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PH 401

PH 401. Dr. Cecilia Vogel approx of notes. Recall Hamiltonian. In 1-D H=p 2 /2m + V(x) but in 3-D p 2 =px 2 +py 2 +pz 2 . and V might be a ftn of x, y, z. 3-D TISE. 3-D TISE where Del-squared =. 3-D Free particle. For a free particle, V=0

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PH 401

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  1. PH 401 Dr. Cecilia Vogel approx of notes

  2. Recall Hamiltonian • In 1-D • H=p2/2m + V(x) • but in 3-D p2=px2+py2+pz2. • and V might be a ftn of x, y, z

  3. 3-D TISE • 3-D TISE • where Del-squared =

  4. 3-D Free particle • For a free particle, V=0 • There are many solutions with the same energy • we can distinguish them by their • eigenvalues of px, py, and pz • Can we find a complete set of stationary states that are also eigenstates of px, py, and pz?

  5. Commutators • Let me digress to define commutators • The commutator of operators A and B • is written [A,B] • and is defined by [A,B]=AB-BA • Often the order that you apply the operators matters • then AB is not equal to BA • but sometimes AB=BA, • then [A,B]=0 • and we say that A and B commute.

  6. Commutator of x and p • For example consider x and px • [x,px]=xpx-pxx • Let this act on an arbitrary state • (xpx-pxx)y = (xpxy-pxxy) • = • So.. [x,px]y=iy …for all y • thus [x,px]=i • they do not commute

  7. Commutator of x and p • generally • [position component,same component of momentum]=i • they do not commute • but [position component,same component of momentum]=0 • they do commute • and [position component,position component] =0 • they commute • and [momentum component,momentum component] =0 • they commute, too

  8. Playing with Commutators • [x,px]=i • so xpx =pxx+ i • Consider [x2,px]= xxpx –pxxx • Let’s rearrange the left to look like the right, so some stuff will cancel • we can replace xpx =pxx+ i • Consider [x2,px]= x(pxx+ i)–pxxx • = xpxx+ ix–pxxx • replace again • = (pxx+ i)x+ ix–pxxx = pxxx+ ix+ ix–pxxx • Finally canceling, [x2,px]= 2 ix

  9. PAL Monday week 7 Prove that [r2,px]=2ix

  10. Simultaneous Eigenstates • Can one have simultaneous eigenstates of A and B? • If [A,B]=0 • then there exists a complete set of simultaneous eigenstates of A and B • every state can be written as a linear combo of these simultaneous eigenstates • If [A,B] y =0 • then y could be a simultaneous eigenstate, even if there isn’t a complete set

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