# A Hamiltonian and constant potential

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1. Apr 5, 2016

### MBPTandDFT

H=p^2/2m+c
What's c? It's of course a shift in energy, but can be thought also as a smoother and smoother real-space local potential that becomes a constant all over the space.
On the other hand, why couldn't one think about it as a constant potential in reciprocal space? It's a shift in energy so it's a constant everywhere...
But how to reconcile this with the fact that the Fourier transform of a constant is a Dirac delta?

2. Apr 5, 2016

### Demystifier

I don't know how to interpret the Fourier transform of a potential in classical physics, but in quantum physics it is a potential in the momentum space. The Dirac delta in the momentum space means that such a potential does not change the momentum of the particle.

3. Apr 5, 2016

### blue_leaf77

You are not Fourier transforming the constant potential alone, but also along with the wavefunction it acts on.
$$\int c \psi(x) e^{ikx} dx = c \tilde{\psi}(k)$$

4. Apr 5, 2016

### secur

Adding a constant to the Hamiltonian doesn't do anything.

Classically it's equivalent to a constant potential energy term, but the potential energy's zero point is freely adjustable, because only the derivative has physical relevance.

In QM it becomes a constant multiplier on the wavefunction, as blue_leaf77 shows. By the way I assume this is a different c from the one added to the H, being exp(the old c). But a wavefunction is normalized anyway, so a constant multiplier has no effect.

Of course if c is complex it shifts the phase, so if there were another state of the wave function to compare, it would have physical relevance. Perhaps you're thinking this is related to anyonic statistics?

Maybe I'm missing something?

Last edited: Apr 5, 2016
5. Apr 6, 2016

### vanhees71

It's in fact very simply to see that a constant added to the Hamiltonian doesn't do anything (with constant I mean a constant times the unit operator). The main purpose of the Hamiltonian is to provide the time evolution of states and eigenvectors of operators representing observables. Take the SchrÃ¶dinger picture, where the state vectors carry the full time dependence via the equation
$\mathrm{i} \partial_t |\psi(t) \rangle = \hat{H} |\psi(t) \rangle.$
Now take
$\hat{H}'=\hat{H}+c \hat{1}, \quad c \in \mathbb{R}.$
Then you have
$\mathrm{i} \partial_t |\psi'(t) \rangle=\hat{H} |\psi'(t) \rangle.$
Now we want to show that for a given initial state $|\psi'(t=0) \rangle=|\psi(t=0) \rangle=|\psi_0 \rangle$ the solution of the evolution equations lead to the same state at any time $t>0$.

Indeed with the ansatz
$$|\psi'(t) \rangle=\exp(-\mathrm{i} t c) |\psi(t) \rangle,$$
you get
$$\mathrm{i} |\psi'(t) \rangle = \exp(-\mathrm{i} t c) (c+\mathrm{i} \partial_t) |\psi(t) \rangle=\exp(-\mathrm{i} t c) (\hat{H}+c \hat{1})|\psi(t) \rangle= \hat{H}' |\psi'(t) \rangle,$$
i.e., $|\psi'(t) \rangle$ differs from $|\psi(t) \rangle$ only by a phase factor and thus represents the same state of the quantum system.

Now to the representation in momentum space. This is most easily solved by using the position representation and transforming to the momentum representation via a fourier transformation:
$$\psi(\vec{x})=\int_{\mathbb{R}^3} \frac{\mathrm{d}^3 \vec{p}}{(2 \pi)^{3/2}} \exp(\mathrm{i} \vec{x} \cdot \vec{p}) \tilde{\psi(\vec{p})} \; \Leftrightarrow \; \tilde{\psi}(\vec{p}) = \int_{\mathbb{R}^3} \frac{\mathrm{d}^3 \vec{x}}{(2 \pi)^{3/2}} \exp(\mathrm{i} \vec{x} \cdot \vec{p}) \psi(\vec{x}).$$
Now the potential-energy part of the Hamiltonian in position space acts as a product on the wave function, i.e.,
$$\hat{V} \psi(\vec{x})=V(\vec{x}) \psi(\vec{x}).$$
From the convolution theorem then you know that in momentum space it acts as a convolution
$$\hat{V} \tilde{\psi}(\vec{p})=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{p} \tilde{V}(\vec{p}-\vec{p}') \tilde{\psi}(\vec{p}').$$
The Fourier transformed potential is defined as
$$\tilde{V}(\vec{p}-\vec{p}')=\langle \vec{p}|\hat{V} |\vec{p}' \rangle=\int_{\mathbb{R}^3} \frac{\mathrm{d}^3 \vec{x}}{(2 \pi)^3} \exp[-\mathrm{i} \vec{x} \cdot (\vec{p}-\vec{p}')] V(\vec{x}).$$
For $V(x)=c=\text{const}$, this leads to
$$\tilde{V}(\vec{p}-\vec{p}')=c \delta^{(3)}(\vec{p}-\vec{p}'),$$
i.e., also in momentum space a constant $V$ is just the multiplication with this constant as it must be, because the unit operator is of course the multiplication of the wave function by simply 1 in any representation.

