Normalising the Wavefunction and Finding Particle Position

[neutrino]

For those who have the book, this is problem 1.4 from Griffiths, 2nd ed.

\psi (x,0) = \left\{ \begin{array}{rcl}<br /> A\frac{x}{a} &amp; \mbox{for} &amp; 0 \leq x \leq a \\<br /> A\frac{b-x}{b-a} &amp; \mbox{for} &amp; a \leq x \leq b \\<br /> 0 &amp; otherwise<br /> \end{array}\right.

a) Normalise the wavefunction.
I found A = \sqrt{\frac{3}{b}} (Am I right?)

c) Where is the particle most likely to be found, at t =0?
Using the above value for A, I evaluated the two integrals, one from 0 to a, and the other from a to b. For the first, I get a probability of a/b, and for the second integral 1- (a/b), as expected. How does this answer the question?

 

[StatusX]

a) Normalise the wavefunction.
I found A = \sqrt{\frac{3}{b}} (Am I right?)

Probably not. The answer should probably involve both a and b (or neither).

c) Where is the particle most likely to be found, at t =0?
Using the above value for A, I evaluated the two integrals, one from 0 to a, and the other from a to b. For the first, I get a probability of a/b, and for the second integral 1- (a/b), as expected. How does this answer the question?

Where is the probability function maximized?

 

[HallsofIvy]

For those who have the book, this is problem 1.4 from Griffiths, 2nd ed.

\psi (x,0) = \left\{ \begin{array}{rcl}<br /> A\frac{x}{a} &amp; \mbox{for} &amp; 0 \leq x \leq a \\<br /> A\frac{b-x}{b-a} &amp; \mbox{for} &amp; a \leq x \leq b \\<br /> 0 &amp; otherwise<br /> \end{array}\right.

a) Normalise the wavefunction.
I found A = \sqrt{\frac{3}{b}} (Am I right?)

Not even close! As StatusX pointed out, it should depend on both a and b. How did you get that?

c)

Where is the particle most likely to be found, at t =0?
Using the above value for A, I evaluated the two integrals, one from 0 to a, and the other from a to b. For the first, I get a probability of a/b, and for the second integral 1- (a/b), as expected. How does this answer the question?

It doesn't! Why would you think it would? Where is the wavefunction a maximum?

 

[neutrino]

Ok here's what I did (I knew I hadn't understood how to do this properly )

\int_{0}^{a} (A\frac{x}{a})^2dx + \int_{a}^{b} (A\frac{b-x}{b-a})^2dx = 1 , for the wavefunction to the left of 0, and to the right of b being zero.
\frac{A^2}{3}(a + (b-a)) = 1

Was this the right method to normalise the wavefunction? (Obviously, not!)

 

[George Jones]

Before doing anything, I think sketching the wavefunction for this question is quite useful. Maybe this is already done in the book - I don't have my copy with me.

 

[neutrino]

George, sketching is problem b.

 

[George Jones]

Looks OK to me.

 

[StatusX]

I apologize, it does appear you were right, and A does not depend on a. The wavefunction superficially appears to be symmetric between a and b, but when you draw it you see they have very different roles, and the dependence on a cancels out when you integrate.

 

[George Jones]

Here's roughly what's happening. The larger a is, the wider and shorter the wavefunction is, and this happens in such a way that area under the square of the wavefunction in independent of a. The width of the wavefunction also depends on b, but the height doesn't This leaves b in the final answer.

However, before doing the calculation, I would never have thought of this.

 

[neutrino]

I feel a bit better now.

Now, how about telling me about this 'maximising' stuff?

 

[StatusX]

Clearly the integral of the wavefunction does not depend on a, because you are just taking half of the area of the rectangle of height 1 stretching from x=0 to x=b. More suprising is that the integral of the square of the wavefunction also does not depend on a, and this can be explained (and generalized) as follows.

Imagine integrating any monotonically increasing function from 0 to 1, and assume f(0)=0. This is equivalent to calculating the area under the curve y=f(x) on this region, and this in turn is equivalent to the area to the right of the curve up to the line x=1, which is just the integral from 0 to f(1) of 1-f^-1(y). When you stretch the region of integration horizontally you are essentially mutiplying this latter integral (over y) by a constant, which explains why the value of the integral scales linearly with the amount of stretching. It isn't hard to show this extends to any function, not necessarily jut monotonically increasing ones.

In this problem, the amount of horizontal stretching is determined by a, and so in particular the integrals will be of the form k(a)+k(b-a)=kb, explaining why the dependence on a drops out. Sorry to dwell on this, but I find it interesting.

 

[George Jones]

Now, how about telling me about this 'maximising' stuff?

