I'm a physics student,my next exam will be quantum mecchanics,

>>8268195
Suck your professor cock and do like ancient Greek

QM is so easy that even chemists do it, come on

>>8268195
https://www.youtube.com/playlist?list=PLPH7f_7ZlzxQVx5jRjbfRGEzWY_upS5K6

>>8268195
>>8268195
>Hey guys teach me QM plox.
Sure thing anon, no problem.

First we have a "ket", that represents an "abstract vector" which we often call a state vector.
[eqn] | \phi \rangle = \int ^{b} _{a} \phi \left ( \vec { r } \right ) d \vec { r } [/eqn]
After that we have the "bra" which is dual to the ket
[eqn] \langle \phi | = \int ^{b} _{a} \phi ^{*} \left ( \vec { r } \right ) d \vec { r } [/eqn]
Using both of these we can form the inner product [eqn] \langle \phi | \phi \rangle = \int ^{b} _{a} \phi ^{*} \left ( \vec { r } \right ) \phi \left ( \vec { r } \right ) d \vec { r } [/eqn] Obviously we enforce the normalisation condition to make the inner product over all space equal to 1.

In QM we are often most interested in the expectation values of observables, [math]\left \langle \hat { Q } \right \rangle [/math], here the name observable could easily be replaced (and imo should be replaced) with the word "measurable", although I'll continue to call it an observable because that's standard nomenclature. Now clearly we want the expectation value of the observable to be real, so: [math]\left \langle \hat { Q } \right \rangle = \left \langle \hat { Q } \right \rangle ^{*} [/math] so an observable is equal to it's complex conjugate, but complex conjugation reverses the order of an inner product, so [math] \langle \phi | \hat { Q } \phi \rangle = \langle \hat { Q } \phi | \phi \rangle [/math] This property is called [math] Hermiticity [/math], so we say that:
>Observables are represented by Hermitian operators.

We will often write the expectation value of an operator as [eqn] \langle \phi | \hat { Q } | \phi \rangle [/eqn] Notice that the operator acts on the state to it's right.

In the next post we'll look at an operator that will take centre stage in QM, the Hamiltonian.

>>8268260

In classical mechanics we define the Hamiltonian as [eqn] H = \frac { p^2 } { 2m } + V(x) [/eqn] We use the canonical quantisation procedure (ie. setting [math] p \to -i/ \hbar \partial _{x} ^2 [/math]) which gives us the Hamiltonian operator, [math] \hat { H } [/math], defined as [eqn] \hat { H } = - \frac { \hbar ^2 } { 2m } \frac { \partial ^2 } { \partial x^2 } + V(x) [/eqn], there's one last bit of machinery that we need, the Commutator of two operators [math] [ A, B ] = AB - BA [/math]. Now we have some ground work done, we can look at some quantum mechanical systems, the one we're going to focus one is the harmonic oscillator, here the potential takes the form [math] V(x) = \frac { 1 } { 2 } m \omega ^2 x^2 [/math] so the Hamiltonian becomes [eqn] \hat { H } = \frac { 1 } { 2m } \left [ p^2 + (m \omega x ) ^2 \right ] [/eqn] We now need to solve the time independent Schroedinger equation, to see this done in full look at your QM text, I'm just going to take the ladder operators and use them. Define [eqn] a _{ \pm } = \frac { 1 } { \sqrt { 2\hbar m \omega } } \left ( \mp i hat { p } + m \omega \hat { x } \right ) [/eqn] When the raising operator [math] \hat { a } _{ + } [/math] acts on a state [math] | \phi \rangle [/math] we get the next highest state, ie [math] \hat { a } _{ + } | \phi _{ n } \rangle = \sqrt { n +1 } | \phi _{n+1} \rangle [/math], likewise with the lowering operator [math] \hat { a } _{-} | \phi _n \rangle = \sqrt { n } | \phi _{n-1} \rangle [/math]. So lets use some of what we've learnt, lets calculate the expectation value of the potential energy in the harmonic oscillator, ie [eqn] \langle \phi | V | \phi \rangle = \langle \phi | \frac { 1 } { 2 } m \omega ^2 \hat {x} ^2 | \phi \rangle [/eqn] Clearly here the only observable is position, so everything else factors out, leaving [eqn] \frac { 1 } { 2 } m \omega ^2 \langle \phi | \hat { x } ^2 | \phi \rangle [/eqn] Cont....

