Is the Hamiltonian \( H = p^2 - x^4 \) Hermitian?

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In summary: Hallo everyone,I have a question, how can I see that the hamiltonian H=p^2-x^4 is not hermitian, with p the momentum operator and x the position operator.Calculate its Hermitian conjugate: rewrite \int \Psi^* (\hat H \Psi) \, dx = \int (H^\dagger \Psi^*) \Psi \, dxand show that H^\dagger \neq H.Thanks for the reply.The
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Hallo everyone, I have a question, how can I see that the hamiltonian H=p^2-x^4 is not hermitian, with p the momentum operator and x the position operator.
 
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  • #2
Calculate its Hermitian conjugate: rewrite
[tex]\int \Psi^* (\hat H \Psi) \, dx = \int (H^\dagger \Psi^*) \Psi \, dx[/tex]
and show that [itex]H^\dagger \neq H[/itex].
 
  • #3
Thanks for the reply. the thing is I don't see why -x^4 is not hermitian. Since we know that p^2 is hermitian (from free particle problem), all we need to do is to investigate -x^4, which has to be hermitian from the integration you have shown. Maybe I miss anything?
 
  • #4
Errr, right, I missed that.
p^2 is hermitian, and so is x. Therefore, x^4 is hermitian, and then the same holds for p^2 - x ^4.
In fact, if H is really a Hamiltonian of a system, it should be hermitian.

So maybe you are missing something, but then so am I :confused:
 
  • #5
Is H defined on the -infinity < x < +infinity?
 
  • #6
logscale said:
... I don't see why -x^4 is not hermitian. Since we know that p^2 is hermitian (from free particle problem), all we need to do is to investigate -x^4, which has to be hermitian from the integration you have shown. Maybe I miss anything?
Hermiticity depends on the space that the operator is acting on. You may have already seen this when (if) you dealt with angular momentum. Neither x nor p need be Hermitian, in general. What is the allowed range of x?
 
  • #7
Hey guys I think I know the solution. All Hamiltonian are subjected to boundary condition. If we allow x to be real, then the eigenfunction to this Hamiltonian will not converge when x goes to infinity (therefore we will have problem to show the Hermiticity of -x^4 in momentum representation, there partial integration will just yield infinity.) So if we extend x into complex plane (PT symmetry), it is clear that -x^4 is not Hermitian, because x is complex. See arXiv:physics/9712001, Bender.

I hope my argument is correct. Thanks
 
  • #8
I will suppress inessential constants, and treat everything as dimensionless, because I think that this is a mathematical issue rather than a physical issue.

If H = -∂x2 - x4 on a < x < b,
then I determine that <ψi|H|ψf>* = <ψf|H|ψi> + jif(b) - jif(a),
where jif = ψf*(∂xψi) - (∂xψf)*ψi.

So, if I simply require jif(a) = jif(b) for all i,f in my Hilbert space, then doesn't that make H Hermitian?

logscale said:
All Hamiltonian are subjected to boundary condition.
Well, I would say that the functions that compose the Hilbert space on which the Hamiltonian is defined are subject to boundary conditions.

logscale said:
... when x goes to infinity (... we will have problem to show the Hermiticity of -x^4 in momentum representation, there partial integration will just yield infinity.)
If x→∞, then of course x4→∞. There is nothing wrong with this, and, in particular, this does not preclude the Hermiticity of x4 nor H. On the contrary, there would be something severely wrong with a finite x4 as x→∞.

logscale said:
So if we extend x into complex plane (PT symmetry), it is clear that -x^4 is not Hermitian, because x is complex.
Extending x into the complex plane is not a PT operation. Usually, PxP = -x, and TxT = x*. I only put the "*" on the time-reversed x because you are suggesting an analytical continuation of the eigenfunctions of x into the complex plane. Normally, TxT = x. In particular, if x is already real-valued, then PTxPT = -x, which is also real-valued.

logscale said:
See arXiv:physics/9712001, Bender.
I started reading that article, but it may be too far over my head. I don't think I agree with it, though. For example, PTpPT = p, but p has negative eigenvalues, so I must be missing part of their argument.
 
  • #9
turin said:
...So, if I simply require jif(a) = jif(b) for all i,f in my Hilbert space, then doesn't that make H Hermitian?

