The authors say that in quantum theory the above expression is ambiguous, because X and P don't commute. Hence there are many nonequivalent quantum choices for H reduces to the same classical object. I am not able to figure this out.

If you review the text, you'll see the authors have provided the commutation relation in equation 10.70 as:

$$[X^i,P_j] = i\delta_j^i$$

which tells you X and P are non-commutative and the commutation operation produce an imaginary number (and this equation might be understood as $\dfrac{1}{X}P-P\dfrac{1}{X} = i$).

The equation you reference is actually written as:

$$H = \dfrac{1}{2}g^{ij}(X)P_iP_j$$

The position variable X is important since momentum is classically understood as $mass \times velocity$ so $p^2 = m^2v^2$and kinetic energy is $\dfrac{1}{2}mv^2$. So the $g^{ij}(X)$ is taking place of an inverse mass term in order to stay consistent with the classically defined energy equation.

The problem arises if someone is trying to solve for an unknown value in the equation. Imagine that you know H and X and want to solve for P, you might get some answer for P and assume everything is hunky dory, however, if for some reason you have the value of P you just calculated as well as the value of H, and then try to solve for X, you run into problems, because the X and P are non-commutative, you will not get the same value of X as you started out with, and as defined you have to include an imaginary component.

This is what is meant by ambiguity, knowing two components of the equation will not give you a definite value for the third, only a range of values.

As far as the relationship to the Laplacian, if one looks at the Schrodinger equation one can see that the kinetic energy operator term is:

$$-\dfrac{\hbar}{2m}\nabla^2$$

so if one compares this to the equation you reference, it should be clear that the momentum symbols (P) are taking the place of the Laplacian.

The authors say that in quantum theory the above expression is ambiguous, because X and P don't commute. Hence there are many nonequivalent quantum choices for H reduces to the same classical object. I am not able to figure this out.

If you review the text, you'll see the authors have provided the commutation relation in equation 10.70 as:

$$[X^i,P_j] = i\delta_j^i$$

which tells you X and P are non-commutative and the commutation operation produce an imaginary number (and this equation might be understood as $\dfrac{1}{X}P-P\dfrac{1}{X} = i$).

The equation you reference is actually written as:

$$H = \dfrac{1}{2}g^{ij}(X)P_iP_j$$

The position variable X is important since momentum is classically understood as $mass \times velocity$ so $p^2 = m^2v^2$and kinetic energy is $\dfrac{1}{2}mv^2$. So the $g^{ij}(X)$ is taking place of an inverse mass term in order to stay consistent with the classically defined energy equation.

The problem arises if someone is trying to solve for an unknown value in the equation. Imagine that you know H and X and want to solve for P, you might get some answer for P and assume everything is hunky dory, however, if for some reason you have the value of P you just calculated as well as the value of H, and then try to solve for X, you run into problems, because the X and P are non-commutative, you will not get the same value of X as you started out with, and as defined you have to include an imaginary component.

This is what is meant by ambiguity, knowing two components of the equation will not give you a definite value for the third, only a range of values.

As far as the relationship to the Laplacian, if one looks at the Schrodinger equation one can see that the kinetic energy operator term is:

$$-\dfrac{\hbar}{2m}\nabla^2$$

so if one compares this to the equation you reference, it should be clear that the momentum symbols (P) are taking the place of the Laplacian operator.

The authors say that in quantum theory the above expression is ambiguous, because X and P don't commute. Hence there are many nonequivalent quantum choices for H reduces to the same classical object. I am not able to figure this out.

If you review the text, you'll see the authors have provided the commutation relation in equation 10.70 as:

$$[X^i,P_j] = i\delta_j^i$$

which tells you X and P are non-commutative and the commutation operation produce an imaginary number (and this equation might be understood as $\dfrac{1}{X}P-P\dfrac{1}{X} = i$).

The equation you reference is actually written as:

$$H = \dfrac{1}{2}g^{ij}(X)P_iP_j$$

The position variable X is important since momentum is classically understood as $mass \times velocity$ so $p^2 = m^2v^2$and kinetic energy is $\dfrac{1}{2}mv^2$. So the $g^{ij}(X)$ is taking place of an inverse mass term in order to stay consistent with the classically defined energy equation.

The problem arises if someone is trying to solve for an unknown value in the equation. Imagine that you know H and X and want to solve for P, you might get some answer for P and assume everything is hunky dory, however, if for some reason you have the value of P you just calculated as well as the value of H, and then try to solve for X, you run into problems, because the X and P are non-commutative, you will not get the same value of X as you started out with, and as defined you have to include an imaginary component.

This is what is meant by ambiguity, knowing two components of the equation will not give you a definite value for the third, only a range of values.

The authors say that in quantum theory the above expression is ambiguous, because X and P don't commute. Hence there are many nonequivalent quantum choices for H reduces to the same classical object. I am not able to figure this out.

If you review the text, you'll see the authors have provided the commutation relation in equation 10.70 as:

$$[X^i,P_j] = i\delta_j^i$$

which tells you X and P are non-commutative and the commutation operation produce an imaginary number (and this equation might be understood as $\dfrac{1}{X}P-P\dfrac{1}{X} = i$).

The equation you reference is actually written as:

$$H = \dfrac{1}{2}g^{ij}(X)P_iP_j$$

The position variable X is important since momentum is classically understood as $mass \times velocity$ so $p^2 = m^2v^2$and kinetic energy is $\dfrac{1}{2}mv^2$. So the $g^{ij}(X)$ is taking place of an inverse mass term in order to stay consistent with the classically defined energy equation.

The problem arises if someone is trying to solve for an unknown value in the equation. Imagine that you know H and X and want to solve for P, you might get some answer for P and assume everything is hunky dory, however, if for some reason you have the value of P you just calculated as well as the value of H, and then try to solve for X, you run into problems, because the X and P are non-commutative, you will not get the same value of X as you started out with, and as defined you have to include an imaginary component.

This is what is meant by ambiguity, knowing two components of the equation will not give you a definite value for the third, only a range of values.

As far as the relationship to the Laplacian, if one looks at the Schrodinger equation one can see that the kinetic energy operator term is:

$$-\dfrac{\hbar}{2m}\nabla^2$$

so if one compares this to the equation you reference, it should be clear that the momentum symbols (P) are taking the place of the Laplacian.