Tensor Products of Operator

[cpsinkule]

In a multi-particle system, the total state is defined by the tensor product of the individual states. Why is it the case that operators, say position of 2 particles, is of the form X⊗I + I⊗Y and not X⊗Y where I are the identities for the respective spaces and X and Y are the position operators of the spaces. Why is it the sum of the two and not the tensor product of the two?

 

[atyy]

You can do the tensor prpduct of operators, eg. http://ocw.mit.edu/courses/physics/...all-2013/lecture-notes/MIT8_05F13_Chap_08.pdf (eq 1.8)

 

[cpsinkule]

You can do the tensor prpduct of operators, eg. http://ocw.mit.edu/courses/physics/...all-2013/lecture-notes/MIT8_05F13_Chap_08.pdf (eq 1.8)

I'm reading shankar at the moment. In the section devoted to identical particles, he claims that, for 2 identical particles, the state after measurement must be an eigenstate of
X⊗I+I⊗Y. I'm not understanding why he concluded that it must be an eigenstate of this. Instead of X*Y

 

[atyy]

I'm not sure, but just to put down a quick guess, that seems to be due to the symmetrization requirement for identical particles (wave function must be symmetric or antisymmetric).

 

[cpsinkule]

I'm not sure, but just to put down a quick guess, that seems to be due to the symmetrization requirement for identical particles (wave function must be symmetric or antisymmetric).

Is it not true that X⊗Y and X⊗I+I⊗Y would both yield symmetric measurements on a 2 particle system with identical particles?

 

[stevendaryl]

The operator $X \otimes I$ corresponds to performing a measurement of $X$ on the first particle, and doing nothing to the second particle, while $I \otimes Y$ corresponds to performing a measurement of $Y$ on the second particle, and doing nothing to the first particle. I'm not sure about position measurements, but for momentum measurements, it certainly makes sense to say that the total momentum $P_{total}$ is obtained by summing the momentum of the first particle, $P_1$ and the momentum of the second particle, $P_2$. So the operator corresponding to total momentum, $\hat{P}_{total}$ should be written as:

$\hat{P}_{total} = \hat{P}_1 + \hat{P}_2$

where
$\hat{P}_1 = \hat{P} \otimes I$
and
$\hat{P}_2 = I \otimes \hat{P}$.

 

[cpsinkule]

The operator $X \otimes I$ corresponds to performing a measurement of $X$ on the first particle, and doing nothing to the second particle, while $I \otimes Y$ corresponds to performing a measurement of $Y$ on the second particle, and doing nothing to the first particle. I'm not sure about position measurements, but for momentum measurements, it certainly makes sense to say that the total momentum $P_{total}$ is obtained by summing the momentum of the first particle, $P_1$ and the momentum of the second particle, $P_2$. So the operator corresponding to total momentum, $\hat{P}_{total}$ should be written as:

$\hat{P}_{total} = \hat{P}_1 + \hat{P}_2$

where
$\hat{P}_1 = \hat{P} \otimes I$
and
$\hat{P}_2 = I \otimes \hat{P}$.

Yea, it makes sense for anything but position. I guess it's strange to me that you add position eigenvalues. I guess I'll just accept it for now and move on and hope it doesn't creep up later in the book.

 

[Fredrik]

Right. The eigenvalue of $\hat P_1\otimes\hat P_2$ would be the product of the two momenta. The eigenvalue of the "total momentum" operator should be the sum of the eigenvalues, not their product.

Edit: This was supposed to be a reply to stevendaryl's post.

 

[stevendaryl]

Yea, it makes sense for anything but position. I guess it's strange to me that you add position eigenvalues. I guess I'll just accept it for now and move on and hope it doesn't creep up later in the book.

Yeah, summing positions doesn't make a lot of sense to me, unless you're trying to have a "center of mass" operator, rather than a position operator. If you have two particles, the first, with mass $m_1$ at position $\vec{r}_1$ and the second, with mass $m_2$, at position $\vec{r}_2$, then the center of mass will be given by:

$\vec{R}_{com} = \frac{1}{m_1 + m_2} (m_1 \vec{r}_1 + m_2 \vec{r}_2)$

For equal-mass particles,

$\vec{R}_{com} = \frac{1}{2} (\vec{r}_1 + \vec{r}_2)$

 

[atyy]

Is it not true that X⊗Y and X⊗I+I⊗Y would both yield symmetric measurements on a 2 particle system with identical particles?

Let Y = I.

Then X1⊗I2 means we measure the spin of particle 1. But since the particles are identical, we don't know which is particle 1.

If we know the operator for the single non-identical particle, we can get operators for identical particles by
http://phy.ntnu.edu.tw/~changmc/Teach/SM/ch01.pdf (Eq 70)
http://yclept.ucdavis.edu/course/242/2Q_Fradkin.pdf (Eq 1.155)

 

[atyy]

Is it not true that X⊗Y and X⊗I+I⊗Y would both yield symmetric measurements on a 2 particle system with identical particles?

I guess you are saying, let Y = X.

In this case, X⊗X and X⊗I+I⊗X should commute. (Sorry for my screwy notation.)

But for better notation (or at least one done by a professional) look at http://yclept.ucdavis.edu/course/242/2Q_Fradkin.pdf, which I linked above, and see the comment on two-body operators around Eq (1.167).

Perhaps Shankar's discussion is a little bit confusing at this point. He talks about measuring the positions of both particles, and maybe he should have talked about measuring the position of one particle. But anyway, his point is just that things should be properly symmetrized.