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Consider a two particle system consisting of two electrons. The complete state of the electron includes its position wave function and also a spinor describing the orientation of its spin: $$\psi(r) \otimes \chi(s).$$ Why does it follows that for the the two particle system that if we have an anti-symmetric spin state of the two electrons such as the singlet state $$\frac{1}{\sqrt{2}}(| \uparrow \rangle \otimes | \downarrow \rangle - | \downarrow \rangle \otimes | \uparrow \rangle)$$ then this has to be joined with a symmetric spatial function (and similarly if we have a symmetric state of two electrons such as $| \uparrow \rangle \otimes |\uparrow \rangle$ then this has to be joined by an anti-symmetric spatial wave function?

Also if two electrons occupy the singlet spin state then the spatial wave function describing the two particle state would be symmetric, but I thought that for identical particles which are fermions (such as electrons), the spatial wave function is always antisymmetric?

Thanks for any assistance.

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    $\begingroup$ An odd state multipled by an even state gives you an odd state. It's been a while but I think it is analogous to that with symmetery, to preserve anti symmetry. Anyway, I am sure someone will sort us both out. $\endgroup$ – user140606 Jan 16 '17 at 18:48
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For fermions, the total wave function, including both the spatial wave function and the spin state, must be antisymmetric under exchange. Since the product of two antisymmetric functions is symmetric (as is the product of two symmetric functions), it is necessary that either the spin is antisymmetric or the spatial wave function is antisymmetric but not both.

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  • $\begingroup$ Thanks for your answer. Do you mean that say the spatial wave function $\psi(x_{1}, x_{2})$ of a two electron system is symmetric under the exchange operator then we require that the spin would be in the singlet state $| 0 0 \rangle =\frac{1}{\sqrt{2}}(| \uparrow \rangle \otimes | \downarrow \rangle - | \downarrow \rangle \otimes | \uparrow \rangle)$ hence we would have a total wave function $$\Psi = |\psi \rangle \otimes |0 0 \rangle?$$ $\endgroup$ – Alex Jan 16 '17 at 20:30
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    $\begingroup$ That's correct: if the spatial wave function for two electrons is symmetric, the spin state must be antisymmetric - ie., the singlet state. If instead the spatial wave function is antisymmetric, the spin state would have to be a triplet state. $\endgroup$ – Harry Levine Jan 16 '17 at 20:33
  • $\begingroup$ Okay thanks. One other query... Would I be right in stating the following: Given that the state of the two particle system is $$\Psi = | \psi \rangle \otimes |00 \rangle$$ could I then project the tensor product onto the position basis by using the linear operator $\langle x_1, x_2 | \otimes I$, thus resulting in the tensor product $$\psi(x_1, x_2) \otimes |00\rangle$$, since the left hand side of the tensor product is a complex number and the right side is a matrix, we can consider the state $$\psi(x_1,x_2)|00 \rangle$$ as simply matrix times a complex scalar? $\endgroup$ – Alex Jan 17 '17 at 10:32

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