Proving that an operator is not Hermitian

McLaren Rulez

When defining the radial momentum operator, we don't use the classical analogue which would be $\underline{x}$.$\underline{p}/r$ where $\underline{x}$ and $\underline{p}$ are operators.

Instead we choose $1/2(\underline{x}$.$\underline{p}/r+\underline{p}$.$\underline{x}/r)$.

If it is because the former expression is not Hermitian (which is what I think), can someone help me prove it? If it is something else, then may I know why we choose this strange "averaged" expression?

Thank you.

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vanhees71

Gold Member
It depends on what you want to do with this special "radial momentum operator". There's no general unique mathematical law to translate expressions from classical to quantum mechanics. Thus, you need to specify the physical situation, in which you need an operator expression.

For sure $\vec{x} \cdot \vec{p}$, is not Hermitean since

$$(\vec{x} \cdot \vec{p})^{\dagger}=\vec{p} \cdot \vec{x}=\delta_{jk} p_j x_k =\delta_{jk} ([p_j,x_k]+x_k p_j)=\delta_{jk} (-\mathrm{i}) \delta_{jk} +\vec{x} \cdot \vec{p}=-3 \mathrm{i}+\vec{x} \cdot \vec{p} \neq \vec{x} \cdot \vec{p}.$$

SpectraCat

There's no general unique mathematical law to translate expressions from classical to quantum mechanics.
What about the postulate that says position representations of QM operators can be created from their CM analogues by substituting $\hat{q}$ for q and $\frac{\hbar}{i}\frac{\partial}{\partial{q}}$ for pq?

Perhaps that is not a "unique mathematical law", but isn't it universally valid for non-relativistic QM?

Note: this is a specialization of the more general postulate that QM operators must satisfy the commutation relations $\left[q,p_{q'}\right]=i\hbar\delta_{q,q'}$ and $\left[q,q'\right]=\left[p_q,p_{q'}\right]=0$.

strangerep

What about the postulate that says position representations of QM operators can be created from their CM analogues by substituting $\hat{q}$ for q and $\frac{\hbar}{i}\frac{\partial}{\partial{q}}$ for pq?

Perhaps that is not a "unique mathematical law", but isn't it universally valid for non-relativistic QM?
There's a no-go theorem of Groenwold & van Hove which says that this can't be done
uniquely for arbitrary operators (i.e., involving arbitrary powers of q , p).
Symmetrization wrt q,p is one the techniques commonly used to bypass this
inconvenience.

For more on Groenwold & van Hove and obstructions to quantization,
try these papers by Gotay:

math-ph/9809011 and math-ph/9809015

McLaren Rulez

It depends on what you want to do with this special "radial momentum operator". There's no general unique mathematical law to translate expressions from classical to quantum mechanics. Thus, you need to specify the physical situation, in which you need an operator expression.

For sure $\vec{x} \cdot \vec{p}$, is not Hermitean since

$$(\vec{x} \cdot \vec{p})^{\dagger}=\vec{p} \cdot \vec{x}=\delta_{jk} p_j x_k =\delta_{jk} ([p_j,x_k]+x_k p_j)=\delta_{jk} (-\mathrm{i}) \delta_{jk} +\vec{x} \cdot \vec{p}=-3 \mathrm{i}+\vec{x} \cdot \vec{p} \neq \vec{x} \cdot \vec{p}.$$
Thank you for the explanation, I can see that its very obviously not Hermitian. Would it be possible to show this non Hermitian-ness in the x basis as an integral? I'm having a fair bit of trouble when it comes to proving whether an operator is Hermitian. Thank you once again.

vanhees71

Gold Member
In the position representation you have

$$\hat{\vec{p}}=-\mathrm{i} \vec{\nabla}, \quad \hat{\vec{x}}=\vec{x}.$$

Now you take to arbitrary wave functions $\psi_{j} \in \mathrm{L}^2(\mathbb{R}^3), \quad j \in \{1,2\}$ and reformulate

$$\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} \; \psi_1^{*}(\vec{x}) \left [\hat{\vec{x}} \cdot \hat{\vec{p}} \; \psi_2(\vec{x}) \right]$$

such that the operator acts on $\psi_1^{*}$ (integration by parts!). Then you'll also see that there's the extra term which violates hermitizity.

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