In Shankar's QM book Chapter 10 pg. 248, it was said that there are other bases possible besides $|x_1x_2\rangle$ for the two particle hilbert space. For example, the momentum basis, consisting of simultaneous eigenkets $|p_1p_2\rangle$ of $P_1$ and $P_2$.

More generally, we can use the simultaneous eigenkets $|w_1w_2\rangle$ of two commuting operators $\Omega_1(X_1,P_1)$ and $\Omega(X_2, P_2)$ to define the $\Omega$ basis, since any function of $X_1$ and $P_1$ commutes with any function of $X_2$ and $P_2$.

Did he miss out saying that the operators $\Omega_1(X_1,P_1)$ and $\Omega_2(X_2,P_2)$ need to be hermitian? Since two commuting Hermitian operators share a same eigenbasis.

Also, can the $\Omega_1$ and $\Omega_2$ operators be for example $X_1$ and $P_2$? Does it need to be the 'same' operator but acting on different particles, e.g. $X_1$ and $X_2$, $P_1$ and $P_2$?


1 Answer 1


A two-particle Hilbert space is just a tensor product $H\otimes H$ for the one-particle Hilbert space $H$.

Essentially by definition of the tensor product, given any two bases $\{b_i\}_{i\in I}$ and $\{c_i\}_{i\in I}$ of $H$ (with $I$ an index set whose cardinality is the dimensionality of $H$), the simple tensors $\{b_i\otimes c_j\}_{i,j\in I}$ form a basis of $H\otimes H$. So in particular you get a basis of $H\otimes H$ by picking two arbitrary self-adjoint operators on $H$ and choosing their eigenbases for $b_i$ and $c_i$.

There is nothing special about the specific situation of $H = L^2(\mathbb{R})$ in quantum mechanics. You can choose any two bases you want, in particular the eigenbases of two different self-adjoint operators.

The remaining problem in your question is what $\Omega_i(X,P)$ is even supposed to mean. You can define a function $f(O)$ of a single self-adjoint operator $O$ via functional calculus without problems, but what is a function of two non-commuting observables like $\Omega_i(X,P)$ supposed to mean? Without further elaboration this is simply bad notation.

One possible interpretation is that the $\Omega_i$ are supposed to be polynomials, and that the ordering ambiguities in applying a polynomial to non-commuting operators are resolved always in the fashion that yields a self-adjoint operator (e.g. the function $f(x,p) = xp$ is turned into the self-adjoint operator $\frac{xp+px}{2}$). In that case it would be superfluous to "restrict" such $\Omega_i(X,P)$ to self-adjoint possibilities, since they would always be self-adjoint.


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