Square of a wave function; way to understand

In summary: The square root of a negative number can be given as i times the square root of the absolute value, so it is i times the square root of (d^2-(i hbar t)/2m). The square root of the absolute value of the square is the absolute value of the complex number, so the square root of the complex number squared is the complex number. That is why the i's disappear. The next step is to simplify the term in the exponent.If you have i in the exponent, you can get rid of it if you take the conjugate of the expression in the exponent. So replace i by -i. The i-s disappear if you do so.I hope
  • #1
Juqon
31
0

Homework Statement


In a book I found the following calculation without the way. How do you get from 1) to 2)?


Homework Equations


(1) [itex]\Psi(x,t)=\frac{A}{2\pi}\sqrt{\frac{\pi}{d²+i\frac{\hbar t}{2m}}} exp{\frac{-\frac{x²}{4}+id²k_{0}(x-\frac{k_{0}\hbar}{2m}t)}{d²+i\frac{\hbar t}{2m}}}[/itex]
Code:
[itex]\Psi(x,t)=\frac{A}{2\pi}\sqrt{\frac{\pi}{d²+i\frac{\hbar t}{2m}}} exp{\frac{-\frac{x²}{4}+id²k_{0}(x-\frac{k_{0}\hbar}{2m}t)}{d²+i\frac{\hbar t}{2m}}}[/itex]

(2) [itex]|\Psi(x,t)²|=\frac{A²}{4\pi\sqrt{d^{4}+\frac{\hbar^{2}t²}{4m²}}} exp{-\frac{(x-\frac{k_{0}\hbar}{m}t)^{2}}{2d²+\frac{\hbar^{2}t²}{2m²d²}}}[/itex]
Code:
[itex]\Psi(x,t)²=\frac{A²}{4\pi\sqrt{d^{4}+\frac{\hbar^{2}t²}{4m²}}} exp{-\frac{(x-\frac{k_{0}\hbar}{m}t)^{2}}{2d²+\frac{\hbar^{2}t²}{2m²d²}}}[/itex]


3. The Attempt at a Solution
[itex](x-\frac{k_{0}\hbar}{m}t)^{2}=x²-2x\frac{\hbar k_{0}}{m}t+\frac{\hbar^{2} k_{0}^{2}}{m^{2}}t^{2}[/itex]
[itex](d²+i\frac{\hbar t}{2m})^{2}=d^{4}+\frac{i \hbar t}{m}+\frac{i² \hbar^{2} t²}{4m²}[/itex]

I have posted the code so that it is easier for you to help me.
 
Last edited:
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  • #2
[itex]|\Psi|^2 = \Psi^* \Psi[/itex], it's not just the regular square. Compute [itex]\Psi^*[/itex] and then compute the product with [itex]\Psi[/itex]. You'll find that (2) is actually incorrect, since [itex]|\Psi|^2 [/itex] is real, not complex.
 
  • #3
Have you copied the formulae correctly? 2. is certainly wrong as it contains the imaginary unit i in spite of being an absolute value.

Do not forget that (ex)2=e2x.

ehild
 
  • #4
I am sorry, its without that i in 2) (I will correct it).

You can find it here: http://books.google.de/books?id=wmY...er quantenmechanik&pg=PA8#v=onepage&q&f=false.
Still, I do not know what "tricks" are used there. I know that (ex)2=e2x, but how do you get the [itex]\frac{i \hbar t}{m}[/itex] in front of the exp away?
What about the next i² or the minus?
You can't get it into the exp function, there would have to be a ln to do that.
What do you do with the id²k_0 in the exponent?
Why do you have a [itex]i\frac{\hbar^{2}t²}{2m²d²}[/itex] down in the exponent of the exp?
And what do I do wrong with the tex so that it doesn't appear? [Edit: [/itex] instead of [\itex]]
 
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  • #5
To get the magnitude-square of a complex number z you need to multiply it by its conjugate, z*. |z|^2=z z*. Conjugate means that all i-s are replaced by -i.
If the complex number is written with real and imaginary parts, z=a+ib, |z|^2=z z*=(a+ib)(a-ib)=a^2+b^2.
Given in the Euler form, z=r, |z|^2=(r)(re-iφ)=r^2

The magnitude of a product or fraction is the product of magnitudes.

|z1z2/z3|=|z1||z2|/|z3|

Change the sign of all i-s to the opposite in Ψ: that is Ψ*, and multiply Ψ and Ψ*. You will see that all the imaginary terms cancel.
Take care: the exponents add up.

ehild
 
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  • #6
Ok, I have tried it anew, but I can get only this far and some i stay.

