ \ = \ \ \int_{-\infty}^{0} (x^{2})(A^2)e^ {-2\lambda \mid x\mid} \, dx \ \ + \ \ \int_{0}^{+\infty} (x^{2})(A^2)e^ {-2\lambda \mid x\mid} \, dx \ \ = \ \$$

$$6: \ \ \ \ \ \ \ = \ \ \color{red}\int_{0}^{+\infty}\color{black} \color{red}(x^{2})\color{black}(A^2)e^ {-2\lambda \mid x\mid} \, dx \ \ + \ \ \int_{0}^{+\infty} (x^{2})(A^2)e^ {-2\lambda \mid x\mid} \, dx \ \ = \ \ \$$

$$7: \ \ \ \ \ \ \ = \ \ \color{red} (2) \cdot \color{black} \int_{0}^{+\infty} (x^{2})(A^2)e^ {-2\lambda \color{red} (x) \color{black} } \, dx \ \ \ \ = \ \ {\color{red}\mathbf{\displaystyle \left( \frac {A^{2}} {2\lambda^{3}}\right )}}$$




~~

 

[cyberdeathreaper]

Thanks. That's the justification I was looking for.

Just to clearify the rule, let me see if I can generalize it...

given:

$$\int_{-\infty}^{+\infty} f(x) dx$$

Then, the integral can be re-written as:

$$2 \int_{0}^{+\infty} f(x) dx$$

if $\int_{-\infty}^{0} f(x) dx = \int_{0}^{+\infty} f(x) dx$

Is this correct?

 

[Galileo]

It holds whenever the the integrand is even wtr 0. So f(x)=f(-x).

If f(x)=-f(-x) the function is odd and the integral will be zero.

 

[dextercioby]

Yeah,that modulus in the exponential surely means a lot.The exponential of real argument in neither odd,nor even,but that modulus changes things.

Daniel.

 

[cyberdeathreaper]

Thanks for the help. I should be set then.