What is the total probability of the particle in state 'N'?

[qm14]

Homework Statement

State vector :
| \psi \rangle = \sum_{n = - \infty}^\infty a_n| \psi \rangle
where
a_n= i\sqrt\frac{3}{5}{\frac{i}{4}}^{|n|/2}e^{-in\pi/2}

Find the probability of the particle in Nth state.
What is the total probability of the particle in 'N' negative state.

Homework Equations

|\langle\psi|\psi\rangle| ^2 = 1

The Attempt at a Solution

Would the probability of the particle being in Nth state be |\langle\psi|\psi\rangle| ^2 =<br /> =&gt; |a_n|^2 = ?

a_n*= -i\sqrt\frac{3}{5}{}\frac{1}{4i}^{|n|/2}e^{+in\pi/2}

If so here is my working...
\sum_{n -\infty}^\infty a_n a_n* \delta_{n m}

\sum_{n -\infty}^\infty ( i\sqrt\frac{3}{5}{\frac{i}{4}}^{|n|/2}e^{-in\pi/2} )(-i\sqrt\frac{3}{5}{}\frac{1}{4i}^{|n|/2}e^{+in\pi/2} )

|a_n|^2 = \gamma \sum_{n -\infty}^\infty(1/4)^{|n|}<br />

where \gamma = 3/5

How do I go around evaluating this summation from -infinity to +ve infinity though the power is |n|.

Evaluating the summation in the range (0,infinty) yields (3/5)+ 0 ...

Thanks.

 

[BvU]

Hello QM, welcome to PF :-)Could that be $| \psi \rangle = \sum_{n = - \infty}^\infty a_n| \psi_n \rangle$ ?

probability of the particle being in Nth state be $|\langle\psi|\psi\rangle| ^2$ ?

On the same footing: no, but close: $|\langle\psi_n|\psi\rangle| ^2$

So for N you get $\sum_{n = - \infty}^\infty a_n| \langle \psi_N|\psi_n \rangle|^2 = \delta(N,n) \ a_n^* a_n\ |\langle \psi_N|\psi_N \rangle|^2 = a_N^* a_N$

Probability for being in N th state with N negative: same procedure (probably dsame answer ?)

Or are they asking for something else: the total probability for the particle being in a state with negative N ? Then you have to sum, but fom minus infinity to 0, not from minus infinity to plus infinity, right ?

 

[qm14]

Hi,

Yes, I have been asked to find the total probability of a particle in a negative energy state, so how is this answer any different than the previous one. Both summations add up to 1 i.e (-infty,0) and (0,infty)..

I thought negative values of N would be discarded due to the^ |N| power, or is my concept of absolute function wrong...

Edit correction... Using N= -infty gives infinite as an answer, am I right in assuming that this indicates the unattainability of negative states?

 

[BvU]

$\sum_{-\infty}^{+\infty}$ should give 1 After all it is a probability.
How do you calculate $|a_n|^2$ and how do you calculate the sum ?

 

[qm14]

I have calculated |a_n|^2 above, I have split summation into two parts..
\sum_{-\infty}^{0} (1/4)^{|n|} = (1/4)^{-\infty} + (1/4)^{0} = (\infty) + 1
Similarly for the range (0,\infty).

 

[qm14]

Anyone...

 

[BvU]

$({1\over 4})^{|-\infty|} = 0$ but in your summation you skipped over all the intermediate terms !
Why don't you try the easier one first, e.g. ${1\over 4}+{1\over 16}+{1\over 64} + ...$

By the way, $|a_n|^2 = \gamma \sum_{n -\infty}^\infty(1/4)^{|n|}$ is not correct ! It is simply $\gamma (1/4)^{|n|}$

And your concept for the absolute function should be $|n| = n$ for $n\ge 0$ and $|n|=-n$ for $n \le 0$.

$\sum_{-\infty}^{+\infty}$ should give 1

And it does (fortunately), but you haven't verified that yet. You should do so now.

 

[qm14]

$({1\over 4})^{|-\infty|} = 0$ but in your summation you skipped over all the intermediate terms !
Why don't you try the easier one first, e.g. ${1\over 4}+{1\over 16}+{1\over 64} + ...$

By the way, $|a_n|^2 = \gamma \sum_{n -\infty}^\infty(1/4)^{|n|}$ is not correct ! It is simply $\gamma (1/4)^{|n|}$

And your concept for the absolute function should be $|n| = n$ for $n\ge 0$ and $|n|=-n$ for $n \le 0$.

And it does (fortunately), but you haven't verified that yet. You should do so now.

I am failing to link together how
$\gamma (1/4)^{|n|}= 1$
Intermediate values of above equation ~ (0.333) for the first few terms.. to meet the unity condition $(1/4)^{|n|}$ must be equivalent to (5/3) since constant is 3/5.

What am I overlooking...

 

[BvU]

Note that there is a summation in front: the SUM over all n must give 1: $\Sigma\ \gamma (1/4)^{|n|}= 1$
${1\over 4}+{1\over 16}+{1\over 64} + ...$ are only the terms $n = 1,2,... \infty$

You are overlooking the term with n = 0 and the terms with n < 0.

The sum of the terms with $n = -\infty ... -3, -2, -1$ gives you the desired probability

$\approx 0.33$ for the first few terms ? Do you know how to calculate the sum of an infinite geometric series ?

 

[qm14]

Note that there is a summation in front: the SUM over all n must give 1: $\Sigma\ \gamma (1/4)^{|n|}= 1$
${1\over 4}+{1\over 16}+{1\over 64} + ...$ are only the terms $n = 1,2,... \infty$

You are overlooking the term with n = 0 and the terms with n < 0.

The sum of the terms with $n = -\infty ... -3, -2, -1$ gives you the desired probability

$\approx 0.33$ for the first few terms ? Do you know how to calculate the sum of an infinite geometric series ?

I have finally realized the flaw in my reasoning. I was not taking the negative values into account, presuming that the summation would blow up at - infinity.

I really appreciate your input, thanks! .

 

[BvU]

Just checking: could you verify $\sum_{-\infty}^{+\infty}\ \gamma (1/4)^{|n|}= 1$ and is your final answer to the first question $\sqrt{3\over 5}\; \left(1\over 4\right)^{|N|}$ and the second -1/3 exactly ?