[tex]
|\Psi^{2}|=\Psi^{*}\cdot\Psi=\frac{A}{2\pi}\sqrt{ \frac{\pi}{d^
2 +i \frac{\hbar t}{2m} } } \cdot\exp\left\{ \frac{-\frac{x^{2}}{4}+idk_{0}\left(x-\frac{k_{0}\hbar}{2m}t\right)}{d^{2}+i\frac{\hbar t}{2m}}\right\} \cdot\frac{A}{2\pi}\sqrt{\frac{\pi}{d^{2}-i\frac{\hbar t}{2m}}}\cdot\exp\left\{ \frac{-\frac{x^{2}}{4}-idk_{0}\left(x-\frac{k_{0}\hbar}{2m}t\right)}{d^{2}-i\frac{\hbar t}{2m}}\right\} [/tex][tex]

[1]=\frac{A^{2}}{4\pi\sqrt{d^{2}+\frac{\hbar^{2}t^{2}}{4m^{2}}}}\cdot\exp\left\{ \frac{-\frac{x^{2}}{4}+idk_{0}\left(x-\frac{k_{0}\hbar}{2m}t\right)}{d^{2}+i\frac{\hbar t}{2m}}+\frac{-\frac{x^{2}}{4}-idk_{0}\left(x-\frac{k_{0}\hbar}{2m}t\right)}{d^{2}-i\frac{\hbar t}{2m}}\right\} [/tex][tex]
[2]=\frac{A^{2}}{4\pi\sqrt{d^{2}+\frac{\hbar^{2}t^{2}}{4m^{2}}}}\cdot\exp\left\{ \frac{\left(-\frac{x^{2}}{4}+idk_{0}\left(x-\frac{k_{0}\hbar}{2m}t\right)\right)\cdot\left(d^{2}-i\frac{\hbar t}{2m}\right)}{\left(d^{2}+i\frac{\hbar t}{2m}\right)\cdot\left(d^{2}-i\frac{\hbar t}{2m}\right)}+\frac{\left(-\frac{x^{2}}{4}-idk_{0}\left(x-\frac{k_{0}\hbar}{2m}t\right)\right)\cdot\left(d^{2}+i\frac{\hbar t}{2m}\right)}{\left(d^{2}-i\frac{\hbar t}{2m}\right)\cdot\left(d^{2}+i\frac{\hbar t}{2m}\right)}\right\} [/tex][tex]
[3]=\frac{A^{2}}{4\pi\sqrt{d^{2}+\frac{\hbar^{2}t^{2}}{4m^{2}}}}\cdot\exp\left\{ \frac{\left(-\frac{x^{2}}{4}+idk_{0}\left(x-\frac{k_{0}\hbar}{2m}t\right)\right)\cdot\left(d^{2}-i\frac{\hbar t}{2m}\right)+\left(-\frac{x^{2}}{4}-idk_{0}\left(x-\frac{k_{0}\hbar}{2m}t\right)\right)\cdot\left(d^{2}+i\frac{\hbar t}{2m}\right)}{\left(d^{4}+\frac{\hbar^{2}t^{2}}{4m^{2}}\right)}\right\} [/tex][tex]
[4] =\frac{A^{2}}{4\pi\sqrt{d^{2}+ \frac{\hbar^{2}t^{2}}{4m^{2}}}}\cdot\exp\left\{ \frac{\left[\left(-\frac{x^{2}}{4}+idk_{0}\left(x-\frac{k_{0}\hbar}{2m}t\right)\right)\cdot\left(-i\frac{\hbar t}{2m}\right)+\left(-\frac{x^{2}}{4}-idk_{0}\left(x-\frac{k_{0}\hbar}{2m}t\right)\right)\cdot\left(i \frac{\hbar t}{2m}\right)\right]\cdot d^{2}}{\left(d^{4}+\frac{\hbar^{2}t^{2}}{4m^{2}} \right )}\right\} [/tex][tex]
[5] =\frac{A^{2}}{4\pi\sqrt{d^{2}+ \frac{\hbar^{2}t^{2}}{4m^{2}}}}\cdot\exp\left\{ \frac{\left[\left(-\frac{x^{2}}{4}+ idk_{0}x-idk_{0}\frac{k_{0}\hbar}{2m}t\right)\cdot\left(-i\frac{\hbar t}{2m}\right)+\left(-\frac{x^{2}}{4}-idk_{0}x+idk_{0}\frac{k_{0}\hbar}{2m}t\right)\cdot\left(i\frac{\hbar t}{2m}\right)\right]\cdot d^{2}}{\left(d^{4}+ \frac{\hbar^{2}t^{2}}{4m^{2}}\right)}\right\}
[/tex][tex]
[6] =\frac{A^{2}}{4\pi\sqrt{d^{2}+ \frac{\hbar^{2}t^{2}}{4m^{2}}}}\cdot\exp\left\{ \frac{\left[+\frac{x}{4}-idk_{0}x+idk_{0} \frac{k_{0}\hbar}{2m}t-\frac{x^{2}}{4}-idk_{0}x+idk_{0}\frac{k_{0}\hbar}{2m}t\right]\cdot\left[d^{2}+i\frac{\hbar t}{2m}\right]}{\left(d^{4}+\frac{\hbar^{2}t^{2}}{4m^{2}}\right)}\right\} [/tex][tex]
[7]=\frac{A^{2}}{4\pi\sqrt{d^{2}+\frac{\hbar^{2}t^{2}}{4m^{2}}}}\cdot\exp\left\{ \frac{\left[2idk_{0}\left(-x+\frac{k_{0}\hbar}{2m}t\right)\right]\cdot\left[d^{2}+i\frac{\hbar t}{2m}\right]}{\left(d^{4}+\frac{\hbar^{2}t^{2}}{4m^{2}}\right)}\right\} [/tex]
 

