Can we change the spin of electron by applying magnetic field from $\uparrow$ to $\downarrow$ configuration?

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  • $\begingroup$ @Brionius but isn't the spin intrinsic property of electron? $\endgroup$
    – user48826
    Jan 13 '15 at 19:27
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    $\begingroup$ the spin quantum number is intrinsic, not its orientation (up or down) $\endgroup$
    – SuperCiocia
    Jan 13 '15 at 19:32
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    $\begingroup$ An electron can exist in two spin eigenstates, or a linear combination of the two states. If you start in the state $|↑>$, a magnetic field and the correct EM pulse rotates the spin through mixed states $a|↑> + b|↓>$ to the final state $|↓>$. $\endgroup$
    – Brionius
    Jan 13 '15 at 20:32

(I understood your question as "how is it possible to change the electron spin at all in presence of a magnetic field", hope that is alright.)

I expand a bit on the electron paramagnetic resonance.

In the presence of an externally applied magnetic field the normally degenerate energy levels split up due to the zeeman effect. So let's say that you have one electron in the innermost shell without external magnetic field. In this case it's energetically not important if the spin is pointing up or down. Either way the electron has the same energy.

When you turn on the magnetic field the degeneracy is broken: the state where the electron points in the direction of the magnetic field has a higher energy and the state where the electron spin points in the other direction has a lower energy. So instead of one energy level you got two. Since the state where the electron spin is pointing down has lower energy, the electron prefers to sit in this state.

Between those energy levels you have a energy difference $\Delta E$. When the electron which sits in the lower state absorbs a photon with energy $\hbar \omega =\Delta E$ then the electron jumps from the lower state with spin $\uparrow$ to the upper level with spin $\downarrow$. So you have changed the spin of the electron. (Conversely, the spin changes from $\downarrow$ to $\uparrow$ when the electron emits a photon of the according energy.)

but isn't the spin intrinsic property of electron?

Intrinsic means that the electron doesn't have to do anything to have this property. (At least that's the way I translate "intrinsic" ^^.) So in the classical picture the electron would have to be a charged rotating sphere to get this spin property. In QM however the electron just "has" spin and it doesn't rotate. Nonetheless, intrinsic doesn't mean that this property can't change. (As SuperCocia said.)

  • $\begingroup$ Actually you have the energy splitting of the electron in a B field backwards. Spins in the direction of the magnetic field have energy -mB where m is the magnetic moment and spins antiparallel to the field have energy +mB because it takes energy to flip them. $\endgroup$ Apr 25 '15 at 18:02

A magnetic field consists of photons. Photons are spin $1$ particles, which means for a given $z$ axis a measurement of spin can yield $+1$ or $-1$. If a photon collides with an electron, we know that spin must be conserved. Let's assume the electron is in a spin up state $\uparrow_e= + \frac{1}{2}$ and the photon in a spin down state $\downarrow_P= -1$. There are two possibilities if these two meet:

$\uparrow_e \rightarrow \uparrow_e$ and $\downarrow_P \rightarrow \downarrow_P$ which means no spin flip


$\uparrow_e \rightarrow \downarrow_e$ and $\downarrow_P \rightarrow \uparrow_P$, which is possible because $ + \frac{1}{2} - 1 = -\frac{1}{2} $

  • $\begingroup$ Is there any way to be sure that the electron in the second case after the collision is the same electron that went into the collision? If you and I collide and exchange mass and/or momentum, it's not clear that the two things exiting the collision are still (in any meaningful way) you and me. $\endgroup$
    – Tom Mercer
    Mar 12 at 0:48

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