Determine the particles most probable position

[Addez123]

Homework Statement

x > 0:
$$\Psi(x) = 2\sqrt{a^3}xe^{-ax}$$

x < 0:
$$\Psi(x) = 0$$

Relevant Equations

Physics

To get expected value I use
$$E = \int \Psi^* Q \Psi dx$$
where Q = x

$$4a^3 \int xe^{ax} \cdot x \cdot xe^{-ax} dx = 4a^3 \int_0^{\inf} x^3 dx$$
which is undefined.

But the answer is suppose to be 1/a.

 

[topsquark]

Homework Statement:: x > 0:
$$\Psi(x) = 2\sqrt{a^3}xe^{-ax}$$

x < 0:
$$\Psi(x) = 0$$
Relevant Equations:: Physics

To get expected value I use
$$E = \int \Psi^* Q \Psi dx$$
where Q = x

$$4a^3 \int xe^{ax} \cdot x \cdot xe^{-ax} dx = 4a^3 \int_0^{\inf} x^3 dx$$
which is undefined.

But the answer is suppose to be 1/a.

What is $\Psi ^*$ again? (Hint: a is real.)

-Dan

 

[Addez123]

What is $\Psi ^*$ again? (Hint: a is real.)

-Dan

Isn't the conjugate simply reversing the minus sign on the exponential?
Conjugate of $xe^{-iax}$ is $xe^{iax}$ according to wolfram alpha too.

 

[topsquark]

Isn't the conjugate simply reversing the minus sign on the exponential?
Conjugate of $xe^{-iax}$ is $xe^{iax}$ according to wolfram alpha too.

The argument of your exponential is -ax, not -iax.

-Dan

 

[Addez123]

The argument of your exponential is -ax, not -iax.

-Dan

That is true!
So since it has no imaginary part the conjugate is simply itself? No changes?

 

[topsquark]

That is true!
So since it has no imaginary part the conjugate is simply itself? No changes?

If a is real then $a = a + 0i$. So $a^* = a - 0i = a$.

-Dan

 

[Addez123]

If a is real then $a = a + 0i$. So $a^* = a - 0i = a$.

-Dan

Great!

But I still get the wrong answer.
$$4a^3 \int_0^{inf} x^3 \cdot e^{-2ax} = 4a^3 \cdot \frac {3}{8a^4} = \frac {3}{2a}$$
not 1/a as the answer suggests.

 

[topsquark]

Great!

But I still get the wrong answer.
$$4a^3 \int_0^{inf} x^3 \cdot e^{-2ax} = 4a^3 \cdot \frac {3}{8a^4} = \frac {3}{2a}$$
not 1/a as the answer suggests.

Ahhhh... I see the problem now.

You are looking for the most probable position. The expectation value is the average position.

What you want to do is find the where the maximum value of the probability density $\Psi ^* \Psi$ is.

-Dan

 

[PeroK]

Note that for a non-negative real-valued function $|\Psi(x)|^2$ is a maximum at $x_0$ iff $\Psi(x)$ is a maximum at $x_0$.

 

[Addez123]

Note that for a non-negative real-valued function $|\Psi(x)|^2$ is a maximum at $x_0$ iff $\Psi(x)$ is a maximum at $x_0$.

Good tip! I solved the $\Psi^* \Psi$ and got correct value though. Solved it again with your solution and it gave correct result too!

 

[PeroK]

Good tip! I solved the $\Psi^* \Psi$ and got correct value though. Solved it again with your solution and it gave correct result too!

It's a good trick. In this case it doesn't save much algebra, but sometimes it can make things a lot simpler.