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  • #7
I think you went wrong on step 3 to 4 because [itex]d^2[/itex] is not a factor which you can take outside the brackets.
 
  • #8
BruceW said:
I think you went wrong on step 3 to 4 because [itex]d^2[/itex] is not a factor which you can take outside the brackets.
Agreed.

Going back to Equation 3, look at the numerator of the exp{...} expression. It must be real, so all imaginary terms must cancel. You have two possible approaches at this point:

1. Expand (multiply out) all terms in the numerator, and see the cancellation of the imaginary terms for yourself

OR

2. Perhaps you can see, by inspection of Equation 3, the imaginary terms in exp{...} will cancel. If so, you can just include the real terms when you expand the terms. Take note, [itex]imaginary \cdot imaginary = real[/itex], so it's not a matter of just dropping all imaginary terms and then doing the expansion. If this is not clear to you, proceed with Method 1 above.
 
  • #9
Ok, thanks, I have corrected that. Really, nice how the mixed i terms cancel. I get this far and now, how do I get to (2) from my first post?

(3)[tex] =\frac{A^{2}}{4\pi\sqrt{d^{2}+\frac{\hbar^{2}t^{2}}{4m^{2}}}}\cdot\exp\left\{ \frac{\left(-\frac{x^{2}}{4}+idk_{0}\left(x-\frac{k_{0}\hbar}{2m}t\right)\right)\cdot\left(d^{2}-i\frac{\hbar t}{2m}\right)+\left(-\frac{x^{2}}{4}-idk_{0}\left(x-\frac{k_{0}\hbar}{2m}t\right)\right)\cdot\left(d^{2}+i\frac{\hbar t}{2m}\right)}{\left(d^{4}+\frac{\hbar^{2}t^{2}}{4m^{2}}\right)}\right\}
[/tex](4)[tex] =\frac{A^{2}}{4\pi\sqrt{d^{2}+\frac{\hbar^{2}t^{2}}{4m^{2}}}}\cdot\exp\left\{ \frac{\left(-\frac{x^{2}}{4}d^{2}+id^{3}k_{0}\left(x-\frac{k_{0}\hbar}{2m}t\right)+\frac{x^{2}}{4}i \frac{\hbar t}{2m}-ii\frac{\hbar t}{2m}dk_{0}\left(x-\frac{k_{0}\hbar}{2m}t\right)\right)-\frac{x^{2}}{4}d^{2}-id^{3}k_{0}\left(x-\frac{k_{0}\hbar}{2m}t\right)-\frac{x^{2}}{4}i\frac{\hbar t}{2m}-ii\frac{\hbar t}{2m}dk_{0}\left(x-\frac{k_{0}\hbar}{2m}t\right)}{\left(d^{4}+\frac{\hbar^{2}t^{2}}{4m^{2}}\right)}\right\}
[/tex](5)[tex] =\frac{A^{2}}{4\pi\sqrt{d^{2}+\frac{\hbar^{2}t^{2}}{4m^{2}}}}\cdot\exp\left\{ \frac{\left(-\frac{x^{2}}{2}d^{2}+2\frac{\hbar t}{2m}dk_{0}\left(x-\frac{k_{0}\hbar}{2m}t\right)\right)}{\left(d^{4}+\frac{\hbar^{2}t^{2}}{4m^{2}}\right)}\right\}
[/tex](6)[tex] =\frac{A^{2}}{4\pi\sqrt{d^{2}+\frac{\hbar^{2}t^{2}}{4m^{2}}}}\cdot\exp\left\{ \frac{\left(-x^{2}d^{2}+2\frac{\hbar t}{m}dk_{0}\left(x-\frac{k_{0}\hbar}{2m}t\right)\right)}{\left(2d^{4}+\frac{\hbar^{2}t^{2}}{2m}\right)}\right\}
[/tex](7)[tex]=\frac{A^{2}}{4\pi\sqrt{d^{2}+\frac{\hbar^{2}t^{2}}{4m^{2}}}}\cdot\exp\left\{ \frac{\left(-x^{2}d^{2}+2\frac{\hbar t}{m} dk_{0}x-\frac{\hbar^{2}t^{2}}{m^{2}} dk_{0}^{2}\right)}{\left(2d^{4}+\frac{\hbar^{2}t^{2}}{2m}\right)}\right\}[/tex]
 
