Okay, let's try to figure out this quantum mechanics problem. So, the question is about finding the probability that a measurement of an observable P yields 0 at time t. The state of the system is given by a column matrix with elements (-1, 2, 1). The observable P is a 3x3 matrix with specific elements. The choices are A through D, with D being 1/3, which seems plausible but I need to check.

First, I remember that in quantum mechanics, the probability of measuring a particular eigenvalue is the square of the absolute value of the inner product between the eigenstate corresponding to that eigenvalue and the given state vector. So, I need to find the eigenvectors of P corresponding to the eigenvalue 0 and then compute the probability.

Wait, but maybe there's a simpler way. The observable P is given as a matrix, and its eigenvalues and eigenvectors can be found. But since the question is about the probability of measuring 0, perhaps 0 is an eigenvalue of P. If 0 is an eigenvalue, the eigenvectors corresponding to 0 will form the subspace where the probability is 1 if the state is in that subspace. But since the state here is (-1, 2, 1), which is a specific vector, the probability will be the squared norm of the projection of this vector onto the eigenspace of P for eigenvalue 0.

Alternatively, maybe I can check if P is Hermitian. Because observables in quantum mechanics are represented by Hermitian matrices. Let me verify if P is Hermitian. The given matrix P has elements:

First row: 0, 1/√2, 0

Second row: 1/√2, 0, 1/√2

Third row: 0, 1/√2, 0

Since all the elements are real and the matrix is symmetric (P = P^T), it is indeed Hermitian. So P is a valid observable.

Now, to find the probability of measuring 0, I need to find the eigenvectors of P corresponding to eigenvalue 0 and then compute the squared magnitude of the projection of the state vector onto that eigenspace.

Alternatively, if 0 is an eigenvalue, the eigenvectors would satisfy P|v> = 0. Let's try to find the eigenvectors for eigenvalue 0.

So, let's set up the equation P|v> = 0. Let the vector |v> be (x, y, z)^T.

Multiplying P by |v>:

First component: 0*x + (1/√2)*y + 0*z = y/√2 = 0 → y = 0.

Second component: (1/√2)*x + 0*y + (1/√2)*z = (x + z)/√2 = 0 → x + z = 0.

Third component: 0*x + (1/√2)*y + 0*z = y/√2 = 0 → y = 0, which is consistent with the first equation.

So, from the equations, y must be 0, and x = -z. Therefore, the eigenvectors corresponding to eigenvalue 0 are all vectors of the form (x, 0, -x)^T. We can write this as x*(1, 0, -1)^T. So the eigenspace is one-dimensional, spanned by the vector (1, 0, -1)^T.

Now, the probability of measuring 0 is the squared magnitude of the projection of the state vector onto this eigenvector. The state vector is (-1, 2, 1)^T.

First, let's check if the state vector is orthogonal to the eigenvector (1, 0, -1). The inner product is (-1)(1) + 2*0 + 1*(-1) = -1 + 0 -1 = -2. Since it's not zero, the state vector is not in the eigenspace, so the projection will have some magnitude.

The projection of the state vector onto the eigenvector (1,0,-1) is given by:

proj = [ ( (-1)(1) + 2*0 + 1*(-1) ) / ||(1,0,-1)||^2 ] * (1,0,-1)

First, compute the inner product: (-1)(1) + 2*0 + 1*(-1) = -1 -1 = -2.

The norm squared of the eigenvector (1,0,-1) is 1^2 + 0^2 + (-1)^2 = 1 + 0 + 1 = 2.

So the scalar multiplier is (-2)/2 = -1.

Therefore, the projection vector is -1*(1,0,-1) = (-1, 0, 1)^T.

The squared magnitude of this projection is (-1)^2 + 0^2 + 1^2 = 1 + 0 + 1 = 2. Then, the probability is 2 divided by the squared norm of the original state vector.

Wait, hold on. The probability is the squared magnitude of the projection divided by the squared magnitude of the original state vector. Because the state vectors are not necessarily normalized. The state vector given is (-1, 2, 1), whose squared norm is (-1)^2 + 2^2 + 1^2 = 1 + 4 + 1 = 6. So the squared norm is 6. Then, the projection squared norm is 2. Therefore, the probability is 2/6 = 1/3.

Wait, but let me check again. The projection formula is:

proj = ( <v|e> / <e|e> ) * e, where |e> is the eigenvector.

Then, the squared magnitude of the projection is |<v|e>|^2 / <e|e>.

So here, <v|e> is the inner product of the state vector and the eigenvector. The eigenvector is (1,0,-1). The state vector is (-1,2,1). The inner product is (-1)(1) + 2*0 + 1*(-1) = -1 + 0 -1 = -2. The magnitude squared is (-2)^2 = 4. The denominator is <e|e> = 1 + 0 + 1 = 2. Therefore, the squared magnitude of the projection is 4/2 = 2. Then, the probability is (2) divided by the squared norm of the state vector. The squared norm of the state vector is 6, so 2/6 = 1/3. So the probability is 1/3.

Looking at the choices, D is 1/3. So the answer should be D.

But let me double-check if I made any mistake. Let's verify the eigenvectors again. The eigenvalue equation P|v> = 0 gives y = 0, x + z = 0. So eigenvectors are (x, 0, -x), which is correct. The eigenspace is one-dimensional, spanned by (1,0,-1). The projection of the state vector onto this eigenvector is indeed (-1, 0, 1). The squared norm of this projection is 1 + 0 + 1 = 2. The original state vector has squared norm 6. So 2/6 = 1/3. That seems right.

Alternatively, maybe there's a different approach. Another way to compute the probability is to compute the expectation of the projection operator onto the eigenspace. But that might be more complicated. Alternatively, since the state vector is (-1,2,1), and the eigenvectors are (1,0,-1), we can compute the probability as |(v ⋅ e)/||e||^2|^2 / ||v||^2. Wait, that's similar to the previous approach.

Alternatively, maybe the observable P has other eigenvalues as well. Let's check the eigenvalues of P. For a 3x3 matrix, that's a bit involved, but maybe we can find them.

The matrix P is:

[ 0   1/√2   0 ]

[1/√2 0    1/√2]

[ 0   1/√2   0 ]

To find eigenvalues, we solve det(P - λI) = 0.

The characteristic equation is:

| -λ     1/√2      0     |

|1/√2   -λ     1/√2   |

| 0     1/√2    -λ     |

Compute the determinant:

-λ * | -λ     1/√2   |

          |1/√2    -λ   |

Minus 1/√2 times the determinant of the minor:

|1/√2    1/√2   |

|0       -λ     |

Plus 0 times something (which is zero).

So expanding along the first row:

-λ * [ (-λ)(-λ) - (1/√2)(1/√2) ] - (1/√2) * [ (1/√2)(-λ) - (1/√2)(0) ] + 0.

Simplify:

-λ [ λ² - (1/2) ] - (1/√2) [ -λ/√2 - 0 ] 

= -λ³ + (λ)/2 + (1/√2)(λ/√2)

= -λ³ + (λ)/2 + (λ)/2

= -λ³ + λ

Set determinant to zero: -λ³ + λ = 0 → λ(-λ² + 1) = 0 → λ(λ² -1) = 0 → λ = 0, ±1.

So eigenvalues are 0, 1, -1. Therefore, the possible measurement outcomes are 0, 1, -1. But the question asks for the probability of measuring 0.

So we need to find the eigenvectors for eigenvalue 0, which we did earlier, and the probability is 1/3.

Therefore, the answer is D, 1/3.
Thus, the correct answer is $\boxed{D}$.
</think>