This holds only if ^R * vR - ^L * vL = 0 (or ^R * vR = 0, ^L * vL = 0)

- why should it be true?

02 Oct 2009, 14:06

Smith

Joined: 23 Aug 2010, 13:12 Posts: 33

Re: Exercise 23.4

Penrose says he is using the position, left ket or right ket, to distinguish the two particles -- not the "L" or "R" label, which can apparently be thought of as distinguishing different spatial versions of the wavefunction. If we remove the L and R, so the spatial parts are identical, we *might* imagine canceling the RR versus LL terms. However, the derivation is still not valid because Laura equates ^RvL with vL^R. Changing the order of the terms is not allowed -- it amounts to interchanging the particles' wavefunctions. Frankly, I dont see how Penrose's claim that the singlet state can be expressed as a product can possibly be true. Obviously, I dont get what he is trying to convey.

26 Sep 2010, 12:45

fallingup

Joined: 13 Mar 2008, 14:06 Posts: 42 Location: Ithaca NY

Re: Exercise 23.4

What I was talking about in that solution is that the RH particle - that's R - is not both spin-up and spin-down at the same time. SImilarly the LH particle.

In other words, ^R> vR> = <^R|vR> = 0

The multiplication here is the inner product in a complex vector space. The inner product of ^R> and vR> is the complex conjugate of ^R> dotted with vR>, which is the same as <^R|vR>, or the amplitude of ^R>, given vR>.

For example, if the complex vector space has only 1 dimension, so it's only the complex numbers, the inner product of x and y is xbar times y. If it had 2 dimensions, the inner product (x1,x2)(y1,y2) = x1bar times y1 + x2bar times y2.

THe complex vector space inner product isn't commutative, that's right. inner product of x and y is the complex conjugate of the inner product of y and x.