  • #10
You're incredibly close now. You just need to rewrite the function inside the exponential to make it the same as the one in (2)
 
  • #11
There is d^2 in the numerator of the exponent in the original wave function.

[tex]-\frac{x^{2}}{4}+id^2k_{0}\left(x-\frac{k_{0}\hbar}{2m}t\right)[/tex]

ehild
 
  • #12
[7][tex]
=\frac{A^{2}}{4\pi\sqrt{d^{2}+\frac{\hbar^{2}t^{2}}{4m^{2}}}}\cdot\exp\left\{ \frac{\left(-x^{2}d^{2}+2\frac{\hbar t}{m}d^{2}k_{0}x-\frac{\hbar^{2}t^{2}}{m^{2}}d^2k_{0}^{2}\right)}{2d^{4}+\frac{\hbar^{2}t^{2}}{2m}}\right\}[/tex]
[8][tex]
\frac{A^{2}}{4\pi\sqrt{d^{2} +\frac{\hbar^{2}t^{2}}{4m^{2}}}} \cdot\exp\left\{ -\frac{\left(x^{2}-2 \frac{\hbar t}{m}k_{0}x+ \frac{\hbar^{2}t^{2}}{m^{2}}k_{0}^{2} \right)}{2d^{2} +\frac{\hbar^{2}t^{2}}{2md^{2}}} \right\}[/tex][9][tex]=\frac{A^{2}}{4\pi\sqrt{d^{2}+\frac{\hbar^{2}t^{2}}{4m^{2}}}}\cdot\exp\left\{ -\frac{ \left (x-\frac{\hbar k_{0}}{m}t\right)^{2}}{2d^{2}+\frac{\hbar^{2}t^{2}} {2md^{2}}} \right\}[/tex]

Thanks everbody! I finally got it.
 
  • #13
nice work. Its quite a long calculation to do.
 
  • #14
wow...

may i enquire where in physics do you need such calculation?

i have done a bit of Quantum mechanics but i don't recall such stuff
 
  • #15
In quantum mechanics, but it was not directly a task, just a way to find the Gaussian packet and read the expectation value for x and also the Delta_X², and since it was in the book I wanted to do it myself. But it seems there are easier ways to do this.
 
  • #16
wow ok lol isee thanks
 

1. What is the square of a wave function?

The square of a wave function is a mathematical operation performed on a wave function, which is a mathematical representation of a quantum mechanical system. It involves taking the complex conjugate of the wave function and multiplying it by itself, resulting in a real-valued probability distribution.

2. How does the square of a wave function relate to probability?

The square of a wave function represents the probability of finding a particle in a particular state. The probability is proportional to the magnitude of the squared wave function at a given point in space. This allows us to calculate the likelihood of a particle being in a certain location or having a certain momentum.

3. What is the significance of the square of a wave function in quantum mechanics?

In quantum mechanics, the square of a wave function is a fundamental concept that helps us understand the behavior of particles on a microscopic level. It allows us to make predictions about the behavior of particles and their interactions with other particles, which is essential for understanding the physical world at a fundamental level.

4. Can the square of a wave function be negative?

No, the square of a wave function is always a positive value. This is because it is a probability distribution, and probabilities cannot be negative. If the square of a wave function were negative, it would imply a negative probability, which is not physically possible.

5. How can we visualize the square of a wave function?

The square of a wave function can be visualized as a graph, where the x-axis represents the position of a particle and the y-axis represents the probability of finding the particle at that position. The shape of the graph is determined by the properties of the wave function, such as its amplitude and wavelength.

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