Question

Spin-1 particle as a bound state of two particles - how to write state ket?

Answer 1

There are a few ways you can formally approach this problem.

First note that the singlet state is an element of the Hilbert space of the tensor product of two spin-1/2 particles 11 and 22, i.e. |s⟩∈H=H1⊗H2|s⟩∈H=H1⊗H2.

To describe the system in the full space, you can form the density matrix ρ=|s⟩⟨s|∈D(H)ρ=|s⟩⟨s|∈D(H).

1st way: Since you care only about the first system, you can take the partial trace over system 22 yielding the reduced density matrix of system 11:

ρ1=tr2(ρ)∈D(H1).ρ1=tr2(ρ)∈D(H1).

This density matrix will give you the correct predictions if you choose to only do measurements on 11. In this case, you can show that

ρ1=12|↑⟩1⟨↑|1+12|↓⟩1⟨↓|1,ρ1=12|↑⟩1⟨↑|1+12|↓⟩1⟨↓|1,

a mixed state.

Now the operator Sz1Sz1 has the spectral decomposition

Sz1=+ℏ2π↑1+(−ℏ2)π↓1,Sz1=+ℏ2π↑1+(−ℏ2)π↓1,

where π↑1π↑1 is the projector onto the spin-z-up axis of the 1st particle: π↑1=|↑⟩1⟨↑|1π↑1=|↑⟩1⟨↑|1 and similarly for the other projector.


According to the Born rule, the probability of getting ℏ/2ℏ/2 is

P(sz1=ℏ/2)=tr1(ρ1π↑1)=∑k⟨k|1[12|↑⟩1⟨↑|1+12|↓⟩1⟨↓|1]|↑⟩1⟨↑|1|k⟩1=12∑k⟨k|1|↑⟩1⟨↑|1|k⟩1=12⟨↑|1∑k|k⟩1⟨k|1|↑⟩1=12↑|1|↑⟩1=12,P(sz1=ℏ/2)=tr1(ρ1π↑1)=∑k⟨k|1[12|↑⟩1⟨↑|1+12|↓⟩1⟨↓|1]|↑⟩1⟨↑|1|k⟩1=12∑k⟨k|1|↑⟩1⟨↑|1|k⟩1=12⟨↑|1∑k|k⟩1⟨k|1|↑⟩1=12↑|1|↑⟩1=12,

and similarly for P(sz1=−ℏ/2)P(sz1=−ℏ/2). This is called a projective valued measurement.

2nd way: You can think of a measurement on the full system, except that you do nothing on system 2. So the relevant operator is the tensor product of the Sz1Sz1operator with the identity operator on 2: Sz1⊗I2Sz1⊗I2, which I'll call the FULL spin operator.

The spectral decomposition is

Sz1⊗I2=+ℏ2π↑1⊗I2+(−ℏ2)π↓1⊗I2,Sz1⊗I2=+ℏ2π↑1⊗I2+(−ℏ2)π↓1⊗I2,

i.e. it has only 2 eigenvalues, but now each eigenspace is 2-dimensional.

So according to the Born rule, the probability of measuring ℏ/2ℏ/2 for the FULL spin operator is given by

P(sfullz1=ℏ/2)=tr(ρ(π↑1⊗I2))P(sz1full=ℏ/2)=tr(ρ(π↑1⊗I2))

which you can work out to be 1/21/2 again.

Note: this is the formal way of showing these calculations, following the axioms of quantum mechanics.

Expressions like |⟨↑|1Sz1|s⟩|2|⟨↑|1Sz1|s⟩|2, should be regarded as heuristic expressions that 'give' the right answer, but they are not mathematically correct because |↑⟩1|↑⟩1 is an element of H1H1, Sz1Sz1 an element of L(H1)L(H1), but |s⟩|s⟩ an element of HH, so it just doesn't quite make sense.



hope it helps you

please mark as brainiest answer

¶potter¶

Answer 2

HEY MATE ⭐⭐⭐
HERE IS THE ANSWER ✌
_________________

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Since the two particles can combine in S=0,1S=0,1 states, orbital angular momentum can take the values
S=0⟶ℓ=1,
S=0⟶ℓ=1,
S=1⟶ℓ=0,1,2.
S=1⟶ℓ=0,1,2.
If the two particles are identical, the state ket must be antisymmetric, so it's only possible to have
S=1,ℓ=1.
S=1,ℓ=1.
But I can't understand how to write |α⟩|α⟩.

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