Continue to Site

Welcome to EDAboard.com

Welcome to our site! EDAboard.com is an international Electronics Discussion Forum focused on EDA software, circuits, schematics, books, theory, papers, asic, pld, 8051, DSP, Network, RF, Analog Design, PCB, Service Manuals... and a whole lot more! To participate you need to register. Registration is free. Click here to register now.

help me understand a 2-degree freedom quantum system's shrodinger equation

Status
Not open for further replies.

Fractional-N

Full Member level 1
Full Member level 1
Joined
Oct 15, 2007
Messages
97
Helped
2
Reputation
4
Reaction score
2
Trophy points
1,288
Activity points
2,071
Hi,
I'm reading Goswami's book, "quantum mechanics",
I can't understand chapter 9 , "systems of two degrees of freedom"
it is said in page 122 that <x|p> is transformation coefficient from momentum basis to position basis
BUT
in chapter 9 , as it is shown in the pic below, it is said that <x|p> is equal to psi(x) !!! why?
I can't understand why equation (9.1) is correct ?

 
Last edited:

I'll prefix the following words by admitting upfront I'm a little (lot!) rusty on Dirac notation, but let me add my interpretation... :)

<a|b> is the :

[some alternative "phrasings"]
1. 'projection' of |b> onto |a>,
2. the inner product of states |a> & |b>
3. the probability of |b> being in state |a>

In this sense, both the definitions you mention are correct. The inner product is akin to the dot product - an operation that returns a complex coefficient indicating the "degree of overlap" between two vectors. If these vectors represent two states [can this be generalised to different basis? Apparently..(?)], then the inner product could be interpreted as a "transformation coefficient".

The definition in the above picture is also OK. \psi is the wavefunction, indicating the complex probability amplitude of the particle at position (x,y). The probability amplitude is the "practical interpretation" of the quantity returned via the inner product (which can be converted into a physically observable probability density by taking |\psi|^2), as seen by 'alternative phrasing #3' above.

Hope that helps!
 

thanks
The probability amplitude is the "practical interpretation" of the quantity returned via the inner product


I thinks it's not (exactly) correct. the inner product <x|p> is the probability amplitude of being in the base state |p> and ending up in the base state |x>. OK. but for a general wave-function \[\psi\](x), which is a linear combination of the base states ( in position space the base states are |x>, and in momentum space the base states are |p>, (continues of course) ) , we have to integrate over all base states; so we have the formula
<x|\[\psi\]> = \[\int\]<x|p> <p|\[\psi\]> dp



so I think the inner product <x|p> is just the probability amplitude of being in one base state and ending up in another base state ; for \[\psi\] we have to integrate over all base states, so the inner product could not be equal to \[\psi\]. (unfortunately the book says it can! )
am I wrong ?!
 

Status
Not open for further replies.

Similar threads

Part and Inventory Search

Welcome to EDABoard.com

Sponsor

Back
Top