# Confusing dirac notation

## The Attempt at a Solution

Ok so basically a bit confused about notation..

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

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• dirac.pdf
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diazona
Homework Helper
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 ar |ur> ?
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.

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

Ok thanks well im 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 isnt the sum of probabilities of (qr|psi) = sum of all r of <qr|qr> = r? confused....!

diazona
Homework Helper
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.

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
Homework Helper
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 )

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?

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

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?

Last edited:
diazona
Homework Helper
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 $\langle\psi|$,
$$|\psi\rangle\langle\psi| = \int\psi^{*}(x)|x\rangle \mathrm{d}x \int\psi(x')\langle x'|\mathrm{d}x$$
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.

OK i see. What do i do then? Thanks

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

vela
Staff Emeritus
Homework Helper
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
Staff Emeritus
Homework Helper
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:

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

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
Homework Helper
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.