>>8268326
Small errata in for that previous post, the ladder operators should be defined as: [eqn] a _{ \pm } = \frac { 1 } { \sqrt { 2\hbar m \omega } } \left ( \mp i \hat { p } + m \omega \hat { x } \right ) [/eqn]
But looking back at our definition of the ladder operators we see that we can express [math] \hat { x } [/math] as [eqn] \hat { x } = \sqrt { \frac { \hbar } { 2 m \omega } } (a_+ + a_-) [/eqn] Which obviously implies that [eqn] \hat { x } ^2 = \frac { \hbar } { 2 m \omega } \left ( a_{+} ^2 + a_{+} a_{-} + a_{-}a_{+} + a_{-} ^2 \right ) [/eqn] So substituting that back into the earlier expectation value gives us [eqn] \frac { \hbar \omega } { 4 } \left ( \langle \phi _n | a_{+} ^2 | \phi _n \rangle + \langle \phi _n | a_{+} a_{-} | \phi _n \rangle + \langle \phi _n | a_{-}a_{+} | \phi _n \rangle + \langle \phi _n | a_{-} ^2 | \phi _n \rangle \right )[/eqn] Notice that the first and last operators give [math] \langle \phi _n | \phi _{n+2} \rangle [/math] and [math] \langle \phi _n | \phi _{n-2} \rangle [/math] respectively, but since wavefunctions are orthonormal, these terms must be 0, so that leaves us with [eqn] \frac { \hbar \omega } { 4 } \left ( \langle \phi _n | a_{+} a_{-} | \phi _n \rangle + \langle \phi _n | a_{-}a_{+} | \phi _n \rangle \right ) [/eqn] but these terms are trivial to evaluate, and give us [eqn] \langle V \rangle = \frac { \hbar \omega } { 4 } ( 2 n + 1 ) [/eqn] There's more to operators and expectation values, but you should now have all the necessary information to do some reading of your own. cont....

>>8268260
your definitions of ket and bra are wrong. why are you integrating?

>>8268366

Not OP but screenshotting this just in case it's ever relevant.

T Physical Chemistry student

>>8268374
I believe he is defining state vectors using their continuous probability densities.

>>8268260
Where the fuck did you get those integrals for bras and kets

>>8268462
That would just be <x|phi> where x is a position eigenstate.

Those integrals are not the same as the state |phi>.

>>8268981
To clear this up a bit, a quantum state |phi> doesn't represent a function that you can plot, it just represents a state of being that some quantum object is in. An example is |x> which is the state where the object is at position x or |p> which has momentum p.

The state |phi> might not have a well defined position or momentum but you can overlap it with a state with a well defined position <x|phi> find the probability density of finding it at x. The value of <x|phi> over all x is the spatial wavefunction phi(x).

A quantum state which you can't plot over space at all is a spin state, your options are |up> and |down>. This most naturally maps to a two component vector where up is (1,0) and down is (0,1). A superposition of these states will take the form of a complex 2-vector with length 1. All discrete quantum states can be represented like this with a high enough dimensional vector.

>>8268981
>>8269021
So basically what >>8268260 defined as |phi> was actually the integral of <x|phi> between positions a and b.

Their definition of <phi|phi> is also wrong, it should be over all space.

>>8269030
>defined as |phi> was actually the integral of <x|phi> between positions a and b.
Not really, there's a certain license to treat |phi> as its own object.

>Their definition of <phi|phi> is also wrong, it should be over all space.

Its really only over the space you're working in, so for hydrogen its from 0 to infinity, with a square well its from 0 to a (or sometimes -a to a ) for the harmonic oscillator its from -infinity to infinity.

>>8269060
|phi> is it's own object yeah, <x|phi> is a function that outputs a value for some x. The same function in the integral in that first post.

>Its really only over the space you're working in
Sure but just calling it a,b without specifying this is a bit misleading.

>>8269078
>Sure but just calling it a,b without specifying this is a bit misleading.

How is it? Surely including general limits in an integral is the least misleading (hell even the standard) way to present the boundary values determined by the physics of the system in question.

>>8269092
Just calling the integral limits a and b impies that they're arbitrary which could be confusing for a beginner. Setting it over all space is correct for all spatial wavefunctions. Yeah you can reduce it for some problems but that's only because the wf is zero in certain regions and a student should be able to figure that out fine.

Still my main problem was with the definitions of |phi> and <phi|, which were just wrong.

>>8268230
>QM is so easy that even physics do it
ftfy math wanabe