Expectation values of x and x^2

In summary, it seems that I made a mistake when calculating the expectation values for x and x^2. I should be getting a non-zero value for one of the expectation values.
  • #1
cyberdeathreaper
46
0
Given the wave function:
[tex] \psi (x,t) = Ae^ {-\lambda \mid x\mid}e^ {-(i ) \omega t} [/tex]
where A, [itex] \lambda [/itex], and [itex] \omega [/itex] 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
[tex] <x> = \int_{-\infty}^{+\infty} x(A^2)e^ {-2\lambda \mid x\mid} dx [/tex]
and
[tex] <x^2> = \int_{-\infty}^{+\infty} (x^2)(A^2)e^ {-2\lambda \mid x\mid} dx [/tex]
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?
 
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  • #2
cyberdeathreaper said:
Given the wave function:
[tex] \psi (x,t) = Ae^ {-\lambda \mid x\mid}e^ {-(i ) \omega t} [/tex]
where A, [itex] \lambda [/itex], and [itex] \omega [/itex] 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
[tex] <x> = \int_{-\infty}^{+\infty} x(A^2)e^ {-2\lambda \mid x\mid} dx [/tex]
and
[tex] <x^2> = \int_{-\infty}^{+\infty} (x^2)(A^2)e^ {-2\lambda \mid x\mid} dx [/tex]
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>:

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

[tex] 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 \ \ = \ \ [/tex]

[tex] 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)} [/tex]


For <x2>:

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

[tex] 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 \ \ = \ \ [/tex]

[tex] 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 \ \ = \ \ \ [/tex]

[tex] 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 )}} [/tex]




~~
 
Last edited:
  • #3
Thanks. That's the justification I was looking for.

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

given:

[tex] \int_{-\infty}^{+\infty} f(x) dx [/tex]

Then, the integral can be re-written as:

[tex] 2 \int_{0}^{+\infty} f(x) dx [/tex]

if [itex] \int_{-\infty}^{0} f(x) dx = \int_{0}^{+\infty} f(x) dx [/itex]

Is this correct?
 
Last edited:
  • #4
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.
 
  • #5
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.
 
  • #6
Thanks for the help. I should be set then.
 

Related to Expectation values of x and x^2

1. What is an expectation value?

An expectation value is a mathematical concept used in quantum mechanics to represent the average value of a measurement of a physical quantity. It is calculated by multiplying the possible outcomes of a measurement by their respective probabilities and summing them up.

2. How is the expectation value of x calculated?

The expectation value of x is calculated by taking the integral of x multiplied by the probability density function over all possible values of x. This can be written as ∫x*|ψ(x)|^2 dx, where ψ(x) is the wavefunction.

3. What is the importance of expectation values in quantum mechanics?

Expectation values play a crucial role in quantum mechanics as they provide a way to predict the average results of measurements on a quantum system. They also allow us to calculate other important quantities, such as the uncertainty in a measurement.

4. How does the expectation value of x^2 differ from the expectation value of x?

The expectation value of x^2 is a measure of the spread or variance of a measurement, while the expectation value of x represents the average value. The difference between the two can give insight into the uncertainty or randomness of a physical system.

5. Can the expectation value of x and x^2 be measured directly?

No, the expectation values of x and x^2 cannot be measured directly as they are mathematical quantities. However, they can be calculated using experimental data from measurements of a physical quantity and compared to the theoretical values predicted by quantum mechanics.

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