# Normalization Factor

1. Aug 8, 2007

### benedwards2020

1. The problem statement, all variables and given/known data

A quantum system has a measurable property represented by the observable S with possible eigenvalues nh, where n = -2, -1, 0, 1, 2. The corresponding eigenstates have normalized wavefunctions $$\psi_{n}$$. The system is prepared in the normalized superposition state given by

$$\psi = \frac{1}{N}(\psi_{-2}+2\psi_{-1}+4\psi_{1}-6\psi_{2}$$

Where N is a normalization factor

(i) Calculate N

(ii) Write down the probability for each of the following measurements os S: -h, 0, 2h

3. The attempt at a solution

Given that the wavefunction is normalised, the sum of the squared moduli of the coefficients equals 1, so

$$\left(\frac{1}{N}\right)^2 = \left(\frac{1}{N}\right)^2\left(+\frac{2}{N}\right)^2+\left(\frac{4}{N}\right)^2+\left(\frac{-6}{N}\right)^2$$

Which equals

$$\frac{1}{N}^2=\frac{1}{N}+\frac{4}{N}+\frac{16}{N}+\frac{36}{N}$$

I'm getting a bit lost from here though

2. Aug 8, 2007

### Timo

The condition you are looking for is $$\left< \psi | \psi \right> = 1$$ (or any other value you'd normalize to), I don't understand what you're saying about the "squared moduli of the coefficients" (possible due to my lack in english), but your equations defenitely does not reflect the condition I stated (an addend independent of N is missing).

3. Aug 8, 2007

### George Jones

Staff Emeritus
The modulus of a complex number is its "length". What Ben is saying is that if a wavefunction iis written as a linear combination of orthonormal wavefunctions, then the squared "length" of the wavefunction is the sum of the squares of the "lengths" of the coefficients for the linear combination.

Ben:

1) you should understand why this is true;

2) from where did the left side of your second-last equation come;

3) are you sure about the denominators on the right side of your last equation?

4. Aug 8, 2007

### benedwards2020

Hmm... I'm probably missing some vital piece of knowledge here.... My books aren't very explicit in describing this situation... In fact Im finding the whole quantum physics stuff a bit hard to follow... But anyhow

For the points you raise...

(i) I understand your point about the squared length of the wavefunction being equal to the sum of the squares

(ii) I see what you mean about the left hand side of the 2nd equation... I should have it down as

$$\frac{1}{N}\left(\frac{1}{N}+\frac{4}{N}+\frac{16}{N}+\frac{36}{N}\right) = 1$$

(ii) I'm not sure I understand about the denominators on the right hand side. If we have

$$\frac{1}{N}\left(1+4+16+36) = 1$$

Isn't this the same as saying

$$\frac{1+4+16+36}{N} = 1$$?

If I solve for N here I get 57

Last edited: Aug 8, 2007
5. Aug 8, 2007

### George Jones

Staff Emeritus
In your last post, how did you go from the first equation to the second equation?

6. Aug 8, 2007

### benedwards2020

I just multiplied out the brackets as you would normally... Something tells me I'm wrong here....

7. Aug 8, 2007

### George Jones

Staff Emeritus
What does

$$\frac{2}{3} \times \frac{5}{7}$$

equal?

8. Aug 8, 2007

### benedwards2020

$$\frac{2 \times 5}{3 \times 7}=\frac{10}{21}$$

Am I right in saying that

$$\frac{1}{N}\left(\frac{1}{N}+\frac{4}{N}+\frac{16} {N}+\frac{36}{N}\right) = 1$$

but wrong in how I've multiplied it out?

9. Aug 8, 2007

### George Jones

Staff Emeritus
What does

$$\frac{1}{N} \times \frac{36}{N}$$

equal?

10. Aug 9, 2007

### benedwards2020

Ah.. I see what you mean..

$$\frac{1}{N} \times \frac{36}{N} = \frac{1}{N^2}$$

Therefore I should have

$$\frac{1}{N^2}+\frac{4}{N^2}+\frac{16}{N^2}+\frac{36}{N^2} = 1$$

11. Aug 9, 2007

### CompuChip

No

Yes.
So
$$\frac{57}{N^2} = 1$$.
What is $N$ then?

12. Aug 9, 2007

### benedwards2020

Sorry... Of course

$$\frac{1}{N} \times \frac{36}{N} = \frac{36}{N^2}$$

So $$\frac{57}{N^2}$$

has N = 7.5498

13. Aug 9, 2007

### CompuChip

You mean $N \approx 7.5498$, or $N = \sqrt{57}$.
Well done.

14. Aug 9, 2007

### benedwards2020

So the probability for each of the measurements S: -h, 0, 2h
will be simply

P(-h) = -1/(sqrt(57))
P(0) = 0
P(2h) = 4/(sqrt(57))

Is this right?

15. Aug 9, 2007

### Dick

Ouch. Those are amplitudes, not probabilities. Probability can't be negative. What's the relation between amplitude and probability? Wait a minute, they aren't even that. P(-h) should be associated with psi(-1) which has a coefficient of 2, not -1. Similarly for P(2h).

Last edited: Aug 9, 2007
16. Aug 9, 2007

### benedwards2020

Oh dear... Back to the books again I think.... My paper asks for probabilities for each of the measurements and gives an example similar to the answers I just gave.... I can honestly say that quantum stuff really isn't my forte!!

What should I be looking out for when calculating probabilities?

17. Aug 9, 2007

### Dick

P(-h) corresponds to psi(-1), which has a coefficient of 2/sqrt(57), right? As it says in the problem statement, probability is the squared modulus of the coefficient. I.e. 4/57. Check that the sum of ALL the probabilities adds to one. It will help you understand why you normalized the function.

Last edited: Aug 9, 2007
18. Aug 9, 2007

### benedwards2020

Ah... So for P(2h) this corresponds to psi(+2) which has coefficient of -6/sqrt(57) yes? which gives us by modulus square of coefficients 36/57?

19. Aug 9, 2007

### Dick

Now you're cooking. What's the sum of probabilities of all four possible states?

20. Aug 9, 2007

### benedwards2020

Ok, that would be

$$\frac{1}{57}+\frac{4}{57}+\frac{16}{57}+\frac{36}{57}$$

which = 1