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How can one show that $\int_{-\infty}^{\infty}\psi^*(x)(d/dx+\tanh x)(-d/dx+\tanh x)\psi(x) dx=\int_{-\infty}^{\infty} |(d/dx+\tanh x)\psi(x)|^2 dx$, where $\psi$ is normalized?

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The equation in the question(v1) contains a sign error. – Qmechanic Oct 17 '11 at 17:04
up vote 1 down vote accepted

tanh(x) is a Hermitian operator, whereas d/dx is an anti-Hermitian one.

Hermitian operator may be applied to the ψ∗ without the change, the anti-Hermitian changes the sign.


How does one know if a specific operator is Hermitian?

The operator d/dx + tanh(x) may be presented as a sum of two operators: d/dx and tanh(x).

Let's start with the former: tanh(x) It's a multiplicative operator. Means it multiplies the function ψ(x) by another function. Since within the multiplication we are free to reorder multipliers the following holds:

ψ∗ (tanhx ψ) = (tanhx ψ∗) ψ

And, since tanhx is a real function - we may also rewrite it this way:

ψ∗ (tanhx ψ) = (tanhx ψ)∗ ψ

This means that the operator tanhx may be applied to either the function itself, or to its conjugate, without any difference. Hence this operator is Hermitian.

Now let's see what happens with d/dx.

I [ ψ∗ (d/dx) ψ dx] = I [ ψ∗ dψ ] = Boundary [ ψ∗ ψ ] - I [ dψ∗ ψ ]

Integrating by part we're left with a boundary term and another integral. The boundary term vanishes since ψ goes to zero at +/- infinity. (BTW we didn't use the fact that ψ is normalized to 1).


I [ ψ∗ {(d/dx) ψ} dx] = - I [ dψ∗ ψ ] = - I [ {(d/dx)ψ}* ψ dx ]

I [ ψ∗ {(d/dx) ψ} dx] = I [ {(-d/dx)ψ}* ψ dx ]

This means that if the operator (d/dx) is transferred to the conjugate part of the integral - its sign should be changed. Such an operator is anti-Hermitian.

P.S. You should remember those special cases:

  • multiplication by a real function is a Hermitian operator.
  • multiplication by an imaginary function is an anti-Hermitian operator.
  • Derivative operators are either Hermitian or anti-Hermitian, depends on the order of the derivative. 1st derivative - anti-Hermitian, 2nd derivative - Hermitian. And so on.


You should also know that the momentum operator is defined as hbar/i d/dx (in 1 dimension). It's Hermitian because it consists of a d/dx and i. Both are anti-Hermitian.

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How does one know that which operators are hermintian? – Raj Oct 17 '11 at 11:50
@Raj do integration by parts on $\int\psi^*(x)\frac{d}{dx}\psi(x)dx=-\int\frac{d}{dx}(\psi^*(x))\psi(x)dx$. For "multiplication by $\tanh(x)$", you can just "move it around inside the integral sign", so we can say that it acts on $\psi(x)$ or $\psi^*(x)$ without any change of sign. Hence, $\frac{d}{dx}$ is anti-Hermitian, "multiplication by $\tanh(x)$ " is Hermitian. – Peter Morgan Oct 17 '11 at 12:58

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