Position-Space Kinetic Energy Operator: Does Representation Matter?

[Niles]

Hi

This is actually a question regarding some formalism of QM, but I guess this is the place to ask it. Say we are looking at some kinetic energy operator T = T(r, ∇_r), which has the form

$$T = \sum\limits_{i,j} {T_{i,j} \left| \psi_i \right\rangle \left\langle \psi_j \right|}$$

in some arbitrary representation. The matrix elements T_{i, j} are given by

$$T_{i,j} = \int {dr\,\psi _i^* (r)} \,\,T(r,\nabla _r )\,\psi _j^{} (r)$$

My question is: The matrix element T_{i, j} as written above is found in position-space. Does it give the same value regardless of what representation we choose to find it in?


Niles.

 

[Landau]

What do you think?

 

[Niles]

I think it does depend on the basis: It is diagonal in the eigenbasis, but non-diagonal in a non-eigenbasis. Is my reasoning correct?

 

[Landau]

Yes; this is actually just linear algebra (but then in infinite dimensions, so functional analysis). The matrix of a linear operator depends heavily on the basis you choose (one should say "the matrix of T w.r.t. to the basis B").

And just as in good old linear algebra, the matrix of T w.r.t. the basis B is diagonal if and only if B consists entirely of eigenvectors of T. This follows directly from the definition.

 

[Niles]

Great, thanks. When I have an operator written in this form

$$T = \sum\limits_{i,j} {T_{i,j} \left| \psi_i \right\rangle \left\langle \psi_j \right|}$$

with

$$T_{i,j} = \int {dr\,\psi _i^* (r)} \,\,T(r,\nabla _r )\,\psi _j^{} (r)$$

then is it correct to say that the operator is written in position-space?

 

[element4]

Great, thanks. When I have an operator written in this form

$$T = \sum\limits_{i,j} {T_{i,j} \left| \psi_i \right\rangle \left\langle \psi_j \right|}$$

with

$$T_{i,j} = \int {dr\,\psi _i^* (r)} \,\,T(r,\nabla _r )\,\psi _j^{} (r)$$

then is it correct to say that the operator is written in position-space?

No, then it is written in $$\left\{\left| \psi_i \right\rangle\right\}$$ basis and $$T_{i,j}$$ is the matrix representation of $$T$$ in this basis (using linear algebra terminology). You need to express $$T$$ in terms of $$T(r,\nabla_r)$$ and $$\left| r \right\rangle$$, for $$T$$ to be written in "position-space".

 

[Niles]

No, then it is written in $$\left\{\left| \psi_i \right\rangle\right\}$$ basis and $$T_{i,j}$$ is the matrix representation of $$T$$ in this basis (using linear algebra terminology). You need to express $$T$$ in terms of $$T(r,\Delta_r)$$ and $$\left| r \right\rangle$$, for $$T$$ to be written in "position-space".

But wait a minute: We just agreed that the matrix elements are basis dependent, and we have found our elements in real space. Then how can we still be in $\left\{\left| \psi_i \right\rangle\right\}$?

Another thing: I have always interpreted $\left\{\left| r \right\rangle\right\}$ and $\left\{\left| \psi_i \right\rangle\right\}$ to be somewhat equivalent, i.e. representation-free states. So I am not sure what you mean when you say I have to express it in terms of $\left\{\left| r \right\rangle\right\}.$

 

[fzero]

But wait a minute: We just agreed that the matrix elements are basis dependent, and we have found our elements in real space. Then how can we still be in $\left\{\left| \psi_i \right\rangle\right\}$?

Another thing: I have always interpreted $\left\{\left| r \right\rangle\right\}$ and $\left\{\left| \psi_i \right\rangle\right\}$ to be somewhat equivalent, i.e. representation-free states. So I am not sure what you mean when you say I have to express it in terms of $\left\{\left| r \right\rangle\right\}.$

$$| r \rangle$$ and $$| \psi_i \rangle$$ are not really equivalent. $$| r \rangle$$ are the eigenstates of the position operator, while $$| \psi_i \rangle$$ are some other basis. By inserting the identity operator in the form

$$\sum_m | \chi_m \rangle \langle \chi_m | ~\text{or}~ \int dr | r \rangle \langle r | ,$$

into the matrix element, we can obtain formulas for the transformation of the matrix elements under a change of basis to some new states $$|\chi_m\rangle$$ or to the position basis.

