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diazona

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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.2.4Let [itex]\psi(x)[/itex] be a properly normalised wavefunction and [itex]Q[/itex] an operator on wavefunctions. Let

[itex]\{q_r\}[/itex] be the spectrum of [itex]Q[/itex] and [itex]\{u_r(x)\}[/itex] be the corresponding correctly normalised eigenfunctions.

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

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

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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

- #4

diazona

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

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But for the second part isnt the sum of probabilities of (qr|psi) = sum of all r of <qr|qr> = r? confused....!

- #6

diazona

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Anyway, if you use [itex]|q_r\rangle[/itex] to denote the state corresponding to the value [itex]q_r[/itex] and function [itex]u_r(x)[/itex], and [itex]|\psi\rangle[/itex] is the state corresponding to the wavefunction [itex]\psi(x)[/itex], then yes, the probability that a measurement of [itex]Q[/itex] on a system in the state [itex]|\psi\rangle[/itex] will yield the value [itex]q_r[/itex] is indeed [itex]|\langle q_r|\psi\rangle|^2[/itex]. 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.

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

But |psi><psi| = 1 therefore

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

- #8

diazona

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You may be thinking of the completeness criterion, which says that

[tex]\sum_i |\phi_i\rangle\langle\phi_i| = 1[/tex]

if and only if the states [itex]\{|\phi_i\rangle\}[/itex] form a complete basis - in other words, if and only if

- #9

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|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?

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

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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?

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- #12

diazona

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Well, you're right that

|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?

[tex]|\psi\rangle = \int\psi(x)|x\rangle \mathrm{d}x[/tex]

but if you use that to compute [itex]|\psi\rangle\langle\psi|[/itex], you have to do one integral for [itex]|\psi\rangle[/itex] and another for [itex]\langle\psi|[/itex],

[tex]|\psi\rangle\langle\psi| = \int\psi^{*}(x)|x\rangle \mathrm{d}x \int\psi(x')\langle x'|\mathrm{d}x[/tex]

This does not come out to be equal to 1.

No, I don't think that's right.Oh i see where i went wrong. Am i right in thinking |psi><psi| = |x><x|? Why does this hold?

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OK i see. What do i do then? Thanks

- #14

hunt_mat

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[tex]

\int_{V}|\psi |^{2}d^{n}x=1

[/tex]

Where you integrate over all space, the genergic probability will mean an integration over a subset.

- #15

vela

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

[tex]|\langle q_r|\psi\rangle|^2 = \langle \psi | q_r \rangle\langle q_r | \psi \rangle[/tex]

And then use diazona's hint:

You may be thinking of the completeness criterion, which says that

[tex]\sum_i |\phi_i\rangle\langle\phi_i| = 1[/tex]

if and only if the states [itex]\{|\phi_i\rangle\}[/itex] form a complete basis - in other words, if and only ifanyarbitrary state can be expressed as a linear combination of the [itex]|\phi_i\rangle[/itex]'s. (That equation may be useful to you in solving this problem )

- #16

vela

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[itex]\langle \psi | \psi \rangle[/itex] and [itex]|\psi\rangle\langle\psi|[/itex] are two different mathematical objects.

|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?

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

[tex]

1 = \langle \psi | \psi \rangle = \langle \psi |\hat{1}| \psi \rangle

= \langle \psi |\left(\int dx\,|x\rangle\langle x|\right)| \psi \rangle

= \int dx\,\langle \psi |x\rangle\langle x| \psi \rangle

= \int dx\,\psi^*(x)\psi(x)

[/tex]

- #17

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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?

- #18

diazona

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

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