Proving that an operator is not Hermitian

In summary, when defining the radial momentum operator in quantum mechanics, the classical analogue of \underline{x}.\underline{p}/r is not used. Instead, an "averaged" expression of 1/2(\underline{x}.\underline{p}/r+\underline{p}.\underline{x}/r) is chosen. This is because the former expression is not Hermitian. There is no general unique mathematical law to translate expressions from classical to quantum mechanics, so the choice of operator expression depends on the specific physical situation. In the position representation, the non-Hermitian nature of this operator can be shown by integrating it with arbitrary wave functions.
  • #1
McLaren Rulez
292
3
When defining the radial momentum operator, we don't use the classical analogue which would be [itex]\underline{x}[/itex].[itex]\underline{p}/r[/itex] where [itex]\underline{x}[/itex] and [itex]\underline{p}[/itex] are operators.

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

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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  • #2
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 [itex]\vec{x} \cdot \vec{p}[/itex], is not Hermitean since

[tex](\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}.[/tex]
 
  • #3
vanhees71 said:
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 [itex]\hat{q}[/itex] for q and [itex]\frac{\hbar}{i}\frac{\partial}{\partial{q}}[/itex] 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 [itex]\left[q,p_{q'}\right]=i\hbar\delta_{q,q'}[/itex] and [itex]\left[q,q'\right]=\left[p_q,p_{q'}\right]=0[/itex].
 
  • #4
SpectraCat said:
What about the postulate that says position representations of QM operators can be created from their CM analogues by substituting [itex]\hat{q}[/itex] for q and [itex]\frac{\hbar}{i}\frac{\partial}{\partial{q}}[/itex] 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
 
  • #5
vanhees71 said:
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 [itex]\vec{x} \cdot \vec{p}[/itex], is not Hermitean since

[tex](\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}.[/tex]

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.
 
  • #6
In the position representation you have

[tex]\hat{\vec{p}}=-\mathrm{i} \vec{\nabla}, \quad \hat{\vec{x}}=\vec{x}.[/tex]

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

[tex]\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][/tex]

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

FAQ: Proving that an operator is not Hermitian

1. How do you determine if an operator is Hermitian or not?

To determine if an operator is Hermitian, you need to check if it satisfies the Hermitian property, which states that the operator's adjoint is equal to its complex conjugate. This can be done by taking the adjoint of the operator and checking if it is equal to its complex conjugate.

2. What is the significance of an operator being Hermitian?

If an operator is Hermitian, it means that it has real eigenvalues and its eigenstates are orthogonal. This makes it easier to work with in quantum mechanics, as it simplifies calculations and allows for the use of important properties such as the spectral theorem.

3. Can an operator be both Hermitian and non-Hermitian?

No, an operator cannot be both Hermitian and non-Hermitian. It either satisfies the Hermitian property or it does not. However, it is possible for an operator to be self-adjoint, meaning its adjoint is equal to itself, without being Hermitian.

4. What are some common examples of non-Hermitian operators?

Some common examples of non-Hermitian operators include the momentum operator in quantum mechanics, the Laplace operator in mathematics, and the raising and lowering operators in quantum harmonic oscillators.

5. How can you prove that an operator is not Hermitian?

To prove that an operator is not Hermitian, you can take its adjoint and check if it is equal to its complex conjugate. If it is not, then the operator is not Hermitian. Additionally, you can also check if the operator satisfies other important properties such as being self-adjoint or having real eigenvalues, which would also indicate that it is not Hermitian.

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