Spin-1 particle as a bound state of two particles - how to write state ket?
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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¶
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¶
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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⟩.
✅✅✅✅✅✅✅✅✅
HERE IS THE ANSWER ✌
_________________
⬇⬇⬇⬇⬇
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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