6. Apr 6, 2016

### MBPTandDFT

Consider a constant potential in real space c. Its second quantization hamiltonian is $\hat{H}=\int d^3x \hat{\psi}^{\dagger}(x)c\hat{\psi}(x)$. Now write this same term in momentum space. It is again $\hat{H}=\sum_{k} \hat{a}^{\dagger}(k)c\hat{a}(k)$, meaning $v(x)=const$ corresponds to $v(k)=const$, right? And what about the Dirac delta function, which is the Fourier transform of a constant?

7. Apr 6, 2016

### blue_leaf77

You mean if $V(x) = c\delta(x-a)$? In this case simply calculate the Fourier transform of $c\delta(x-a)$, which is easy, and substitute the resulting $V(p)$ into vanhees' formula above
$$\hat{V} \tilde{\psi}(\vec{p})=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{p} \tilde{V}(\vec{p}-\vec{p}') \tilde{\psi}(\vec{p}').$$

8. Apr 6, 2016

### vanhees71

Of course $\hat{H}=c \hat{1}$. The matrix element must thus be a $\delta$ function in all representations. That's the case for both the position and the momentum representation, because
$$\langle \vec{x}|c \hat{1}|\vec{x}' \rangle=c \delta^{(3)}(\vec{x}-\vec{x}'), \quad \langle \vec{p}|c \hat{1}|\vec{p}' \rangle = c \delta^{(3)}(\vec{p}-\vec{p}'),$$
and thus in any representation it boils down by multiplying the corresponding wave function by the number $c$.

9. Apr 6, 2016

### MBPTandDFT

No, what I mean is that a constant in real space, $V(x)=c$, corresponds to the same constant in reciprocal space, $V(k)=c$. Hence I'm puzzled because a trivial application of Fourier would result in a delta function in reciprocal space, not a constant.

10. Apr 6, 2016

### vanhees71

The point is that for a general potential $V(\vec{x})$ in position space, in momentum space you have to use the matrix element
$$V(\vec{p},\vec{p}')=\langle \vec{p}|\hat{V}|\vec{p}' \rangle=\tilde{V}(\vec{p}-\vec{p}').$$
In position space the matrix element is
$$V(\vec{x},\vec{x}')=\langle \vec{x}|\hat{V}|\vec{x}' \rangle=V(x) \delta^{(3)}(\vec{x}-\vec{x}'),$$
and in the application to the wave function this immediately implies that it simply reduces to
$$\hat{V} \psi(\vec{x})=\langle \vec{x}|\hat{V} \psi \rangle=V(\vec{x}) \langle \vec{x}|\psi \rangle=V(\vec{x}) \psi(\vec{x}).$$
So everything is consistent.

11. Apr 6, 2016

### MBPTandDFT

Cannot get the conclusion..

12. Apr 6, 2016

### vanhees71

Ok, with some more steps
$$\hat{V} \psi(\vec{x}):=\langle \vec{x}|\hat{V}|\psi \rangle = \int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x}' \langle \vec{x}|\hat{V}|\vec{x}' \rangle \langle \vec{x}'|\psi \rangle=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x}' V(\vec{x}) \delta^{(3)}(\vec{x}-\vec{x}') \psi(\vec{x}')=V(\vec{x}) \psi(\vec{x}).$$
Of course, I've assumed the whole time that you have the usual case of a potential $\hat{V}=V(\hat{\vec{x}})$.

13. Apr 6, 2016

### MBPTandDFT

Your derivation is clear. Just I cannot get what you are proving with it...

14. Apr 6, 2016

### vanhees71

I thought the question was, how to describe a constant Hamiltonian in the momentum representation via Fourier transformation. Of course, using the bra-ket formalism is much more straight forward. It's of course multiplying the wave function in any representation just by this constant value.

15. Apr 6, 2016

### MBPTandDFT

Finally I got it: a locality in real space is usually transformed to a non-locality in reciprocal space, and this exactly balances the fact that a constant is transformed to a delta: in the end one ends up with a constant Hamiltonian term in both representations.