Scanning Tunnelling Microscope (STM)

EIRE2003
Messages
107
Reaction score
0
''The Scanning Tunnelling Microscope probes the density of states of a material''...

What does this actually mean?
 
Physics news on Phys.org
The signal from an STM is proportional to the tunnelling current between the tip of the probe and the surface that is being scanned.
Now, tunnelling current is of course proportional to the tunnelling propability which in turn is determined by -among other things- the number of states that are allowed to tunnell for a specfic bias, i.e. it is proportional to the density of states.
 
EIRE2003 said:
''The Scanning Tunnelling Microscope probes the density of states of a material''...

What does this actually mean?

It means that the STM is doing "spectroscopy" rather than imaging. It scans something resembling an "I vs. V", at a particular location of a material, where I is the current, and V is the bias potential between the STM tip and the material. The density of states (assuming you know what this is), is proportional to the the derivative of this curve, i.e. dI/dV vs. V.

Zz.
 
Ok, am I right in suggesting that the 'density of states' is, how closely packed the energy levels in an atom are, or is it how closely packed the energy levels in the material as a whole? That is, the energy gap between the valence band and the conduction band?
 
Your first definition of density of states is the more correct one. The density of states determines how closely packed (in energy space) the energy levels in a quantum mechanical system are.
 
EIRE2003 said:
Ok, am I right in suggesting that the 'density of states' is, how closely packed the energy levels in an atom are, or is it how closely packed the energy levels in the material as a whole? That is, the energy gap between the valence band and the conduction band?

Er... DOS is dN/dE, where N is the number of states per unit volume.

You can't really consider the "energy levels in an atom" when you're doing a STM on a solid. I'm sure I've mentioned this many times. A solid doesn't behave the same way as isolated atoms. That's why "solid state physics" isn't the same field of study as "atomic/molecular physics". An atom's "DOS" is meaningless because an atom has discrete energy levels, and thus a dN/dE operation would be strange. The DOS you get from an STM of a solid, such as a superconductor, gives a continuous result!

Zz.
 
Last edited:
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. Towards the end of the first lecture for the Qiskit Global Summer School 2025, Foundations of Quantum Mechanics, Olivia Lanes (Global Lead, Content and Education IBM) stated... Source: https://www.physicsforums.com/insights/quantum-entanglement-is-a-kinematic-fact-not-a-dynamical-effect/ by @RUTA
I am reading WHAT IS A QUANTUM FIELD THEORY?" A First Introduction for Mathematicians. The author states (2.4 Finite versus Continuous Models) that the use of continuity causes the infinities in QFT: 'Mathematicians are trained to think of physical space as R3. But our continuous model of physical space as R3 is of course an idealization, both at the scale of the very large and at the scale of the very small. This idealization has proved to be very powerful, but in the case of Quantum...
Thread 'Lesser Green's function'
The lesser Green's function is defined as: $$G^{<}(t,t')=i\langle C_{\nu}^{\dagger}(t')C_{\nu}(t)\rangle=i\bra{n}C_{\nu}^{\dagger}(t')C_{\nu}(t)\ket{n}$$ where ##\ket{n}## is the many particle ground state. $$G^{<}(t,t')=i\bra{n}e^{iHt'}C_{\nu}^{\dagger}(0)e^{-iHt'}e^{iHt}C_{\nu}(0)e^{-iHt}\ket{n}$$ First consider the case t <t' Define, $$\ket{\alpha}=e^{-iH(t'-t)}C_{\nu}(0)e^{-iHt}\ket{n}$$ $$\ket{\beta}=C_{\nu}(0)e^{-iHt'}\ket{n}$$ $$G^{<}(t,t')=i\bra{\beta}\ket{\alpha}$$ ##\ket{\alpha}##...

Similar threads

Back
Top