Understanding Dirac Notation: Tips for Completing Confusing Homework Problems

[bon]

Homework Statement

Please see attached

Homework Equations

The Attempt at a Solution

Ok so basically a bit confused about notation..

does |psi> = sum over all r of a_r |u_r> ?
any help would be great..thanks

 

[diazona]

Here's the problem statement from the PDF file:

2.4 Let \psi(x) be a properly normalised wavefunction and Q an operator on wavefunctions. Let
\{q_r\} be the spectrum of Q and \{u_r(x)\} be the corresponding correctly normalised eigenfunctions.

Write down an expression for the probability that a measurement of Q will yield the value q_r. Show that \sum_r P(q_r|\psi) = 1. Show further that the expectation of Q is \langle Q\rangle \equiv \int_{-\infty}^{\infty}\psi^{*}\hat{Q}\psi\mathrm{d}x.

For future reference, it's much preferred to include the problem directly in the post, rather than making people download a file to just read it.

Ok so basically a bit confused about notation..

does |psi> = sum over all r of a_r |u_r> ?
any help would be great..thanks

This problem makes no use of Dirac notation at all. I'd suggest just thinking about eigenfunctions and eigenvalues, don't worry about states.

 

[bon]

Here's the problem statement from the PDF file:

For future reference, it's much preferred to include the problem directly in the post, rather than making people download a file to just read it.

This problem makes no use of Dirac notation at all. I'd suggest just thinking about eigenfunctions and eigenvalues, don't worry about states.

Thanks. Sorry - I just didn't know how to write mathematics on this forum...

Im still not sure how to progress... Please can you give me some pointers?

Thanks

 

[diazona]

OK, well what do you know about the probability of getting a particular result for a measurement?

 

[bon]

Ok thanks well I am supposed to do it in dirac notation..soi am i right in thinknig that the probability we are looking for in the first part is |<qr|psi>|^2? I get this to be the mod squared of the integral of Ur*(x)psi(x) dx.

But for the second part isn't the sum of probabilities of (qr|psi) = sum of all r of <qr|qr> = r? confused...!

 

[diazona]

Really? Dirac notation is not just a different way of writing functions, it actually deals with different mathematical objects (states, not functions). I'm surprised to hear that you're required to use it for a problem which does not talk about states at all.

Anyway, if you use |q_r\rangle to denote the state corresponding to the value q_r and function u_r(x), and |\psi\rangle is the state corresponding to the wavefunction \psi(x), then yes, the probability that a measurement of Q on a system in the state |\psi\rangle will yield the value q_r is indeed |\langle q_r|\psi\rangle|^2. And you're correct about the integral that gives you the value of that expression.

For the next part, can you explain in more detail how you're getting that result? I can't follow your argument well enough to identify what's wrong with it.

 

[bon]

Oh ok sure - I'm just saying that if |<qr|psi>|^2 is the probability that a measurement of Q will yield value qr, then for the next part i.e. sum over all r of P(qr|psi) = sum over all r of |<qr|psi>|^2.

But |<qr|psi>|^2. = |<qr|psi><psi|qr>|
But |psi><psi| = 1 therefore

probability = sum over all r of <qr|qr> which surely = r?

 

[diazona]

No, it's not true that |\psi\rangle\langle\psi| = 1.

You may be thinking of the completeness criterion, which says that
\sum_i |\phi_i\rangle\langle\phi_i| = 1
if and only if the states \{|\phi_i\rangle\} form a complete basis - in other words, if and only if any arbitrary state can be expressed as a linear combination of the |\phi_i\rangle's. (That equation may be useful to you in solving this problem )

 

[bon]

Ooh ok that kind of makes sense but let me put my point another way:

|psi> = integral over all x dx psi(x) |x>

so |psi><psi| = integral over all x dx psi*(x) psi(x) = 1 since psi(x) is a normalised wavefunction...

What's wrong with this argument?

 

[bon]

Oh i see where i went wrong. Am i right in thinking |psi><psi| = |x><x|? Why does this hold?

 

[bon]

In which case i seem to be left with the sum over all r of the integral dx of |Ur(x)|^2 |psi(x)|^2. How can i show this = 1?

 

[diazona]

Ooh ok that kind of makes sense but let me put my point another way:

|psi> = integral over all x dx psi(x) |x>

so |psi><psi| = integral over all x dx psi*(x) psi(x) = 1 since psi(x) is a normalised wavefunction...

What's wrong with this argument?

Well, you're right that
|\psi\rangle = \int\psi(x)|x\rangle \mathrm{d}x
but if you use that to compute |\psi\rangle\langle\psi|, you have to do one integral for |\psi\rangle and another for \langle\psi|,
|\psi\rangle\langle\psi| = \int\psi^{*}(x)|x\rangle \mathrm{d}x \int\psi(x&#039;)\langle x&#039;|\mathrm{d}x
This does not come out to be equal to 1.

Oh i see where i went wrong. Am i right in thinking |psi><psi| = |x><x|? Why does this hold?

No, I don't think that's right.

 

[bon]

OK i see. What do i do then? Thanks

 

[hunt_mat]

You know that
<br /> \int_{V}|\psi |^{2}d^{n}x=1<br />
Where you integrate over all space, the genergic probability will mean an integration over a subset.

 

[vela]

Oh ok sure - I'm just saying that if |<qr|psi>|^2 is the probability that a measurement of Q will yield value qr, then for the next part i.e. sum over all r of P(qr|psi) = sum over all r of |<qr|psi>|^2.

But |<qr|psi>|^2. = |<qr|psi><psi|qr>|

You actually just need to swap the order of the factors on the RHS:

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

And then use diazona's hint:

You may be thinking of the completeness criterion, which says that
\sum_i |\phi_i\rangle\langle\phi_i| = 1
if and only if the states \{|\phi_i\rangle\} form a complete basis - in other words, if and only if any arbitrary state can be expressed as a linear combination of the |\phi_i\rangle's. (That equation may be useful to you in solving this problem )

 

[vela]

Ooh ok that kind of makes sense but let me put my point another way:

|psi> = integral over all x dx psi(x) |x>

so |psi><psi| = integral over all x dx psi*(x) psi(x) = 1 since psi(x) is a normalised wavefunction...

What's wrong with this argument?

\langle \psi | \psi \rangle and |\psi\rangle\langle\psi| are two different mathematical objects.

The normalization condition is \langle \psi | \psi \rangle = 1. If you insert the complete set \int dx\,|x\rangle\langle x|=\hat{1}, you recover the normalization condition in terms of the wave function:

<br /> 1 = \langle \psi | \psi \rangle = \langle \psi |\hat{1}| \psi \rangle <br /> = \langle \psi |\left(\int dx\,|x\rangle\langle x|\right)| \psi \rangle <br /> = \int dx\,\langle \psi |x\rangle\langle x| \psi \rangle <br /> = \int dx\,\psi^*(x)\psi(x)<br />

 

[bon]

Great this all makes a lot more sense now! Thank you!

Trying to do the last part of the question now about <Q>...

do i start with the definition that <Q> = sum over all r of qr times P(qr given psi)..

not sure where to go from there...any ideas?

 

[diazona]

Well, what do you know about the probability P(q_r|\psi)? Can you express it another way?

You might need to use the method of inserting an integral over complete basis that vela showed you in post #16.