Your $$T(r,\nabla_r)$$ is the expression for the operator in position space. It's analogous to saying that $$\hat{p} = -i\hbar \nabla_r$$ is the position space representation of the momentum operator. If you want to be perfectly accurate about what we mean when we write $$\hat{T}$$ in the position basis, we really mean

$$\hat{T} = \int dr | r\rangle T(r,\nabla_r) \langle r|. ~~~(*)$$

When you write

$$T_{i,j} = \langle \psi_i | \hat{T} | \psi_j \rangle = \int {dr\,\psi _i^* (r)} \,\,T(r,\nabla _r )\,\psi _j^{} (r),$$

You're actually using the formula (*) and noting that the expressions

$$\langle r | \psi_j \rangle = \psi_j (r)$$

have been identified as the wavefunctions.

Note that we can also consider the matrix elements of $$\hat{T}$$ in position space:

$$\langle r | \hat{T} | r' \rangle = T(r,\nabla _r ) \delta(r-r'),$$

which can be easily derived from (*).

 

[Niles]

1) So the matrix elements of the T-operator in e.g. the momentum basis are given by

$$T_{i,j} = \langle \psi_{k_1} | \hat{T} | \psi_{k_2} \rangle = \int {dr\,\psi_{k_1}^* (r)} \,\,T(r,\nabla _r )\,\psi _{k_1}^{} (r).$$

2) Ok, now let's say I want to write $\hat{T}$ in the momentum basis. What I do is

$$\hat T = \hat 1 \times \hat T \times \hat 1 = \sum\limits_{k_1 ,k_2 } {\left| {k_1 } \right\rangle \left\langle {k_1 } \right|\hat T\left| {k_2 } \right\rangle \left\langle {k_2 } \right|},$$

where we find the matrix elements as in #1.

3) If the above in #1 and #2 is correct, I have to ask: Why is it that we are integration over r when finding matrix elements? I assume it is because our operators are usually given in real space (e.g. $\hat{p} = -i\hbar \nabla_r$ as you said), but would we get the same if we had integrated over k?Niles.

 

[fzero]

1) So the matrix elements of the T-operator in e.g. the momentum basis are given by

$$T_{i,j} = \langle \psi_{k_1} | \hat{T} | \psi_{k_2} \rangle = \int {dr\,\psi_{k_1}^* (r)} \,\,T(r,\nabla _r )\,\psi _{k_1}^{} (r).$$

2) Ok, now let's say I want to write $\hat{T}$ in the momentum basis. What I do is

$$\hat T = \hat 1 \times \hat T \times \hat 1 = \sum\limits_{k_1 ,k_2 } {\left| {k_1 } \right\rangle \left\langle {k_1 } \right|\hat T\left| {k_2 } \right\rangle \left\langle {k_2 } \right|},$$

where we find the matrix elements as in #1.

Those are both correct.

3) If the above in #1 and #2 is correct, I have to ask: Why is it that we are integration over r when finding matrix elements? I assume it is because our operators are usually given in real space (e.g. $\hat{p} = -i\hbar \nabla_r$ as you said), but would we get the same if we had integrated over k?

Since x and p are conjugate variables, we can change to a momentum representation. We could have written

$$\hat{T} = \int dp |p\rangle T(-i\hbar \nabla_p, p) \langle p |,$$

or instead derived this expression from an explicit change of basis.

This is actually a simple expression as long as the kinetic energy has the standard form $$\hat{T} = \hat{p}^2/(2m)$$. However the momentum basis is not usually used because the potential energy becomes a very complicated differential operator.

 

[Niles]

Just to be absolutely positive: Would

$$T_{i,j} = \langle \psi_{i} | \hat{T} | \psi_{j} \rangle = \int {dr\,\psi_{i}^* (r)} \,\,T(r,\nabla _r )\,\psi _{j}^{} (r).$$

and $$T_{i,j} = \langle \psi_{i} | \hat{T} | \psi_{j} \rangle = \int {dp\,\psi_{i}^* (p)} \,\,T(-i\hbar \nabla_p,p)\,\psi _{j}^{} (p).$$

yield the same matrix element? (I personally think yes, since they are basically both found in the same basis, more specifically the $| \psi_i \rangle$ basis).

 

[fzero]

Yes, $$\psi_j(r)$$ and $$\psi_j(p)$$ are just Fourier transforms of one another:

$$\langle p | \psi_j \rangle = \int dx \langle p | x \rangle \langle x | \psi_i \rangle .$$

 

[Niles]

Thanks, it was very kind of you and everybody else to help.Niles.