The thing is we are still treating x as real, which if we use WKB to approximate the asymtotic eigenstate, you will notice that it just blows to infinity when x goes to infinity.That is what I am trying to say, if the Hamiltonian produces non L2 integrable function, then it is a not a well defined operator, hence the problem of hermiticity does not even have a place to start with.

turin said:
Extending x into the complex...particular, if x is already real-valued, then PTxPT = -x, which is also real-valued.

Your statement makes me ponder on another question: does one replaces the operator as real valued variable when going from quantum mechanics to classical mechanics? Or is complex variable also allowed? The problem is if we treat x and p as complex, then H=p^2-x^4 is not PT symmetry, as [H,PT][tex]\neq[/tex]0
 
  • #10
logscale said:
... if we use WKB to approximate the asymtotic eigenstate, you will notice that it just blows to infinity when x goes to infinity.
By "it" you apparently mean the "wavenumber", that is, how quickly the function oscillates in x. I thought that you were talking about x4 . Anyway, the WKB method assumes the existence of the eigenfunction of H. So, if the WKB method does not give you a valid eigenfunction, then all that means is that you cannot use the WKB method to determine a valid eigenfunction. I don't think that you can use this argument to demonstrate the nonexistence of valid eigenfunctions.

logscale said:
... if the Hamiltonian produces non L2 integrable function, then it is a not a well defined operator, hence the problem of hermiticity does not even have a place to start with.
I don't know what you mean by "the Hamiltonian produces [a] ... function". The QM Hamiltonian acts on a wavefunction in order to generate time translation of that wavefunction, but QM posits the existence of the wavefunction independently of the Hamiltonian (i.e. you are free to specify an initial state). I will assume that you meant that the wavefunction on which the Hamiltonian acts must be L2(ℝn) (where I've been assuming that n = 1 for this problem). In that case, either you have been VERY frustrated with QM, or you have just recently decided to impose this condition on the "Hilbert" space of QM. See if you can figure out why I think this.
Hints:
- Why did I put quotation marks around "Hilbert"?
- What is momentum space?
- What is the norm of a position eigenstate?

BTW, integrability is typically unrelated to the divergence of the wavenumber. For example, exp(i/x) is L2(-R,R), where R is any positive real number, even though it contains the point of explosion.

logscale said:
... does one replaces the operator as real valued variable when going from quantum mechanics to classical mechanics? Or is complex variable also allowed?
CM variables may be complexified (e.g. ict in early formulations of relativity). You just have to decide how to interpret results in terms of observables. The correspondence between CM and QM does involve a complex phase, one way or another.

logscale said:
The problem is if we treat x and p as complex, then H=p^2-x^4 is not PT symmetry, as [H,PT][tex]\neq[/tex]0
Basically, I don't think we should be talking about "complex-valued" x and p.
 

FAQ: Is the Hamiltonian \( H = p^2 - x^4 \) Hermitian?

What does it mean for a Hamiltonian to be non-hermitian?

A Hamiltonian is considered non-hermitian if it does not obey the mathematical property of hermiticity, which requires the Hamiltonian to be equal to its own adjoint (conjugate transpose). In other words, the Hamiltonian matrix is not symmetric.

How is non-hermiticity of a Hamiltonian relevant in physics?

Non-hermitian Hamiltonians can arise in various physical systems, such as open quantum systems or systems with time-dependent perturbations. They can lead to interesting phenomena, such as non-conservation of energy and non-unitary time evolution.

Can a non-hermitian Hamiltonian still produce real eigenvalues?

Yes, a non-hermitian Hamiltonian can still have real eigenvalues. However, the eigenvalues may not necessarily be positive, as they are for hermitian Hamiltonians. In fact, a non-hermitian Hamiltonian can have complex eigenvalues, which can have physical implications.

How are non-hermitian Hamiltonians treated in quantum mechanics?

In quantum mechanics, non-hermitian Hamiltonians are typically treated using the framework of complex quantum mechanics. This involves using a modified inner product and a complex-valued probability interpretation. Alternatively, non-hermitian Hamiltonians can also be analyzed using the theory of pseudo-Hermitian operators.

Are there any applications of non-hermitian Hamiltonians in technology?

Yes, non-hermitian Hamiltonians have been used in various technological applications, such as in optical systems and electronic devices. For example, non-hermitian Hamiltonians have been used to design lasers with enhanced output power and to study the dynamics of coupled optical resonators.

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