The probability that the particle will be found at any particular x between 0 and b is exactly zero. The question asks for for an x such that the particle is more likely to be found "in the neighbourhood" of this x than "in the neighbourhood" of any other x.

The probabilty density is given by the square of the wavefunction, so the particle is most likey to be found in the neighbourhood of the maximum of (the square of) the wavefunction.

 

[neutrino]

The probability that the particle will be found at any particular x between 0 and b is exactly zero.

Did you mean 1?

The question asks for for an x such that the particle is more likely to be found "in the neighbourhood" of this x than "in the neighbourhood" of any other x.

The probabilty density is given by the square of the wavefunction, so the particle is most likey to be found in the neighbourhood of the maximum of (the square of) the wavefunction.

Thanks for the reply, George. (Sorry for being late with a reply)...I did plot |\psi|^2. It is discontinuous at the point a, but the graph also reaches a max at \left(a, \frac{3}{b})\right. So, is the answer to the question a (around a)?

 

[nrqed]

Did you mean 1?

No, George *did* mean zero. Because |Psi(x)|^2 is a probability density so |Psi(x)|^2 dx represents the probability of finding the particle between x and x+dx. So strictly speaking, when people say at what x is the probability max, they really mean "for what value of x is it most likely to find the particle between x and x+dx?". But it is standard sloppiness to talk about "values of x for which the probability is max".

Thanks for the reply, George. (Sorry for being late with a reply)...I did plot |\psi|^2. It is discontinuous at the point a, but the graph also reaches a max at \left(a, \frac{3}{b})\right. So, is the answer to the question a (around a)?

Well, the answer is simply "at x=a".

Patrick
EDIT: Again, the real question should have been "for what value of x is it most probable to find the particle between x and x+dx?". And the answer is "at x=a" (meaning that the probability is maximum for finding the particle between a and a+dx).

 

[maverick280857]

At t = 0, you already have the wavefunction. If you sketch it, you will see that it is a triangular section centered at x = 0 stretching from x = 0 to x = b and zero everywhere else. It is clear then that the wavefunction is maximum at x = a and so is the probability of finding it (probability "maximum").

 

[borisleprof]

Your result for A is right.
You have to understand the meaning of Psi.
If, at the instant t, a measurement is made to locate the particle associated with the wave function Psi(x,t), the the probability P(x,t)dx that the particle will be found at a coordinate x and x+dx is equal to
Psi*(x,t)Psi(x,t)dx. Where Psi*(x,t) represent the complex conjugate of Psi(x,t) (Quantum physics, R. Eisberg, R. Resnick)

In the interval [0, a] Psi*Psi is increasing and in the interval [a,b] it's decrease to zero. So the maximum probability Psi*(x,t)Psi(x,t)dx is obtained around x = a;
So the particle is more likely to be found at x = a and x = a+dx.
Answer given also by nrqed.
bye

 

[Niles]

At t = 0, you already have the wavefunction. If you sketch it, you will see that it is a triangular section centered at x = 0 stretching from x = 0 to x = b and zero everywhere else. It is clear then that the wavefunction is maximum at x = a and so is the probability of finding it (probability "maximum").

I am dealing with this question too.

Why is it that the expectation value of x, <x>, is NOT equal to the point at which the particle is most likely to be found at t=0? And how is it that I can just use Psi, and not |Psi|²?

 

[Dick]

I am dealing with this question too.

Why is it that the expectation value of x, <x>, is NOT equal to the point at which the particle is most likely to be found at t=0? And how is it that I can just use Psi, and not |Psi|²?

If a particle has a 50% probability of being found at x=0 and a 50% probability of being found at x=1, then <x>=0.5. On the other hand, there's no chance it will actually be found at x=0.5. You can use Psi because the maximum of |Psi| is located in the same place as the maximum of |Psi|^2. Think about it.

 

[Niles]

Does this mean that <x> is the point, where the probability of finding the particle to the right of <x> equals the probability of finding the particle to the left of <x>, which is 0.5?

About using |Psi|: Since |Psi| is the norm of Psi, then |Psi| and |Psi|^2 will have maximum at the same x - can this explanation be used?

 

[HallsofIvy]

No. it doesn't. <x> is the mean or expected value of the probability distribution. The value of x such that the probability is 1/2 of being less than x and 1/2 of being more than x is the median. They are not necessarily the saime.

 

[Niles]

The value of x such that the probability is 1/2 of being less than x and 1/2 of being more than x is the median.

Using this definiton, then ideally shouldn't half the data points of some measurement lie on the left side of the median, and the other half lie on the right side?

 

[Dick]

Using this definiton, then ideally shouldn't half the data points of some measurement lie on the left side of the median, and the other half lie on the right side?

Yes, that's what the median is. But the median isn't necessarily the same as the mean, <x>.

 

[Niles]

Great, thanks.